Develop Reference

How do you change a variable in a tkinter command without making it bulky with functions? [duplicate]

I have a list of objects and I want to remove all objects that are empty except for one, using filter and a lambda expression.
For example if the input is:
[Object(name=""), Object(name="fake_name"), Object(name="")]
...then the output should be:
[Object(name=""), Object(name="fake_name")]
Is there a way to add an assignment to a lambda expression? For example:
flag = True
input = [Object(name=""), Object(name="fake_name"), Object(name="")]
output = filter(
(lambda o: [flag or bool(o.name), flag = flag and bool(o.name)][0]),
input
)

The assignment expression operator := added in Python 3.8 supports assignment inside of lambda expressions. This operator can only appear within a parenthesized (...), bracketed [...], or braced {...} expression for syntactic reasons. For example, we will be able to write the following:
import sys
say_hello = lambda: (
message := "Hello world",
sys.stdout.write(message + "\n")
)[-1]
say_hello()
In Python 2, it was possible to perform local assignments as a side effect of list comprehensions.
import sys
say_hello = lambda: (
[None for message in ["Hello world"]],
sys.stdout.write(message + "\n")
)[-1]
say_hello()
However, it's not possible to use either of these in your example because your variable flag is in an outer scope, not the lambda's scope. This doesn't have to do with lambda, it's the general behaviour in Python 2. Python 3 lets you get around this with the nonlocal keyword inside of defs, but nonlocal can't be used inside lambdas.
There's a workaround (see below), but while we're on the topic...
In some cases you can use this to do everything inside of a lambda:
(lambda: [
['def'
for sys in [__import__('sys')]
for math in [__import__('math')]
for sub in [lambda *vals: None]
for fun in [lambda *vals: vals[-1]]
for echo in [lambda *vals: sub(
sys.stdout.write(u" ".join(map(unicode, vals)) + u"\n"))]
for Cylinder in [type('Cylinder', (object,), dict(
__init__ = lambda self, radius, height: sub(
setattr(self, 'radius', radius),
setattr(self, 'height', height)),
volume = property(lambda self: fun(
['def' for top_area in [math.pi * self.radius ** 2]],
self.height * top_area))))]
for main in [lambda: sub(
['loop' for factor in [1, 2, 3] if sub(
['def'
for my_radius, my_height in [[10 * factor, 20 * factor]]
for my_cylinder in [Cylinder(my_radius, my_height)]],
echo(u"A cylinder with a radius of %.1fcm and a height "
u"of %.1fcm has a volume of %.1fcm³."
% (my_radius, my_height, my_cylinder.volume)))])]],
main()])()
A cylinder with a radius of 10.0cm and a height of 20.0cm has a volume of 6283.2cm³.
A cylinder with a radius of 20.0cm and a height of 40.0cm has a volume of 50265.5cm³.
A cylinder with a radius of 30.0cm and a height of 60.0cm has a volume of 169646.0cm³.
Please don't.
...back to your original example: though you can't perform assignments to the flag variable in the outer scope, you can use functions to modify the previously-assigned value.
For example, flag could be an object whose .value we set using setattr:
flag = Object(value=True)
input = [Object(name=''), Object(name='fake_name'), Object(name='')]
output = filter(lambda o: [
flag.value or bool(o.name),
setattr(flag, 'value', flag.value and bool(o.name))
][0], input)
[Object(name=''), Object(name='fake_name')]
If we wanted to fit the above theme, we could use a list comprehension instead of setattr:
[None for flag.value in [bool(o.name)]]
But really, in serious code you should always use a regular function definition instead of a lambda if you're going to be doing outer assignment.
flag = Object(value=True)
def not_empty_except_first(o):
result = flag.value or bool(o.name)
flag.value = flag.value and bool(o.name)
return result
input = [Object(name=""), Object(name="fake_name"), Object(name="")]
output = filter(not_empty_except_first, input)

You cannot really maintain state in a filter/lambda expression (unless abusing the global namespace). You can however achieve something similar using the accumulated result being passed around in a reduce() expression:
>>> f = lambda a, b: (a.append(b) or a) if (b not in a) else a
>>> input = ["foo", u"", "bar", "", "", "x"]
>>> reduce(f, input, [])
['foo', u'', 'bar', 'x']
>>>
You can, of course, tweak the condition a bit. In this case it filters out duplicates, but you can also use a.count(""), for example, to only restrict empty strings.
Needless to say, you can do this but you really shouldn't. :)
Lastly, you can do anything in pure Python lambda: http://vanderwijk.info/blog/pure-lambda-calculus-python/

Normal assignment (=) is not possible inside a lambda expression, although it is possible to perform various tricks with setattr and friends.
Solving your problem, however, is actually quite simple:
input = [Object(name=""), Object(name="fake_name"), Object(name="")]
output = filter(
lambda o, _seen=set():
not (not o and o in _seen or _seen.add(o)),
input
)
which will give you
[Object(Object(name=''), name='fake_name')]
As you can see, it's keeping the first blank instance instead of the last. If you need the last instead, reverse the list going in to filter, and reverse the list coming out of filter:
output = filter(
lambda o, _seen=set():
not (not o and o in _seen or _seen.add(o)),
input[::-1]
)[::-1]
which will give you
[Object(name='fake_name'), Object(name='')]
One thing to be aware of: in order for this to work with arbitrary objects, those objects must properly implement __eq__ and __hash__ as explained here.

There's no need to use a lambda, when you can remove all the null ones, and put one back if the input size changes:
input = [Object(name=""), Object(name="fake_name"), Object(name="")]
output = [x for x in input if x.name]
if(len(input) != len(output)):
output.append(Object(name=""))

UPDATE:
[o for d in [{}] for o in lst if o.name != "" or d.setdefault("", o) == o]
or using filter and lambda:
flag = {}
filter(lambda o: bool(o.name) or flag.setdefault("", o) == o, lst)
Previous Answer
OK, are you stuck on using filter and lambda?
It seems like this would be better served with a dictionary comprehension,
{o.name : o for o in input}.values()
I think the reason that Python doesn't allow assignment in a lambda is similar to why it doesn't allow assignment in a comprehension and that's got something to do with the fact that these things are evaluated on the C side and thus can give us an increase in speed. At least that's my impression after reading one of Guido's essays.
My guess is this would also go against the philosophy of having one right way of doing any one thing in Python.

TL;DR: When using functional idioms it's better to write functional code
As many people have pointed out, in Python lambdas assignment is not allowed. In general when using functional idioms your better off thinking in a functional manner which means wherever possible no side effects and no assignments.
Here is functional solution which uses a lambda. I've assigned the lambda to fn for clarity (and because it got a little long-ish).
from operator import add
from itertools import ifilter, ifilterfalse
fn = lambda l, pred: add(list(ifilter(pred, iter(l))), [ifilterfalse(pred, iter(l)).next()])
objs = [Object(name=""), Object(name="fake_name"), Object(name="")]
fn(objs, lambda o: o.name != '')
You can also make this deal with iterators rather than lists by changing things around a little. You also have some different imports.
from itertools import chain, islice, ifilter, ifilterfalse
fn = lambda l, pred: chain(ifilter(pred, iter(l)), islice(ifilterfalse(pred, iter(l)), 1))
You can always reoganize the code to reduce the length of the statements.

