How to add with tuples - python

I have pseudo-code like this:
if( b < a)
return (1,0)+foo(a-b,b)
I want to write it in python. But can python add tuples? What is the best way to code something like that?

I'd go for
>>> map(sum, zip((1, 2), (3, 4)))
[4, 6]
or, more naturally:
>>> numpy.array((1, 2)) + numpy.array((3, 4))
array([4, 6])

Do you want to do element-wise addition, or to append the tuples? By default python does
(1,2)+(3,4) = (1,2,3,4)
You could define your own as:
def myadd(x,y):
z = []
for i in range(len(x)):
z.append(x[i]+y[i])
return tuple(z)
Also, as #delnan's comment makes it clear, this is better written as
def myadd(xs,ys):
return tuple(x + y for x, y in izip(xs, ys))
or even more functionally:
myadd = lambda xs,ys: tuple(x + y for x, y in izip(xs, ys))
Then do
if( b < a) return myadd((1,0),foo(a-b,b))

tuple(map(operator.add, a, b))
In contrast to the answer by highBandWidth, this approach requires that the tuples be of the same length in Python 2.7 or earlier, instead raising a TypeError. In Python 3, map is slightly different, so that the result is a tuple of sums with length equal to the shorter of a and b.
If you want the truncation behavior in Python 2, you can replace map with itertools.imap:
tuple(itertools.imap(operator.add, a, b))

If you want + itself to act this way, you could subclass tuple and override the addition:
class mytup(tuple):
def __add__(self, other):
if len(self) != len(other):
return NotImplemented # or raise an error, whatever you prefer
else:
return mytup(x+y for x,y in izip(self,other))
The same goes for __sub__, __mul__, __div__, __gt__ (elementwise >) etc. More information on these special operators can be found e.g. here (numeric operations) and here (comparisions)
You can still append tuples by calling the original tuple addition: tuple.__add__(a,b) instead of a+b. Or define an append() function in the new class to do this.

Related

Python - Lambda function invalid syntax error [duplicate]

