I'm trying to define a function of 3 unknown variables (k,m,c) but let's say I have to define it 100 times due to a different frequency f each time. Can I do this in a for loop in python and store all the functions in a list such that they are callable later?
This is what I have so far
index = -1
f_t = []
for f in inputs[:,0]:
index = index +1
def F_analytic(k, m, c):
F_t = k*m*c*f
return F_t
f_t.append([])
f_t[index].append(F_analytic)
but what I get in turn is a list of functions which are not callable:
Out[58]:
[[<function __main__.F_analytic>],
[<function __main__.F_analytic>],
[<function __main__.F_analytic>],
[<function __main__.F_analytic>],
[<function __main__.F_analytic>],
...
...
[<function __main__.F_analytic>],
[<function __main__.F_analytic>],
[<function __main__.F_analytic>]]
and the error is:
TypeError: 'list' object is not callable
Any help?
Thank you!
① You are nesting lists although you just want a flat list (of functions). See the answer of #BartoszKP on that.
② You want to create functions based on local variables. You can do this using lambdas, as #Harsh is proposing, or you can do it using defaulted variables:
def F_analytic(k, m, c, f=f): # notice the f=f here!
F_t = k*m*c*f
return F_t
③ You should consider whether having a list of functions really is what you want (as #Wooble already pointed out).
No, what you have is a list of lists of functions, with each inner list having 1 function.
Replace
f_t.append([])
f_t[index].append(F_analytic)
with
f_t.append(F_analytic)
(Although honestly, this whole approach seems rather suspicious; any reason you don't want one function with 4 parameters instead of 100 with 3?)
This is because you are creating a list of lists, you can call each function object like this:
f_t[0][0](1,2,3)
Alternatively, just create a list of functions:
f_t = []
for f in inputs[:,0]:
index = index +1
def F_analytic(k, m, c):
F_t = k*m*c*f
return F_t
f_t.append(F_analytic)
And as noted by other answers and comments, this all seems a rather strange approach. Do you really need to do this? :)
Sorry i dont think i answered the question properly, try this approach:
f_t = []
for f in input[:,0]:
f_t.append(lambda k, m, c: k*m*c*f)
You can call the functions easily by:
lets say you want to call the first one:
f_t[0](k,m,c)
Where k,m,c are your variables
Let's clean this code up:
#instead of an index counter, python provide `enumerate`
# so that you can have the counter as part of your loop
#index = -1
f_t = []
for index, f in enumerate(inputs[:,0]):
# index = index +1
def F_analytic(k, m, c):
#these two lines are redundant. You're just returning an expression
# so this whole function can be a lambda
F_t = k*m*c*f
return F_t
#why are you creating an inner list for each index position,
# and then appending the function to it?
#even if you wanted to have the inner list, you could do it in one step
f_t.append([])
f_t[index].append(F_analytic)
If we make the specified changes, we end up with:
f_t = [(lambda k, m, c: k*m*c*f) for f in inputs[:,0]]
Of course a much simpler approach would be simply to calculate the inputs to the product, including f, and multiply directly.
Related
I know that there are way simpler ways to calculate the square of a number and store it in an array, but for the sake of another problem. I need to understand why nothing happens in this code and its structure (is the return(a) necessary ?):
s=[1,2,3,4,5]
def square(x):
return x*x
def iterate(b):
sol=[]
for b in s:
a=square(b)
return(a)
sol.append(a)
print(sol)
The goal is to store the square in sol : sol = [1,4,9,16,25]. But the code runs without printing anything. What make the following code work and not the previous one ?
s=[1,2,3,4,5]
def square(x):
return x*x
sol=[]
for b in s:
a=square(b)
sol.append(a)
print(sol)
(My problem involves curve fitting, and this structure doesnt fit my needs)
The problem is that you define iterate within square but you never call iterate. It would be better to have iterate be a separate function that calls square:
values = [1,2,3,4,5] # do not call your variable set - it is a Python keyword
def square(x):
return x*x
def iterate(values):
solution = []
for value in values:
value_squared = square(value)
solution.append(value_squared)
return solution
You could also do this without defining iterate using a list comprehension:
[square(value) for value in values]
Edit:
To answer your other questions, here is your code:
s=[1,2,3,4,5]
def square(x):
return x*x
def iterate(b):
sol=[]
for b in s:
a=square(b)
return(a)
sol.append(a)
print(sol)
In square, you never call iterate so this part of the code never runs.