The pythonic way to track state during iteration is with generators. The itertools way is quite hard to understand IMHO and trying to hack lambdas to do this is plain silly. I'd try:
def keep_last_empty(input):
last = None
for item in iter(input):
if item.name: yield item
else: last = item
if last is not None: yield last
output = list(keep_last_empty(input))
Overall, readability trumps compactness every time.

If instead of flag = True we can do an import instead, then I think this meets the criteria:
>>> from itertools import count
>>> a = ['hello', '', 'world', '', '', '', 'bob']
>>> filter(lambda L, j=count(): L or not next(j), a)
['hello', '', 'world', 'bob']
Or maybe the filter is better written as:
>>> filter(lambda L, blank_count=count(1): L or next(blank_count) == 1, a)
Or, just for a simple boolean, without any imports:
filter(lambda L, use_blank=iter([True]): L or next(use_blank, False), a)

No, you cannot put an assignment inside a lambda because of its own definition. If you work using functional programming, then you must assume that your values are not mutable.
One solution would be the following code:
output = lambda l, name: [] if l==[] \
else [ l[ 0 ] ] + output( l[1:], name ) if l[ 0 ].name == name \
else output( l[1:], name ) if l[ 0 ].name == "" \
else [ l[ 0 ] ] + output( l[1:], name )

If you need a lambda to remember state between calls, I would recommend either a function declared in the local namespace or a class with an overloaded __call__. Now that all my cautions against what you are trying to do is out of the way, we can get to an actual answer to your query.
If you really need to have your lambda to have some memory between calls, you can define it like:
f = lambda o, ns = {"flag":True}: [ns["flag"] or o.name, ns.__setitem__("flag", ns["flag"] and o.name)][0]
Then you just need to pass f to filter(). If you really need to, you can get back the value of flag with the following:
f.__defaults__[0]["flag"]
Alternatively, you can modify the global namespace by modifying the result of globals(). Unfortunately, you cannot modify the local namespace in the same way as modifying the result of locals() doesn't affect the local namespace.

You can use a bind function to use a pseudo multi-statement lambda. Then you can use a wrapper class for a Flag to enable assignment.
bind = lambda x, f=(lambda y: y): f(x)
class Flag(object):
def __init__(self, value):
self.value = value
def set(self, value):
self.value = value
return value
input = [Object(name=""), Object(name="fake_name"), Object(name="")]
flag = Flag(True)
output = filter(
lambda o: (
bind(flag.value, lambda orig_flag_value:
bind(flag.set(flag.value and bool(o.name)), lambda _:
bind(orig_flag_value or bool(o.name))))),
input)

Kind of a messy workaround, but assignment in lambdas is illegal anyway, so it doesn't really matter. You can use the builtin exec() function to run assignment from inside the lambda, such as this example:
>>> val
Traceback (most recent call last):
File "<pyshell#31>", line 1, in <module>
val
NameError: name 'val' is not defined
>>> d = lambda: exec('val=True', globals())
>>> d()
>>> val
True

first , you dont need to use a local assigment for your job, just check the above answer
second, its simple to use locals() and globals() to got the variables table and then change the value
check this sample code:
print [locals().__setitem__('x', 'Hillo :]'), x][-1]
if you need to change the add a global variable to your environ, try to replace locals() with globals()
python's list comp is cool but most of the triditional project dont accept this(like flask :[)
hope it could help

Clean callbacks in python [duplicate]

I've heard it said that multiline lambdas can't be added in Python because they would clash syntactically with the other syntax constructs in Python. I was thinking about this on the bus today and realized I couldn't think of a single Python construct that multiline lambdas clash with. Given that I know the language pretty well, this surprised me.
Now, I'm sure Guido had a reason for not including multiline lambdas in the language, but out of curiosity: what's a situation where including a multiline lambda would be ambiguous? Is what I've heard true, or is there some other reason that Python doesn't allow multiline lambdas?

Guido van Rossum (the inventor of Python) answers this exact question himself in an old blog post.
Basically, he admits that it's theoretically possible, but that any proposed solution would be un-Pythonic:
"But the complexity of any proposed solution for this puzzle is immense, to me: it requires the parser (or more precisely, the lexer) to be able to switch back and forth between indent-sensitive and indent-insensitive modes, keeping a stack of previous modes and indentation level. Technically that can all be solved (there's already a stack of indentation levels that could be generalized). But none of that takes away my gut feeling that it is all an elaborate Rube Goldberg contraption."

Look at the following:
map(multilambda x:
y=x+1
return y
, [1,2,3])
Is this a lambda returning (y, [1,2,3]) (thus map only gets one parameter, resulting in an error)? Or does it return y? Or is it a syntax error, because the comma on the new line is misplaced? How would Python know what you want?
Within the parens, indentation doesn't matter to python, so you can't unambiguously work with multilines.
This is just a simple one, there's probably more examples.

This is generally very ugly (but sometimes the alternatives are even more ugly), so a workaround is to make a braces expression:
lambda: (
doFoo('abc'),
doBar(123),
doBaz())
It won't accept any assignments though, so you'll have to prepare data beforehand.
The place I found this useful is the PySide wrapper, where you sometimes have short callbacks. Writing additional member functions would be even more ugly. Normally you won't need this.
Example:
pushButtonShowDialog.clicked.connect(
lambda: (
field1.clear(),
spinBox1.setValue(0),
diag.show())

A couple of relevant links:
For a while, I was following the development of Reia, which was initially going to have Python's indentation based syntax with Ruby blocks too, all on top of Erlang. But, the designer wound up giving up on indentation sensitivity, and this post he wrote about that decision includes a discussion about problems he ran into with indentation + multi-line blocks, and an increased appreciation he gained for Guido's design issues/decisions:
http://www.unlimitednovelty.com/2009/03/indentation-sensitivity-post-mortem.html
Also, here's an interesting proposal for Ruby-style blocks in Python I ran across where Guido posts a response w/o actually shooting it down (not sure whether there has been any subsequent shoot down, though):
http://tav.espians.com/ruby-style-blocks-in-python.html

Let me present to you a glorious but terrifying hack:
import types
def _obj():
return lambda: None
def LET(bindings, body, env=None):
'''Introduce local bindings.
ex: LET(('a', 1,
'b', 2),
lambda o: [o.a, o.b])
gives: [1, 2]
Bindings down the chain can depend on
the ones above them through a lambda.
ex: LET(('a', 1,
'b', lambda o: o.a + 1),
lambda o: o.b)
gives: 2
'''
if len(bindings) == 0:
return body(env)
env = env or _obj()
k, v = bindings[:2]
if isinstance(v, types.FunctionType):
v = v(env)
setattr(env, k, v)
return LET(bindings[2:], body, env)
You can now use this LET form as such:
map(lambda x: LET(('y', x + 1,
'z', x - 1),
lambda o: o.y * o.z),
[1, 2, 3])
which gives: [0, 3, 8]