In Python 2, I can write:
In [5]: points = [ (1,2), (2,3)]
In [6]: min(points, key=lambda (x, y): (x*x + y*y))
Out[6]: (1, 2)
But that is not supported in 3.x:
File "<stdin>", line 1
min(points, key=lambda (x, y): (x*x + y*y))
^
SyntaxError: invalid syntax
The straightforward workaround is to index explicitly into the tuple that was passed:
>>> min(points, key=lambda p: p[0]*p[0] + p[1]*p[1])
(1, 2)
This is very ugly. If the lambda were a function, I could do
def some_name_to_think_of(p):
x, y = p
return x*x + y*y
But because the lambda only supports a single expression, it's not possible to put the x, y = p part into it.
How else can I work around this limitation?
No, there is no other way. You covered it all. The way to go would be to raise this issue on the Python ideas mailing list, but be prepared to argue a lot over there to gain some traction.
Actually, just not to say "there is no way out", a third way could be to implement one more level of lambda calling just to unfold the parameters - but that would be at once more inefficient and harder to read than your two suggestions:
min(points, key=lambda p: (lambda x,y: (x*x + y*y))(*p))
Python 3.8 update
Since the release of Python 3.8, PEP 572 — assignment expressions — have been available as a tool.
So, if one uses a trick to execute multiple expressions inside a lambda - I usually do that by creating a tuple and just returning the last component of it, it is possible to do the following:
>>> a = lambda p:(x:=p[0], y:=p[1], x ** 2 + y ** 2)[-1]
>>> a((3,4))
25
One should keep in mind that this kind of code will seldom be more readable or practical than having a full function. Still, there are possible uses - if there are various one-liners that would operate on this point, it could be worth to have a namedtuple, and use the assignment expression to effectively "cast" the incoming sequence to the namedtuple:
>>> from collections import namedtuple
>>> point = namedtuple("point", "x y")
>>> b = lambda s: (p:=point(*s), p.x ** 2 + p.y ** 2)[-1]
According to http://www.python.org/dev/peps/pep-3113/ tuple unpacking are gone, and 2to3 will translate them like so:
As tuple parameters are used by lambdas because of the single
expression limitation, they must also be supported. This is done by
having the expected sequence argument bound to a single parameter and
then indexing on that parameter:
lambda (x, y): x + y
will be translated into:
lambda x_y: x_y[0] + x_y[1]
Which is quite similar to your implementation.
I don't know any good general alternatives to the Python 2 arguments unpacking behaviour. Here's a couple of suggestion that might be useful in some cases:
if you can't think of a name; use the name of the keyword parameter:
def key(p): # more specific name would be better
x, y = p
return x**2 + y**3
result = min(points, key=key)
you could see if a namedtuple makes your code more readable if the list is used in multiple places:
from collections import namedtuple
from itertools import starmap
points = [ (1,2), (2,3)]
Point = namedtuple('Point', 'x y')
points = list(starmap(Point, points))
result = min(points, key=lambda p: p.x**2 + p.y**3)
While the destructuring arguments was removed in Python3, it was not removed from comprehensions. It is possible to abuse it to obtain similar behavior in Python 3.
For example:
points = [(1,2), (2,3)]
print(min(points, key=lambda y: next(x*x + y*y for (x,y) in [y])))
In comparison with the accepted answer of using a wrapper, this solution is able to completely destructure the arguments while the wrapper only destructures the first level. That is, you can do
values = [(('A',1),'a'), (('B',0),'b')]
print(min(values, key=lambda y: next(b for ((a,b),c) in (y,))))
In comparison to the accepted answer using an unwrapper lambda:
values = [(('A',1),'a'), (('B',0),'b')]
print(min(points, key=lambda p: (lambda a,b: (lambda x,y: (y))(*a))(*p)))
Alternatively one can also use a list instead of a tuple.
values = [(('A',1),'a'), (('B',0),'b')]
print(min(points, key=lambda y: next(b for (a,b),c in [y])))
This is just to suggest that it can be done, and should not be taken as a recommendation. However, IMO, this is better than the hack of using using multiple expressions in a tuple and returning the last one.
I think the better syntax is x * x + y * y let x, y = point, let keyword should be more carefully chosen.
The double lambda is the closest version.
lambda point: (lambda x, y: x * x + y * y)(*point)
High order function helper would be useful in case we give it a proper name.
def destruct_tuple(f):
return lambda args: f(*args)
destruct_tuple(lambda x, y: x * x + y * y)
Consider whether you need to unpack the tuple in the first place:
min(points, key=lambda p: sum(x**2 for x in p))
or whether you need to supply explicit names when unpacking:
min(points, key=lambda p: abs(complex(*p)**2)
Based on Cuadue suggestion and your comment on unpacking still being present in comprehensions, you can use, using numpy.argmin :
result = points[numpy.argmin(x*x + y*y for x, y in points)]
Another option is to write it into a generator producing a tuple where the key is the first element. Tuples are compared starting from beginning to end so the tuple with the smallest first element is returned. You can then index into the result to get the value.
min((x * x + y * y, (x, y)) for x, y in points)[1]
There may be a real solution to this, using PyFunctional!
Although not currently supported, I've submitted a tuple arg unpacking feature request to support:
(
seq((1, 2), (3, 4))
.map(unpack=lambda a, b: a + b)
) # => [3, 7]
Since questions on Stack Overflow are not supposed to contain the answer in the question, nor have explicit "update" sections, I am converting OP's original "updates" to a proper answer and making it community wiki.
OP originally claimed that this solution was "extending the idea in the answer". I cannot discern which answer that meant, or which idea. The idea is functionally the same as anthony.hl's answer, but that came years later. Considering the state of answers at the time, I think this qualifies as OP's original work.)
Make a wrapper function that generalizes the process of unpacking the arguments, like so:
def star(f):
return lambda args: f(*args)
Now we can use this to transform the lambda we want to write, into one that will receive the argument properly:
min(points, key=star(lambda x,y: (x*x + y*y))
We can further clean this up by using functools.wraps:
import functools
def star(f):
#functools.wraps(f)
def f_inner(args):
return f(*args)
return f_inner

sorted dictionary with lambda / Python3 vs Python 2 [duplicate]