If you add a call to iterate within square, you will end up in an infinite loop. This is because within iterate you call square, but you always iterate over your list s. This means that inside iterate, square(b) will always be square(1).
Within iterate you use the global variable s but it would be better to restructure your code so that you take s as input.
If you are learning about inner functions, you could define iterate and within this define square:
values = [1,2,3,4,5]
def iterate(values):
def _square(x):
return x*x
solution = []
for value in values:
value_squared = _square(value)
solution.append(value_squared)
return solution
I would like to design a function f(x) whose input could be
one object
or a list of objects
In the second case, f(x) should return a list of the corresponding results.
I am thinking of designing it as follow.
def f(x):
if isinstance(x, list):
return [f(y) for y in x]
# some calculation
# from x to result
return result
Is this a good design? What would be the canonical way to do this?
No, it's not good design.
Design the function to take only one datatype. If the caller has only one item, it's trivial for them to wrap that in a list before calling.
result = f([list x])
Or, have the function only accept a single value and the caller can easily apply that function to a list:
result = map(f, [x, y, z])
They can easily map over the function when they have a list(example):
def f(x):
return x + 1 #calcuation
lst = map(f, [1, 2, 3])
print(lst) # [2, 3, 4]
And remember: The function should do one thing and do it well :)
I'd avoid it. My biggest issue with it is that sometimes you're returning a list, and sometimes you're returning an object. I'd make it work on a list or an object, and then have the user deal with either wrapping the object, of calling the function in a list comprehension.
If you really do need to have it work on both I think you're better off using:
def func(obj):
if not isinstance(obj, list):
obj = [obj]
# continue
That way you're always returning a list.
Actually the implementation may be valid (but with room for improvement). The problem is that you're creating an ambigous and unexpected behaviour. The best way would be to have 2 different functions f(x) and f_on_list() or something like this, where the second apply the first to a list.
Is there any non-explicit for way to call a member n times upon an object?
I was thinking about some map/reduce/lambda approach, but I couldn't figure out a way to do this -- if it's possible.
Just to add context, I'm using BeautifulSoup, and I'm extracting some elements from an html table; I extract some elements, and then, the last one.
Since I have:
# First value
print value.text
# Second value
value = value.nextSibling
print value.text
# Ninth value
for i in xrange(1, 7):
value = value.nextSibling
print value.text
I was wondering if there's any lambda approach -- or something else -- that would allow me to do this:
# Ninth value
((value = value.nextSibling) for i in xrange(1, 7))
print value.text
P.S.: No, there's no problem whatsoever with the for approach, except I really enjoy one-liner solutions, and this would fit really nice in my code.
I have a strong preference for the loop, but you could use reduce:
>>> class Foo(object):
... def __init__(self):
... self.count = 0
... def callme(self):
... self.count += 1
... return self
...
>>> a = Foo()
>>> reduce(lambda x,y:x.callme(),range(7),a)
<__main__.Foo object at 0xec390>
>>> a.count
7
You want a one-liner equivalent of this:
for i in xrange(1, 7):
value = value.nextSibling
This is one line:
for i in xrange(1, 7): value = value.nextSibling
If you're looking for something more functional, what you really want is a compose function, so you can compose callme() (or attrgetter('my_prop'), or whatever) 7 times.
In case of BS you can use nextSiblingGenerator() with itertools.islice to get the nth sibling. It would also handle situations where there is no nth element.
from itertools import islice
nth = 7
next(islice(elem.nextSiblingGenerator(), nth, None), None)
Disclaimer: eval is evil.
value = eval('value' + ('.nextSibling' * 7))
Ah! But reduce is not available in Python3, at least not as a built in.