[Edit Edit] Since this question is somehow still active 12 years after being asked. I will continue the tradition of amending my answer every 4 years or so.
Firstly, the question was how does multi-line lambda clash with Python. The accepted answer shows how with a simple example. The highly rated answer I linked below some years ago answers the question of "Why is it not a part of Python"--this answer is perhaps more satisfying to those who believe that the existing examples of "clashing" are not enough to make multi-line lambda impossible to implement in Python.
In previous iterations of this answer I discussed how to implement multi-line lambda into Python as is. I've since removed that part, because it was a flurry of bad practices. You may see it in the edit history of this answer if you wish.
However the answer to "Why not?", being "because Rossum said so" can still be a source of frustration. So lets see if it could be engineered around the counter example given by user balpha:
map(lambda x:
y=x+1 # <-- this line defines the outmost indent level*
for i in range(12):
y+=12
return y
, [1,2,3])
#*By convention it is always one-indent past the 'l' in lambda
As for the return value we have that the following is non-permissible in python:
def f():
return 3
, [1,2,3]
So by the same logic, "[1,2,3]" should not be part of the return value. Let's try it this way instead:
map(lambda x:
y=x+1 # part of lambda block
for i in range(12): # part of lambda block
y+=12 # part of lambda block
return y, [1,2,3]) # part of lambda block
This one's trickier, but since the lambda block has a clearly defined beginning (the token 'lambda') yet no clear ending, I would argue anything that is on the same line as part of a lambda block is also part of the lambda block.
One might imagine some features that can identify closing parenthesis or even inference based on the number of tokens expected by the enclosing element. In general, the above expression does not seem totally impossible to parse, but it may be a bit of a challenge.
To simplify things, you could separate all characters not intended to be part of the block:
map(lambda x:
y=x+1 # part of lambda block
for i in range(12): # part of lambda block
y+=12 # part of lambda block
return y # part of lambda block
, [1,2,3]) # argument separator, second argument, and closing paren for map
Back to where we were but this time it is unambiguous, because the last line is behind the lowest indent-depth for the lambda block.
Single line lambda would be a special case (identified by the lack of an immediate newline after the color), that behaves the same as it does now.
This is not to say that it necessarily should be a part of Python--but it is a quick illustration that is perhaps is possible with some changes in the language.
[Edit] Read this answer. It explains why multi-line lambda is not a thing.
Simply put, it's unpythonic. From Guido van Rossum's blog post:
I find any solution unacceptable that embeds an indentation-based block in the middle of an expression. Since I find alternative syntax for statement grouping (e.g. braces or begin/end keywords) equally unacceptable, this pretty much makes a multi-line lambda an unsolvable puzzle.

I'm guilty of practicing this dirty hack in some of my projects which is bit simpler:
lambda args...:( expr1, expr2, expr3, ...,
exprN, returnExpr)[-1]
I hope you can find a way to stay pythonic but if you have to do it this less painful than using exec and manipulating globals.

Let me also throw in my two cents about different workarounds.
How is a simple one-line lambda different from a normal function? I can think only of lack of assignments, some loop-like constructs (for, while), try-except clauses... And that's it? We even have a ternary operator - cool! So, let's try to deal with each of these problems.
Assignments
Some guys here have rightly noted that we should take a look at lisp's let form, which allows local bindings. Actually, all the non state-changing assignments can be performed only with let. But every lisp programmer knows that let form is absolutely equivalent to call to a lambda function! This means that
(let ([x_ x] [y_ y])
(do-sth-with-x-&-y x_ y_))
is the same as
((lambda (x_ y_)
(do-sth-with-x-&-y x_ y_)) x y)
So lambdas are more than enough! Whenever we want to make a new assignment we just add another lambda and call it. Consider this example:
def f(x):
y = f1(x)
z = f2(x, y)
return y,z
A lambda version looks like:
f = lambda x: (lambda y: (y, f2(x,y)))(f1(x))
You can even make the let function, if you don't like the data being written after actions on the data. And you can even curry it (just for the sake of more parentheses :) )
let = curry(lambda args, f: f(*args))
f_lmb = lambda x: let((f1(x),), lambda y: (y, f2(x,y)))
# or:
f_lmb = lambda x: let((f1(x),))(lambda y: (y, f2(x,y)))
# even better alternative:
let = lambda *args: lambda f: f(*args)
f_lmb = lambda x: let(f1(x))(lambda y: (y, f2(x,y)))
So far so good. But what if we have to make reassignments, i.e. change state? Well, I think we can live absolutely happily without changing state as long as task in question doesn't concern loops.
Loops
While there's no direct lambda alternative for loops, I believe we can write quite generic function to fit our needs. Take a look at this fibonacci function:
def fib(n):
k = 0
fib_k, fib_k_plus_1 = 0, 1
while k < n:
k += 1
fib_k_plus_1, fib_k = fib_k_plus_1 + fib_k, fib_k_plus_1
return fib_k
Impossible in terms of lambdas, obviously. But after writing a little yet useful function we're done with that and similar cases:
def loop(first_state, condition, state_changer):
state = first_state
while condition(*state):
state = state_changer(*state)
return state
fib_lmb = lambda n:\
loop(
(0,0,1),
lambda k, fib_k, fib_k_plus_1:\
k < n,
lambda k, fib_k, fib_k_plus_1:\
(k+1, fib_k_plus_1, fib_k_plus_1 + fib_k))[1]
And of course, one should always consider using map, reduce and other higher-order functions if possible.
Try-except and other control structs
It seems like a general approach to this kind of problems is to make use of lazy evaluation, replacing code blocks with lambdas accepting no arguments:
def f(x):
try: return len(x)
except: return 0
# the same as:
def try_except_f(try_clause, except_clause):
try: return try_clause()
except: return except_clause()
f = lambda x: try_except_f(lambda: len(x), lambda: 0)
# f(-1) -> 0
# f([1,2,3]) -> 3
Of course, this is not a full alternative to try-except clause, but you can always make it more generic. Btw, with that approach you can even make if behave like function!
Summing up: it's only natural that everything mentioned feels kinda unnatural and not-so-pythonically-beautiful. Nonetheless - it works! And without any evals and other trics, so all the intellisense will work. I'm also not claiming that you shoud use this everywhere. Most often you'd better define an ordinary function. I only showed that nothing is impossible.

Let me try to tackle #balpha parsing problem. I would use parentheses around the multiline lamda. If there is no parentheses, the lambda definition is greedy. So the lambda in
map(lambda x:
y = x+1
z = x-1
y*z,
[1,2,3]))
returns a function that returns (y*z, [1,2,3])
But
map((lambda x:
y = x+1
z = x-1
y*z)
,[1,2,3]))
means
map(func, [1,2,3])
where func is the multiline lambda that return y*z. Does that work?

(For anyone still interested in the topic.)
Consider this (includes even usage of statements' return values in further statements within the "multiline" lambda, although it's ugly to the point of vomiting ;-)
>>> def foo(arg):
... result = arg * 2;
... print "foo(" + str(arg) + ") called: " + str(result);
... return result;
...
>>> f = lambda a, b, state=[]: [
... state.append(foo(a)),
... state.append(foo(b)),
... state.append(foo(state[0] + state[1])),
... state[-1]
... ][-1];
>>> f(1, 2);
foo(1) called: 2
foo(2) called: 4
foo(6) called: 12
12

Here's a more interesting implementation of multi line lambdas. It's not possible to achieve because of how python use indents as a way to structure code.
But luckily for us, indent formatting can be disabled using arrays and parenthesis.
As some already pointed out, you can write your code as such:
lambda args: (expr1, expr2,... exprN)
In theory if you're guaranteed to have evaluation from left to right it would work but you still lose values being passed from one expression to an other.
One way to achieve that which is a bit more verbose is to have
lambda args: [lambda1, lambda2, ..., lambdaN]
Where each lambda receives arguments from the previous one.
def let(*funcs):
def wrap(args):
result = args
for func in funcs:
if not isinstance(result, tuple):
result = (result,)
result = func(*result)
return result
return wrap
This method let you write something that is a bit lisp/scheme like.
So you can write things like this:
let(lambda x, y: x+y)((1, 2))
A more complex method could be use to compute the hypotenuse
lst = [(1,2), (2,3)]
result = map(let(
lambda x, y: (x**2, y**2),
lambda x, y: (x + y) ** (1/2)
), lst)
This will return a list of scalar numbers so it can be used to reduce multiple values to one.
Having that many lambda is certainly not going to be very efficient but if you're constrained it can be a good way to get something done quickly then rewrite it as an actual function later.