In Python 2, I can write:
In [5]: points = [ (1,2), (2,3)]
In [6]: min(points, key=lambda (x, y): (x*x + y*y))
Out[6]: (1, 2)
But that is not supported in 3.x:
File "<stdin>", line 1
min(points, key=lambda (x, y): (x*x + y*y))
^
SyntaxError: invalid syntax
The straightforward workaround is to index explicitly into the tuple that was passed:
>>> min(points, key=lambda p: p[0]*p[0] + p[1]*p[1])
(1, 2)
This is very ugly. If the lambda were a function, I could do
def some_name_to_think_of(p):
x, y = p
return x*x + y*y
But because the lambda only supports a single expression, it's not possible to put the x, y = p part into it.
How else can I work around this limitation?
No, there is no other way. You covered it all. The way to go would be to raise this issue on the Python ideas mailing list, but be prepared to argue a lot over there to gain some traction.
Actually, just not to say "there is no way out", a third way could be to implement one more level of lambda calling just to unfold the parameters - but that would be at once more inefficient and harder to read than your two suggestions:
min(points, key=lambda p: (lambda x,y: (x*x + y*y))(*p))
Python 3.8 update
Since the release of Python 3.8, PEP 572 — assignment expressions — have been available as a tool.
So, if one uses a trick to execute multiple expressions inside a lambda - I usually do that by creating a tuple and just returning the last component of it, it is possible to do the following:
>>> a = lambda p:(x:=p[0], y:=p[1], x ** 2 + y ** 2)[-1]
>>> a((3,4))
25
One should keep in mind that this kind of code will seldom be more readable or practical than having a full function. Still, there are possible uses - if there are various one-liners that would operate on this point, it could be worth to have a namedtuple, and use the assignment expression to effectively "cast" the incoming sequence to the namedtuple:
>>> from collections import namedtuple
>>> point = namedtuple("point", "x y")
>>> b = lambda s: (p:=point(*s), p.x ** 2 + p.y ** 2)[-1]
According to http://www.python.org/dev/peps/pep-3113/ tuple unpacking are gone, and 2to3 will translate them like so:
As tuple parameters are used by lambdas because of the single
expression limitation, they must also be supported. This is done by
having the expected sequence argument bound to a single parameter and
then indexing on that parameter:
lambda (x, y): x + y
will be translated into:
lambda x_y: x_y[0] + x_y[1]
Which is quite similar to your implementation.
I don't know any good general alternatives to the Python 2 arguments unpacking behaviour. Here's a couple of suggestion that might be useful in some cases:
if you can't think of a name; use the name of the keyword parameter:
def key(p): # more specific name would be better
x, y = p
return x**2 + y**3
result = min(points, key=key)
you could see if a namedtuple makes your code more readable if the list is used in multiple places:
from collections import namedtuple
from itertools import starmap
points = [ (1,2), (2,3)]
Point = namedtuple('Point', 'x y')
points = list(starmap(Point, points))
result = min(points, key=lambda p: p.x**2 + p.y**3)
While the destructuring arguments was removed in Python3, it was not removed from comprehensions. It is possible to abuse it to obtain similar behavior in Python 3.
For example:
points = [(1,2), (2,3)]
print(min(points, key=lambda y: next(x*x + y*y for (x,y) in [y])))
In comparison with the accepted answer of using a wrapper, this solution is able to completely destructure the arguments while the wrapper only destructures the first level. That is, you can do
values = [(('A',1),'a'), (('B',0),'b')]
print(min(values, key=lambda y: next(b for ((a,b),c) in (y,))))
In comparison to the accepted answer using an unwrapper lambda:
values = [(('A',1),'a'), (('B',0),'b')]
print(min(points, key=lambda p: (lambda a,b: (lambda x,y: (y))(*a))(*p)))
Alternatively one can also use a list instead of a tuple.
values = [(('A',1),'a'), (('B',0),'b')]
print(min(points, key=lambda y: next(b for (a,b),c in [y])))
This is just to suggest that it can be done, and should not be taken as a recommendation. However, IMO, this is better than the hack of using using multiple expressions in a tuple and returning the last one.
I think the better syntax is x * x + y * y let x, y = point, let keyword should be more carefully chosen.
The double lambda is the closest version.
lambda point: (lambda x, y: x * x + y * y)(*point)
High order function helper would be useful in case we give it a proper name.
def destruct_tuple(f):
return lambda args: f(*args)
destruct_tuple(lambda x, y: x * x + y * y)
Consider whether you need to unpack the tuple in the first place:
min(points, key=lambda p: sum(x**2 for x in p))
or whether you need to supply explicit names when unpacking:
min(points, key=lambda p: abs(complex(*p)**2)
Based on Cuadue suggestion and your comment on unpacking still being present in comprehensions, you can use, using numpy.argmin :
result = points[numpy.argmin(x*x + y*y for x, y in points)]
Another option is to write it into a generator producing a tuple where the key is the first element. Tuples are compared starting from beginning to end so the tuple with the smallest first element is returned. You can then index into the result to get the value.
min((x * x + y * y, (x, y)) for x, y in points)[1]
There may be a real solution to this, using PyFunctional!
Although not currently supported, I've submitted a tuple arg unpacking feature request to support:
(
seq((1, 2), (3, 4))
.map(unpack=lambda a, b: a + b)
) # => [3, 7]
Since questions on Stack Overflow are not supposed to contain the answer in the question, nor have explicit "update" sections, I am converting OP's original "updates" to a proper answer and making it community wiki.
OP originally claimed that this solution was "extending the idea in the answer". I cannot discern which answer that meant, or which idea. The idea is functionally the same as anthony.hl's answer, but that came years later. Considering the state of answers at the time, I think this qualifies as OP's original work.)
Make a wrapper function that generalizes the process of unpacking the arguments, like so:
def star(f):
return lambda args: f(*args)
Now we can use this to transform the lambda we want to write, into one that will receive the argument properly:
min(points, key=star(lambda x,y: (x*x + y*y))
We can further clean this up by using functools.wraps:
import functools
def star(f):
#functools.wraps(f)
def f_inner(args):
return f(*args)
return f_inner