So here is my try, portable to Python2/3 and based on the OP failed attempt:
[globals().update(value=value.nextSibling) for i in range(7)]
That assumes that value is a global variable. If value happens to be a member variable, then write instead:
[self.__dict__.update(value=value.nextSibling) for i in range(7)]
You cannot use locals() because the list comprehension creates a nested local scope, so the real locals() is not directly available. However, you can capture it with a bit of work:
(lambda loc : [loc.update(x=x.nextSibling) for i in range(7)])(locals())
Or easier if you don't mind duplicating the number of lines:
loc = locals()
[loc.update(value=value.nextSibling) for i in range(7)]
Or if you really fancy one-liners:
loc = locals() ; [loc.update(value=value.nextSibling) for i in range(7)]
Yes, Python can use ; too 8-)
UPDATE:
Another fancy variation, now with map instead of the list comprehension:
list(map(lambda d : d.update(value=value.nextSibling), 7 * [locals()]))
Note the clever use of list multiplication to capture the current locals() and create the initial iterable at the same time.
The most direct way to write it would be:
value = reduce(lambda x, _: x.nextSibling, xrange(1,7), value)
I was trying to do a "strange" (but useful in my case) function that can return a dynamic list whose len depends on the amount of receiver.
For example:
f() returns a dynamic list of None, so I can do the following:
a = f() => a = None
a, b = f() => a=b= None
(a, b) = f() => a=b= None
(a, b, c, d, e, f) = f() => a=b=c=d=e=f= None
I think this might be done via generator comprehension or iterator, but I was blocked on how to get the amount of recevier. Maybe I was in the wrong direction. Would you advise me some tips?
Any helps will be appreciated.
Many Thank,
Tiezhen
This is not possible in Python. The function on the right hand site has no knowledge of the context it was called in. The right hand site is evaluated before any of the name bindings take place.
Unfortunately, Python unpacks returned tuples using the Pythonic "it's easier to ask forgiveness than permission" approach. That is, if you have a statement:
a,b,c = f()
Behind the scenes, it's doing something along the lines of:
try:
a = returned[0]
b = returned[1]
c = returned[2]
except IndexError:
raise ValueError('need more than k values to unpack')
try:
_ = returned[4]
except IndexError:
pass
else:
raise ValueError('too many values to unpack')
So it's discovering dynamically the number of values returned. Unfortunately, that precludes us from being clever and creating a new type for handling variable returns:
class VariableReturn(object):
def __getitem__(self, index):
return ...
In Python 3, you can sort of do what you're asking, but the burden is on the caller, not the function being called. The function should always return the same number of results, but we'll trap the remaining results using extended tuple unpacking, as shown in this StackOverflow question.
Using this approach, you can return as many results as you'd like, but you need to always return at least as many as you need in the maximal case. The rest get packed into a trailing tuple.
a,*others = f()
a,b,*others = f()
a,b,c,*others = f()
If you don't mind using Python 3, you can ignore what you don't need, for example:
a, b, c, d, *_ = (x for x in range(100))
Try this:
def f(n):
return (None, ) * n
For example:
a, b, c = f(3)
... That's about as far as you can get, since in Python there's no way to know how many variables are in the left-hand side of an assignment.
Can't be done.
Functions in Python return one value, only. While it may sometimes look like more, it's still just one value: a tuple. Multiple assignment is then a process of tuple unpacking.
Your question then can be restated: can we create an object that acts like a tuple of varying length, depending on how many values need to be unpacked? And that's simply not made available as an option.
Probably the closest I can think of is to use a generator and get the desired number of items with itertools.islice:
a = itertools.count()
x, y, z = itertools.islice(a, 3) # 0, 1, 2
u, v = itertools.islice(a, 2) # 3, 4
But that's pretty far from what was hoped for.
pretty not nice but perhaps this helps you:
def f(x):
for i in x:
globals()[i] = None
f(['a','b','c'])
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))