In Python 3.8/3.9 there is Assignment Expression, so it could be used in lambda, greatly
expanding functionality
E.g., code
#%%
x = 1
y = 2
q = list(map(lambda t: (
tx := t*x,
ty := t*y,
tx+ty
)[-1], [1, 2, 3]))
print(q)
will print [3, 6, 9]

After Python3.8, there is another method for local binding
lambda x: (
y := x + 1,
y ** 2
)[-1]
For Loop
lambda x: (
y := x ** 2,
[y := y + x for _ in range(10)],
y
)[-1]
If Branch
lambda x: (
y := x ** 2,
x > 5 and [y := y + x for _ in range(10)],
y
)[-1]
Or
lambda x: (
y := x ** 2,
[y := y + x for _ in range(10)] if x > 5 else None,
y
)[-1]
While Loop
import itertools as it
lambda x: (
l := dict(y = x ** 2),
cond := lambda: l['y'] < 100,
body := lambda: l.update(y = l['y'] + x),
*it.takewhile(lambda _: cond() and (body(), True)[-1], it.count()),
l['y']
)[-1]
Or
import itertools as it
from types import SimpleNamespace as ns
lambda x: (
l := ns(y = x ** 2),
cond := lambda: l.y < 100,
body := lambda: vars(l).update(y = l.y + x),
*it.takewhile(lambda _: cond() and (body(), True)[-1], it.count()),
l.y
)[-1]
Or
import itertools as it
lambda x: (
y := x ** 2,
*it.takewhile(lambda t: t[0],
((
pred := y < 100,
pred and (y := y + x))
for _ in it.count())),
y
)[-1]

On the subject of ugly hacks, you can always use a combination of exec and a regular function to define a multiline function like this:
f = exec('''
def mlambda(x, y):
d = y - x
return d * d
''', globals()) or mlambda
You can wrap this into a function like:
def mlambda(signature, *lines):
exec_vars = {}
exec('def mlambda' + signature + ':\n' + '\n'.join('\t' + line for line in lines), exec_vars)
return exec_vars['mlambda']
f = mlambda('(x, y)',
'd = y - x',
'return d * d')

I know it is an old question, but for the record here is a kind of a solution to the problem of multiline lambda problem in which the result of one call is consumed by another call.
I hope it is not super hacky, since it is based only on standard library functions and uses no dunder methods.
Below is a simple example in which we start with x = 3 and then in the first line we add 1 and then in the second line we add 2 and get 6 as the output.
from functools import reduce
reduce(lambda data, func: func(data), [
lambda x: x + 1,
lambda x: x + 2
], 3)
## Output: 6

I was just playing a bit to try to make a dict comprehension with reduce, and come up with this one liner hack:
In [1]: from functools import reduce
In [2]: reduce(lambda d, i: (i[0] < 7 and d.__setitem__(*i[::-1]), d)[-1], [{}, *{1:2, 3:4, 5:6, 7:8}.items()])
Out[3]: {2: 1, 4: 3, 6: 5}
I was just trying to do the same as what was done in this Javascript dict comprehension: https://stackoverflow.com/a/11068265

You can simply use slash (\) if you have multiple lines for your lambda function
Example:
mx = lambda x, y: x if x > y \
else y
print(mx(30, 20))
Output: 30

I am starting with python but coming from Javascript the most obvious way is extract the expression as a function....
Contrived example, multiply expression (x*2) is extracted as function and therefore I can use multiline:
def multiply(x):
print('I am other line')
return x*2
r = map(lambda x : multiply(x), [1, 2, 3, 4])
print(list(r))
https://repl.it/#datracka/python-lambda-function
Maybe it does not answer exactly the question if that was how to do multiline in the lambda expression itself, but in case somebody gets this thread looking how to debug the expression (like me) I think it will help

One safe method to pass any number of variables between lambda items:
print((lambda: [
locals().__setitem__("a", 1),
locals().__setitem__("b", 2),
locals().__setitem__("c", 3),
locals().get("a") + locals().get("b") + locals().get("c")
])()[-1])
Output: 6

because a lambda function is supposed to be one-lined, as its the simplest form of a function, an entrance, then return

Is there a Python equivalent of the Haskell 'let'

Is there a Python equivalent of the Haskell 'let' expression that would allow me to write something like:
list2 = [let (name,size)=lookup(productId) in (barcode(productId),metric(size))
for productId in list]
If not, what would be the most readable alternative?
Added for clarification of the let syntax:
x = let (name,size)=lookup(productId) in (barcode(productId),metric(size))
is equivalent to
(name,size) = lookup(productId)
x = (barcode(productId),metric(size))
The second version doesn't work that well with list comprehensions, though.

You could use a temporary list comprehension
[(barcode(productId), metric(size)) for name, size in [lookup(productId)]][0]
or, equivalently, a generator expression
next((barcode(productId), metric(size)) for name, size in [lookup(productId)])
but both of those are pretty horrible.
Another (horrible) method is via a temporary lambda, which you call immediately
(lambda (name, size): (barcode(productId), metric(size)))(lookup(productId))
I think the recommended "Pythonic" way would just be to define a function, like
def barcode_metric(productId):
name, size = lookup(productId)
return barcode(productId), metric(size)
list2 = [barcode_metric(productId) for productId in list]

Recent python versions allows multiple for clauses in a generator expression, so you can now do something like:
list2 = [ barcode(productID), metric(size)
for productID in list
for (name,size) in (lookup(productID),) ]
which is similar to what Haskell provides too:
list2 = [ (barcode productID, metric size)
| productID <- list
, let (name,size) = lookup productID ]
and denotationally equivalent to
list2 = [ (barcode productID, metric size)
| productID <- list
, (name,size) <- [lookup productID] ]

There is no such thing. You could emulate it the same way let is desugared to lambda calculus (let x = foo in bar <=> (\x -> bar) (foo)).
The most readable alternative depends on the circumstances. For your specific example, I'd choose something like [barcode(productId), metric(size) for productId, (_, size) in zip(productIds, map(lookup, productIds))] (really ugly on second thought, it's easier if you don't need productId too, then you could use map) or an explicit for loop (in a generator):
def barcodes_and_metrics(productIds):
for productId in productIds:
_, size = lookup(productId)
yield barcode(productId), metric(size)

The multiple for clauses in b0fh's answer is the style I have personally been using for a while now, as I believe it provides more clarity and doesn't clutter the namespace with temporary functions. However, if speed is an issue, it is important to remember that temporarily constructing a one element list takes notably longer than constructing a one-tuple.
Comparing the speed of the various solutions in this thread, I found that the ugly lambda hack is slowest, followed by the nested generators and then the solution by b0fh. However, these were all surpassed by the one-tuple winner:
list2 = [ barcode(productID), metric(size)
for productID in list
for (_, size) in (lookup(productID),) ]
This may not be so relevant to the OP's question, but there are other cases where clarity can be greatly enhanced and speed gained in cases where one might wish to use a list comprehension, by using one-tuples instead of lists for dummy iterators.