Pythonic way to do `if x in y where x.attr == val`?

I have a class that represents a polynomial as a collection of terms where each term has a coefficient and an exponent. I am working on the __add__ method of the class and I am wondering what the most effective way to do something like:
def __add__(self, other):
new_terms = []
for term in self.terms:
if there is a term in other with an exponent == term.exponent
new_terms.append(Term(term.coef + other_term.coef, term.exponent))
It strikes me that I'm looking for something such as:
if x in y where x.attr == val
Or in my specific case:
if x in other where x.exponent == term.exponent
Does such a thing exist?
You need to filter your list before doing your contains check. As tobias_k suggested, you can either build a new list, e.g.
[x for x in other if x.exponent == term.exponent]
This works directly in an if statement, because an empty list is False:
if [x for x in other if x.exponent == term.exponent]:
But this does some wasteful work, since it a) has to construct a new list and b) doesn't short-circuit once a result is found. Better is to use the same syntax in a generator expression:
(True for x in other if x.exponent == term.exponent)
Which you can then similarly use in an if statement, but no wasteful work is done:
if next((True for x in other if x.exponent == term.exponent), False):
I think you want [x for x in y if x.attr == val], or use next with the same expression for just the first such value.
In your case, it could look something like this:
def __add__(self, other):
for term in self.terms:
for other_term in (x for x in other.terms
if x.exponent == term.exponent):
term.coefficient += other_term.coefficient
However, this will not work too well. First, __add__ should not modify neither self nor other but instead create a new polynomial. Also, this will ignore any values from other that have a different exponent that any of the terms in self. And third, the performance is pretty lousy, as it loops the list of terms in other for each term in self, giving it quadratic complexity.
Instead, I suggest using a dictionary, mapping exponents in the term to their coefficient. In fact, you could probably just use a collections.Counter for that; it already implements __add__ in the right way. Something like this:
class Poly:
def __init__(self, counts):
self.terms = collections.Counter(counts)
def __add__(self, other):
return Poly(self.terms + other.terms)
def __str__(self):
return " + ".join("%dx^%d" % (c, x) for x, c in self.terms.items())
Example:
>>> Poly({2: 1, 1: 3, 0: 5}) + Poly({3: 1, 1: 2, 0: 3})
8x^0 + 5x^1 + 1x^2 + 1x^3