In Python 3.8, assignment expressions using the := operator were added: PEP 572.
This can be used somewhat like let in Haskell, although iterable unpacking is not supported.
list2 = [
(lookup_result := lookup(productId), # store tuple since iterable unpacking isn't supported
name := lookup_result[0], # manually unpack tuple
size := lookup_result[1],
(barcode(productId), metric(size)))[-1] # put result as the last item in the tuple, then extract on the result using the (...)[-1]
for productId in list1
]
Note that this is scoped like a normal Python assignment, e.g. if used inside a function, the variables bound will be accessible throughout the entire function, not just in the expression.

Only guessing at what Haskell does, here's the alternative. It uses what's known in Python as "list comprehension".
[barcode(productId), metric(size)
for (productId, (name, size)) in [
(productId, lookup(productId)) for productId in list_]
]
You could include the use of lambda:, as others have suggested.

Since you asked for best readability you could consider the lambda-option but with a small twist: initialise the arguments. Here are various options I use myself, starting with the first I tried and ending with the one I use most now.
Suppose we have a function (not shown) which gets data_structure as argument, and you need to get x from it repeatedly.
First try (as per 2012 answer from huon):
(lambda x:
x * x + 42 * x)
(data_structure['a']['b'])
With multiple symbols this becomes less readable, so next I tried:
(lambda x, y:
x * x + 42 * x + y)
(x = data_structure['a']['b'],
y = 16)
That is still not very readable as it repeats the symbolic names. So then I tried:
(lambda x = data_structure['a']['b'],
y = 16:
x * x + 42 * x + y)()
This almost reads as an 'let' expression. The positioning and formatting of the assignments is yours of course.
This idiom is easily recognised by the starting '(' and the ending '()'.
In functional expressions (also in Python), many parenthesis tend to pile up at the end. The odd one out '(' is easily spotted.

class let:
def __init__(self, var):
self.x = var
def __enter__(self):
return self.x
def __exit__(self, type, value, traceback):
pass
with let(os.path) as p:
print(p)
But this is effectively the same as p = os.path as p's scope is not confined to the with block. To achieve that, you'd need
class let:
def __init__(self, var):
self.value = var
def __enter__(self):
return self
def __exit__(self, type, value, traceback):
del var.value
var.value = None
with let(os.path) as var:
print(var.value) # same as print(os.path)
print(var.value) # same as print(None)
Here var.value will be None outside of the with block, but os.path within it.

To get something vaguely comparable, you'll either need to do two comprehensions or maps, or define a new function. One approach that hasn't been suggested yet is to break it up into two lines like so. I believe this is somewhat readable; though probably defining your own function is the right way to go:
pids_names_sizes = (pid, lookup(pid) for pid in list1)
list2 = [(barcode(pid), metric(size)) for pid, (name, size) in pids_names_sizes]

Although you can simply write this as:
list2 = [(barcode(pid), metric(lookup(pid)[1]))
for pid in list]
You could define LET yourself to get:
list2 = [LET(('size', lookup(pid)[1]),
lambda o: (barcode(pid), metric(o.size)))
for pid in list]
or even:
list2 = map(lambda pid: LET(('name_size', lookup(pid),
'size', lambda o: o.name_size[1]),
lambda o: (barcode(pid), metric(o.size))),
list)
as follows:
import types
def _obj():
return lambda: None
def LET(bindings, body, env=None):
'''Introduce local bindings.
ex: LET(('a', 1,
'b', 2),
lambda o: [o.a, o.b])
gives: [1, 2]
Bindings down the chain can depend on
the ones above them through a lambda.
ex: LET(('a', 1,
'b', lambda o: o.a + 1),
lambda o: o.b)
gives: 2
'''
if len(bindings) == 0:
return body(env)
env = env or _obj()
k, v = bindings[:2]
if isinstance(v, types.FunctionType):
v = v(env)
setattr(env, k, v)
return LET(bindings[2:], body, env)

Is there a map without result in python? [duplicate]

This question already has answers here:
Is it Pythonic to use list comprehensions for just side effects?
(7 answers)
Closed 4 months ago.
Sometimes, I just want to execute a function for a list of entries -- eg.:
for x in wowList:
installWow(x, 'installed by me')
Sometimes I need this stuff for module initialization, so I don't want to have a footprint like x in global namespace. One solution would be to just use map together with lambda:
map(lambda x: installWow(x, 'installed by me'), wowList)
But this of course creates a nice list [None, None, ...] so my question is, if there is a similar function without a return-list -- since I just don't need it.
(off course I can also use _x and thus not leaving visible footprint -- but the map-solution looks so neat ...)

You could make your own "each" function:
def each(fn, items):
for item in items:
fn(item)
# called thus
each(lambda x: installWow(x, 'installed by me'), wowList)
Basically it's just map, but without the results being returned. By using a function you'll ensure that the "item" variable doesn't leak into the current scope.

You can use the built-in any function to apply a function without return statement to any item returned by a generator without creating a list. This can be achieved like this:
any(installWow(x, 'installed by me') for x in wowList)
I found this the most concise idom for what you want to achieve.
Internally, the installWow function does return None which evaluates to False in logical operations. any basically applies an or reduction operation to all items returned by the generator, which are all None of course, so it has to iterate over all items returned by the generator. In the end it does return False, but that doesn't need to bother you. The good thing is: no list is created as a side-effect.
Note that this only works as long as your function returns something that evaluates to False, e.g., None or 0. If it does return something that evaluates to True at some point, e.g., 1, it will not be applied to any of the remaining elements in your iterator. To be safe, use this idiom mainly for functions without return statement.

How about this?
for x in wowList:
installWow(x, 'installed by me')
del x

Every expression evaluates to something, so you always get a result, whichever way you do it. And any such returned object (just like your list) will get thrown away afterwards because there's no reference to it anymore.
To clarify: Very few things in python are statements that don't return anything. Even a function call like
doSomething()
still returns a value, even if it gets discarded right away. There is no such thing as Pascal's function / procedure distinction in python.

You might try this:
filter(lambda x: installWow(x, 'installed by me') and False, wowList)
That way, the return result is an empty list no matter what.
Or you could just drop the and False if you can force installWow() to always return False (or 0 or None or another expression that evaluates false).