How do i write a function acting on a list properly?

In Python, I would like to have a function working on different input types. Something like this:
def my_square(x):
return x ** 2
my_square(2) #return 4
my_square(range(10)) #should return a list [0 ... 81]
npa = numpy.zeros(10)
my_square(npa) # should return a numpy array with the squares of zeros
Basically, what is good practice to write functions for both scalars and iterables? Can this be done with *args or *kwargs perhaps?
A typical way to do this is to use numpy.asarray to convert the input of your function to an ndarray. If the input is a scalar, then the result is an array with 0 dimensions, which acts essentially like a Python number type. For instance:
def square(x):
x = np.asarray(x)
return x**2
So that:
>>> square(4)
16
>>> square([1, 2, 3, 4])
array([ 1, 4, 9, 16])
Note that I gave a list as input and received an ndarray as output. If you absolutely must receive the same type of output as you provided as input, you can convert the result before returning it:
def square(x):
in_type = type(x)
x = np.asarray(x)
return in_type(x**2)
But this incurs an additional cost for little benefit.
Since Python is dynamically typed, and the design philosophy of Python is against the idea of wanting a difference to be obvious when the function is called, this isn't considered Pythonic. You can however use the isInstance() method to accomplish what you want, if you must that is. For example:
def mySquare(x):
if isinstance(x, int):
return x**2
elif isinstance(x, range):
return [i ** 2 for i in x]
print(mySquare(2)); //4
print(mySquare(range(10))); //[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]
IDEOne link here.
However, just because you can do something, doesn't mean you should.
Refer to this question for further information on isinstance, and I suggest you take a look at Duck Typing as well.
Additionally, a single dispatch function might also provide what you need, but I am not experienced enough to provide an explanation for this, however, it might be something you want to look into.
As suggested in the comments, just write a scalar version of your function and use map for lists etc and imap for iterables (map will not work on those):
map(myFunc,myList)
and
import itertools
itertools.imap(myFunc,myIterable)
It would be much better practice and much more maintainable to just do:
def foo(n):
return n ** 2
And then build lists, dicts, etc when needed (I'm not super familiar with numpy but I imagine there is something similar you could do):
foo_list = [foo(n) for n in range(10)]
foo_dict = {n: foo(n) for n in range(10)}
And it seems that there is for numpy using numpy.fromiter(). From the docs:
iterable = (foo(n) for n in range(5))
foo_arr = np.fromiter(iterable, np.whatever_numpy_datatype)
you could even make those into functions if you really needed to:
def foo_list(start=0, stop=0):
return [foo(n) for n in range(start, stop)]
def foo_dict(start=0, stop=0):
return {n: foo(n) for n in range(10)}
Another option as it's easier to ask for forgiveness than permission:
def foo(scalar_or_iter):
try:
return [n ** 2 for n in scalar_or_iter]
except TypeError:
return scalar_or_iter ** 2

Python, lambda, find minimum

I have foreach function which calls specified function on every element which it contains. I want to get minimum from thise elements but I have no idea how to write lambda or function or even a class that would manage that.
Thanks for every help.
I use my foreach function like this:
o.foreach( lambda i: i.call() )
or
o.foreach( I.call )
I don't like to make a lists or other objects. I want to iterate trough it and find min.
I manage to write a class that do the think but there should be some better solution than that:
class Min:
def __init__(self,i):
self.i = i
def get_min(self):
return self.i
def set_val(self,o):
if o.val < self.i: self.i = o.val
m = Min( xmin )
self.foreach( m.set_val )
xmin = m.get_min()
Ok, so I suppose that my .foreach method is non-python idea. I should do my Class iterable because all your solutions are based on lists and then everything will become easier.
In C# there would be no problem with lambda function like that, so I though that python is also that powerful.
Python has built-in support for finding minimums:
>>> min([1, 2, 3])
1
If you need to process the list with a function first, you can do that with map:
>>> def double(x):
... return x * 2
...
>>> min(map(double, [1, 2, 3]))
2
Or you can get fancy with list comprehensions and generator expressions, for example:
>>> min(double(x) for x in [1, 2, 3])
2
You can't do this with foreach and a lambda. If you want to do this in a functional style without actually using min, you'll find reduce is pretty close to the function you were trying to define.
l = [5,2,6,7,9,8]
reduce(lambda a,b: a if a < b else b, l[1:], l[0])
Writing foreach method is not very pythonic. You should better make it an iterator so that it works with standard python functions like min.
Instead of writing something like this:
def foreach(self, f):
for d in self._data:
f(d)
write this:
def __iter__(self):
for d in self._data:
yield d
Now you can call min as min(myobj).
I have foreach function which calls specified function on every element which it contains
It sounds, from the comment you subsequently posted, that you have re-invented the built-in map function.
It sounds like you're looking for something like this:
min(map(f, seq))
where f is the function that you want to call on every item in the list.
As gnibbler shows, if you want to find the value x in the sequence for which f(x) returns the lowest value, you can use:
min(seq, key=f)
...unless you want to find all of the items in seq for which f returns the lowest value. For instance, if seq is a list of dictionaries,
min(seq, key=len)
will return the first dictionary in the list with the smallest number of items, not all dictionaries that contain that number of items.
To get a list of all items in a sequence for which the function f returns the smallest value, do this:
values = map(f, seq)
result = [seq[i] for (i, v) in enumerate(values) if v == min(values)]
Okay, one thing you need to understand: lambda creates a function object for you. But so does plain, ordinary def. Look at this example:
lst = range(10)
print filter(lambda x: x % 2 == 0, lst)
def is_even(x):
return x % 2 == 0
print filter(is_even, lst)
Both of these work. They produce the same identical result. lambda makes an un-named function object; def makes a named function object. filter() doesn't care whether the function object has a name or not.
So, if your only problem with lambda is that you can't use = in a lambda, you can just make a function using def.
Now, that said, I don't suggest you use your .foreach() method to find a minimum value. Instead, make your main object return a list of values, and simply call the Python min() function.
lst = range(10)
print min(lst)
EDIT: I agree that the answer that was accepted is better. Rather than returning a list of values, it is better to define __iter__() and make the object iterable.
Suppose you have
>>> seq = range(-4,4)
>>> def f(x):
... return x*x-2
for the minimum value of f
>>> min(f(x) for x in seq)
-2
for the value of x at the minimum
>>> min(seq, key=f)
0
of course you can use lambda too
>>> min((lambda x:x*x-2)(x) for x in range(-4,4))
-2
but that is a little ugly, map looks better here
>>> min(map(lambda x:x*x-2, seq))
-2
>>> min(seq,key=lambda x:x*x-2)
0
You can use this:
x = lambda x,y,z: min(x,y,z)
print(x(3,2,1))

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