You could use a filter and a function that doesn't return a True value. You'd get an empty return list since filter only adds the values which evaluates to true, which I suppose would save you some memory. Something like this:
#!/usr/bin/env python
y = 0
def myfunction(x):
global y
y += x
input = (1, 2, 3, 4)
print "Filter output: %s" % repr(filter(myfunction, input))
print "Side effect result: %d" % y
Running it produces this output:
Filter output: ()
Side effect result: 10

I can not resist myself to post it as separate answer
reduce(lambda x,y: x(y, 'installed by me') , wowList, installWow)
only twist is installWow should return itself e.g.
def installWow(*args):
print args
return installWow

if it is ok to distruct wowList
while wowList: installWow(wowList.pop(), 'installed by me')
if you do want to maintain wowList
wowListR = wowList[:]
while wowListR: installWow(wowListR.pop(), 'installed by me')
and if order matters
wowListR = wowList[:]; wowListR.reverse()
while wowListR: installWow(wowListR.pop(), 'installed by me')
Though as the solution of the puzzle I like the first :)

I tested several different variants, and here are the results I got.
Python 2:
>>> timeit.timeit('for x in xrange(100): L.append(x)', 'L = []')
14.9432640076
>>> timeit.timeit('[x for x in xrange(100) if L.append(x) and False]', 'L = []')
16.7011508942
>>> timeit.timeit('next((x for x in xrange(100) if L.append(x) and False), None)', 'L = []')
15.5235641003
>>> timeit.timeit('any(L.append(x) and False for x in xrange(100))', 'L = []')
20.9048290253
>>> timeit.timeit('filter(lambda x: L.append(x) and False, xrange(100))', 'L = []')
27.8524758816
Python 3:
>>> timeit.timeit('for x in range(100): L.append(x)', 'L = []')
13.719769178002025
>>> timeit.timeit('[x for x in range(100) if L.append(x) and False]', 'L = []')
15.041426660001889
>>> timeit.timeit('next((x for x in range(100) if L.append(x) and False), None)', 'L = []')
15.448063717998593
>>> timeit.timeit('any(L.append(x) and False for x in range(100))', 'L = []')
22.087335471998813
>>> timeit.timeit('next(filter(lambda x: L.append(x) and False, range(100)), None)', 'L = []')
36.72446593800123
Note that the time values are not that precise (for example, the relative performance of the first three options varied from run to run). My conclusion is that you should just use a loop, it's more readable and performs at least as well as the alternatives. If you want to avoid polluting the namespace, just del the variable after using it.

first rewrite the for loop as a generator expression, which does not allocate any memory.
(installWow(x, 'installed by me') for x in wowList )
But this expression doesn't actually do anything without finding some way to consume it. So we can rewrite this to yield something determinate, rather than rely on the possibly None result of installWow.
( [1, installWow(x, 'installed by me')][0] for x in wowList )
which creates a list, but returns only the constant 1. this can be consumed conveniently with reduce
reduce(sum, ( [1, installWow(x, 'installed by me')][0] for x in wowList ))
Which conveniently returns the number of items in wowList that were affected.

Just make installWow return None or make the last statement be pass like so:
def installWow(item, phrase='installed by me'):
print phrase
pass
and use this:
list(x for x in wowList if installWow(x))
x won't be set in the global name space and the list returned is [] a singleton

If you're worried about the need to control the return value (which you need to do to use filter) and prefer a simpler solution than the reduce example above, then consider using reduce directly. Your function will need to take an additional first parameter, but you can ignore it, or use a lambda to discard it:
reduce(lambda _x: installWow(_x, 'installed by me'), wowList, None)

Let me preface this by saying that it seems the original poster was more concerned about namespace clutter than anything else. In that case, you can wrap your working variables in separate function namespace and call it after declaring it, or you can simply remove them from the namespace after you've used them with the "del" builtin command. Or, if you have multiple variables to clean up, def the function with all the temp variables in it, run it, then del it.
Read on if the main concern is optimization:
Three more ways, potentially faster than others described here:
For Python >= 2.7, use collections.deque((installWow(x, 'installed by me') for x in wowList),0) # saves 0 entries while iterating the entire generator, but yes, still has a byproduct of a final object along with a per-item length check internally
If worried about this kind of overhead, install cytoolz. You can use count which still has a byproduct of incrementing a counter but it may be a smaller number of cycles than deque's per-item check, not sure. You can use it instead of any() in the next way:
Replace the generator expression with itertools.imap (when installWow never returns True. Otherwise you may consider itertools.ifilter and itertools.ifilterfalse with None for the predicate): any(itertools.imap(installWow,wowList,itertools.repeat('installed by me')))
But the real problem here is the fact that a function returns something and you do not want it to return anything.. So to resolve this, you have 2 options. One is to refactor your code so installWow takes in the wowList and iterates it internally. Another is rather mindblowing, but you can load the installWow() function into a compiled ast like so:
lines,lineno=inspect.getsourcelines(func) # func here is installWow without the parens
return ast.parse(join(l[4:] for l in lines if l)) # assumes the installWow function is part of a class in a module file.. For a module-level function you would not need the l[4:]
You can then do the same for the outer function, and traverse the ast to find the for loop. Then in the body of the for loop, insert the instalWow() function ast's function definition body, matching up the variable names. You can then simply call exec on the ast itself, and provide a namespace dictionary with the right variables filled in. To make sure your tree modifications are correct, you can check what the final source code would look like by running astunparse.
And if that isn't enough you can go to cython and write a .pyx file which will generate and compile a .c file into a library with python bindings. Then, at least the lost cycles won't be spent converting to and from python objects and type-checking everything repeatedly.

A simple DIY whose sole purpose is to loop through a generator expression:
def do(genexpr):
for _ in genexpr:
pass
Then use:
do(installWow(x, 'installed by me') for x in wowList)

In python 3 there are some ways to use a function with no return(just use a semicolon in jupyter ot ommit the output from the cell):
[*map(print, MY_LIST)]; # form 1 - unpack the map generator to a list
any(map(print, MY_LIST)); # form 2 - force execution with any
list(map(print, MY_LIST)); # form 3 - collect list from generator

Someone needs to answer --
The more pythonic way here is to not worry about polluting the namespace, and using __all__ to define the public variables.
myModule/__init__.py:
__all__ = ['func1', 'func2']
for x in range(10):
print 'init {}'.format(x)
def privateHelper1(x):
return '{}:{}'.format(x,x)
def func1():
print privateHelper1('func1')
def func2():
print privateHelper1('func1')
Then
python -c "import myModule; help(myModule);"
init 0
init 1
init 2
init 3
init 4
init 5
init 6
init 7
init 8
init 9
Help on package mm:
NAME
myModule
FILE
h:\myModule\__init__.py
PACKAGE CONTENTS
FUNCTIONS
func1()
func2()
DATA
__all__ = ['func1', 'func2']

No Multiline Lambda in Python: Why not?

I've heard it said that multiline lambdas can't be added in Python because they would clash syntactically with the other syntax constructs in Python. I was thinking about this on the bus today and realized I couldn't think of a single Python construct that multiline lambdas clash with. Given that I know the language pretty well, this surprised me.
Now, I'm sure Guido had a reason for not including multiline lambdas in the language, but out of curiosity: what's a situation where including a multiline lambda would be ambiguous? Is what I've heard true, or is there some other reason that Python doesn't allow multiline lambdas?

Guido van Rossum (the inventor of Python) answers this exact question himself in an old blog post.
Basically, he admits that it's theoretically possible, but that any proposed solution would be un-Pythonic:
"But the complexity of any proposed solution for this puzzle is immense, to me: it requires the parser (or more precisely, the lexer) to be able to switch back and forth between indent-sensitive and indent-insensitive modes, keeping a stack of previous modes and indentation level. Technically that can all be solved (there's already a stack of indentation levels that could be generalized). But none of that takes away my gut feeling that it is all an elaborate Rube Goldberg contraption."

Look at the following:
map(multilambda x:
y=x+1
return y
, [1,2,3])
Is this a lambda returning (y, [1,2,3]) (thus map only gets one parameter, resulting in an error)? Or does it return y? Or is it a syntax error, because the comma on the new line is misplaced? How would Python know what you want?
Within the parens, indentation doesn't matter to python, so you can't unambiguously work with multilines.
This is just a simple one, there's probably more examples.

This is generally very ugly (but sometimes the alternatives are even more ugly), so a workaround is to make a braces expression:
lambda: (
doFoo('abc'),
doBar(123),
doBaz())
It won't accept any assignments though, so you'll have to prepare data beforehand.
The place I found this useful is the PySide wrapper, where you sometimes have short callbacks. Writing additional member functions would be even more ugly. Normally you won't need this.
Example:
pushButtonShowDialog.clicked.connect(
lambda: (
field1.clear(),
spinBox1.setValue(0),
diag.show())

A couple of relevant links:
For a while, I was following the development of Reia, which was initially going to have Python's indentation based syntax with Ruby blocks too, all on top of Erlang. But, the designer wound up giving up on indentation sensitivity, and this post he wrote about that decision includes a discussion about problems he ran into with indentation + multi-line blocks, and an increased appreciation he gained for Guido's design issues/decisions:
http://www.unlimitednovelty.com/2009/03/indentation-sensitivity-post-mortem.html
Also, here's an interesting proposal for Ruby-style blocks in Python I ran across where Guido posts a response w/o actually shooting it down (not sure whether there has been any subsequent shoot down, though):
http://tav.espians.com/ruby-style-blocks-in-python.html

Let me present to you a glorious but terrifying hack:
import types
def _obj():
return lambda: None
def LET(bindings, body, env=None):
'''Introduce local bindings.
ex: LET(('a', 1,
'b', 2),
lambda o: [o.a, o.b])
gives: [1, 2]
Bindings down the chain can depend on
the ones above them through a lambda.
ex: LET(('a', 1,
'b', lambda o: o.a + 1),
lambda o: o.b)
gives: 2
'''
if len(bindings) == 0:
return body(env)
env = env or _obj()
k, v = bindings[:2]
if isinstance(v, types.FunctionType):
v = v(env)
setattr(env, k, v)
return LET(bindings[2:], body, env)
You can now use this LET form as such:
map(lambda x: LET(('y', x + 1,
'z', x - 1),
lambda o: o.y * o.z),
[1, 2, 3])
which gives: [0, 3, 8]

[Edit Edit] Since this question is somehow still active 12 years after being asked. I will continue the tradition of amending my answer every 4 years or so.
Firstly, the question was how does multi-line lambda clash with Python. The accepted answer shows how with a simple example. The highly rated answer I linked below some years ago answers the question of "Why is it not a part of Python"--this answer is perhaps more satisfying to those who believe that the existing examples of "clashing" are not enough to make multi-line lambda impossible to implement in Python.
In previous iterations of this answer I discussed how to implement multi-line lambda into Python as is. I've since removed that part, because it was a flurry of bad practices. You may see it in the edit history of this answer if you wish.
However the answer to "Why not?", being "because Rossum said so" can still be a source of frustration. So lets see if it could be engineered around the counter example given by user balpha:
map(lambda x:
y=x+1 # <-- this line defines the outmost indent level*
for i in range(12):
y+=12
return y
, [1,2,3])
#*By convention it is always one-indent past the 'l' in lambda
As for the return value we have that the following is non-permissible in python:
def f():
return 3
, [1,2,3]
So by the same logic, "[1,2,3]" should not be part of the return value. Let's try it this way instead:
map(lambda x:
y=x+1 # part of lambda block
for i in range(12): # part of lambda block
y+=12 # part of lambda block
return y, [1,2,3]) # part of lambda block
This one's trickier, but since the lambda block has a clearly defined beginning (the token 'lambda') yet no clear ending, I would argue anything that is on the same line as part of a lambda block is also part of the lambda block.
One might imagine some features that can identify closing parenthesis or even inference based on the number of tokens expected by the enclosing element. In general, the above expression does not seem totally impossible to parse, but it may be a bit of a challenge.
To simplify things, you could separate all characters not intended to be part of the block:
map(lambda x:
y=x+1 # part of lambda block
for i in range(12): # part of lambda block
y+=12 # part of lambda block
return y # part of lambda block
, [1,2,3]) # argument separator, second argument, and closing paren for map
Back to where we were but this time it is unambiguous, because the last line is behind the lowest indent-depth for the lambda block.
Single line lambda would be a special case (identified by the lack of an immediate newline after the color), that behaves the same as it does now.
This is not to say that it necessarily should be a part of Python--but it is a quick illustration that is perhaps is possible with some changes in the language.
[Edit] Read this answer. It explains why multi-line lambda is not a thing.
Simply put, it's unpythonic. From Guido van Rossum's blog post:
I find any solution unacceptable that embeds an indentation-based block in the middle of an expression. Since I find alternative syntax for statement grouping (e.g. braces or begin/end keywords) equally unacceptable, this pretty much makes a multi-line lambda an unsolvable puzzle.

I'm guilty of practicing this dirty hack in some of my projects which is bit simpler:
lambda args...:( expr1, expr2, expr3, ...,
exprN, returnExpr)[-1]
I hope you can find a way to stay pythonic but if you have to do it this less painful than using exec and manipulating globals.

Let me also throw in my two cents about different workarounds.
How is a simple one-line lambda different from a normal function? I can think only of lack of assignments, some loop-like constructs (for, while), try-except clauses... And that's it? We even have a ternary operator - cool! So, let's try to deal with each of these problems.
Assignments
Some guys here have rightly noted that we should take a look at lisp's let form, which allows local bindings. Actually, all the non state-changing assignments can be performed only with let. But every lisp programmer knows that let form is absolutely equivalent to call to a lambda function! This means that
(let ([x_ x] [y_ y])
(do-sth-with-x-&-y x_ y_))
is the same as
((lambda (x_ y_)
(do-sth-with-x-&-y x_ y_)) x y)
So lambdas are more than enough! Whenever we want to make a new assignment we just add another lambda and call it. Consider this example:
def f(x):
y = f1(x)
z = f2(x, y)
return y,z
A lambda version looks like:
f = lambda x: (lambda y: (y, f2(x,y)))(f1(x))
You can even make the let function, if you don't like the data being written after actions on the data. And you can even curry it (just for the sake of more parentheses :) )
let = curry(lambda args, f: f(*args))
f_lmb = lambda x: let((f1(x),), lambda y: (y, f2(x,y)))
# or:
f_lmb = lambda x: let((f1(x),))(lambda y: (y, f2(x,y)))
# even better alternative:
let = lambda *args: lambda f: f(*args)
f_lmb = lambda x: let(f1(x))(lambda y: (y, f2(x,y)))
So far so good. But what if we have to make reassignments, i.e. change state? Well, I think we can live absolutely happily without changing state as long as task in question doesn't concern loops.
Loops
While there's no direct lambda alternative for loops, I believe we can write quite generic function to fit our needs. Take a look at this fibonacci function:
def fib(n):
k = 0
fib_k, fib_k_plus_1 = 0, 1
while k < n:
k += 1
fib_k_plus_1, fib_k = fib_k_plus_1 + fib_k, fib_k_plus_1
return fib_k
Impossible in terms of lambdas, obviously. But after writing a little yet useful function we're done with that and similar cases:
def loop(first_state, condition, state_changer):
state = first_state
while condition(*state):
state = state_changer(*state)
return state
fib_lmb = lambda n:\
loop(
(0,0,1),
lambda k, fib_k, fib_k_plus_1:\
k < n,
lambda k, fib_k, fib_k_plus_1:\
(k+1, fib_k_plus_1, fib_k_plus_1 + fib_k))[1]
And of course, one should always consider using map, reduce and other higher-order functions if possible.
Try-except and other control structs
It seems like a general approach to this kind of problems is to make use of lazy evaluation, replacing code blocks with lambdas accepting no arguments:
def f(x):
try: return len(x)
except: return 0
# the same as:
def try_except_f(try_clause, except_clause):
try: return try_clause()
except: return except_clause()
f = lambda x: try_except_f(lambda: len(x), lambda: 0)
# f(-1) -> 0
# f([1,2,3]) -> 3
Of course, this is not a full alternative to try-except clause, but you can always make it more generic. Btw, with that approach you can even make if behave like function!
Summing up: it's only natural that everything mentioned feels kinda unnatural and not-so-pythonically-beautiful. Nonetheless - it works! And without any evals and other trics, so all the intellisense will work. I'm also not claiming that you shoud use this everywhere. Most often you'd better define an ordinary function. I only showed that nothing is impossible.

Let me try to tackle #balpha parsing problem. I would use parentheses around the multiline lamda. If there is no parentheses, the lambda definition is greedy. So the lambda in
map(lambda x:
y = x+1
z = x-1
y*z,
[1,2,3]))
returns a function that returns (y*z, [1,2,3])
But
map((lambda x:
y = x+1
z = x-1
y*z)
,[1,2,3]))
means
map(func, [1,2,3])
where func is the multiline lambda that return y*z. Does that work?

(For anyone still interested in the topic.)
Consider this (includes even usage of statements' return values in further statements within the "multiline" lambda, although it's ugly to the point of vomiting ;-)
>>> def foo(arg):
... result = arg * 2;
... print "foo(" + str(arg) + ") called: " + str(result);
... return result;
...
>>> f = lambda a, b, state=[]: [
... state.append(foo(a)),
... state.append(foo(b)),
... state.append(foo(state[0] + state[1])),
... state[-1]
... ][-1];
>>> f(1, 2);
foo(1) called: 2
foo(2) called: 4
foo(6) called: 12
12

Here's a more interesting implementation of multi line lambdas. It's not possible to achieve because of how python use indents as a way to structure code.
But luckily for us, indent formatting can be disabled using arrays and parenthesis.
As some already pointed out, you can write your code as such:
lambda args: (expr1, expr2,... exprN)
In theory if you're guaranteed to have evaluation from left to right it would work but you still lose values being passed from one expression to an other.
One way to achieve that which is a bit more verbose is to have
lambda args: [lambda1, lambda2, ..., lambdaN]
Where each lambda receives arguments from the previous one.
def let(*funcs):
def wrap(args):
result = args
for func in funcs:
if not isinstance(result, tuple):
result = (result,)
result = func(*result)
return result
return wrap
This method let you write something that is a bit lisp/scheme like.
So you can write things like this:
let(lambda x, y: x+y)((1, 2))
A more complex method could be use to compute the hypotenuse
lst = [(1,2), (2,3)]
result = map(let(
lambda x, y: (x**2, y**2),
lambda x, y: (x + y) ** (1/2)
), lst)
This will return a list of scalar numbers so it can be used to reduce multiple values to one.
Having that many lambda is certainly not going to be very efficient but if you're constrained it can be a good way to get something done quickly then rewrite it as an actual function later.

In Python 3.8/3.9 there is Assignment Expression, so it could be used in lambda, greatly
expanding functionality
E.g., code
#%%
x = 1
y = 2
q = list(map(lambda t: (
tx := t*x,
ty := t*y,
tx+ty
)[-1], [1, 2, 3]))
print(q)
will print [3, 6, 9]

After Python3.8, there is another method for local binding
lambda x: (
y := x + 1,
y ** 2
)[-1]
For Loop
lambda x: (
y := x ** 2,
[y := y + x for _ in range(10)],
y
)[-1]
If Branch
lambda x: (
y := x ** 2,
x > 5 and [y := y + x for _ in range(10)],
y
)[-1]
Or
lambda x: (
y := x ** 2,
[y := y + x for _ in range(10)] if x > 5 else None,
y
)[-1]
While Loop
import itertools as it
lambda x: (
l := dict(y = x ** 2),
cond := lambda: l['y'] < 100,
body := lambda: l.update(y = l['y'] + x),
*it.takewhile(lambda _: cond() and (body(), True)[-1], it.count()),
l['y']
)[-1]
Or
import itertools as it
from types import SimpleNamespace as ns
lambda x: (
l := ns(y = x ** 2),
cond := lambda: l.y < 100,
body := lambda: vars(l).update(y = l.y + x),
*it.takewhile(lambda _: cond() and (body(), True)[-1], it.count()),
l.y
)[-1]
Or
import itertools as it
lambda x: (
y := x ** 2,
*it.takewhile(lambda t: t[0],
((
pred := y < 100,
pred and (y := y + x))
for _ in it.count())),
y
)[-1]

On the subject of ugly hacks, you can always use a combination of exec and a regular function to define a multiline function like this:
f = exec('''
def mlambda(x, y):
d = y - x
return d * d
''', globals()) or mlambda
You can wrap this into a function like:
def mlambda(signature, *lines):
exec_vars = {}
exec('def mlambda' + signature + ':\n' + '\n'.join('\t' + line for line in lines), exec_vars)
return exec_vars['mlambda']
f = mlambda('(x, y)',
'd = y - x',
'return d * d')

I know it is an old question, but for the record here is a kind of a solution to the problem of multiline lambda problem in which the result of one call is consumed by another call.
I hope it is not super hacky, since it is based only on standard library functions and uses no dunder methods.
Below is a simple example in which we start with x = 3 and then in the first line we add 1 and then in the second line we add 2 and get 6 as the output.
from functools import reduce
reduce(lambda data, func: func(data), [
lambda x: x + 1,
lambda x: x + 2
], 3)
## Output: 6

I was just playing a bit to try to make a dict comprehension with reduce, and come up with this one liner hack:
In [1]: from functools import reduce
In [2]: reduce(lambda d, i: (i[0] < 7 and d.__setitem__(*i[::-1]), d)[-1], [{}, *{1:2, 3:4, 5:6, 7:8}.items()])
Out[3]: {2: 1, 4: 3, 6: 5}
I was just trying to do the same as what was done in this Javascript dict comprehension: https://stackoverflow.com/a/11068265

You can simply use slash (\) if you have multiple lines for your lambda function
Example:
mx = lambda x, y: x if x > y \
else y
print(mx(30, 20))
Output: 30

I am starting with python but coming from Javascript the most obvious way is extract the expression as a function....
Contrived example, multiply expression (x*2) is extracted as function and therefore I can use multiline:
def multiply(x):
print('I am other line')
return x*2
r = map(lambda x : multiply(x), [1, 2, 3, 4])
print(list(r))
https://repl.it/#datracka/python-lambda-function
Maybe it does not answer exactly the question if that was how to do multiline in the lambda expression itself, but in case somebody gets this thread looking how to debug the expression (like me) I think it will help

One safe method to pass any number of variables between lambda items:
print((lambda: [
locals().__setitem__("a", 1),
locals().__setitem__("b", 2),
locals().__setitem__("c", 3),
locals().get("a") + locals().get("b") + locals().get("c")
])()[-1])
Output: 6

because a lambda function is supposed to be one-lined, as its the simplest form of a function, an entrance, then return