I\'m trying to create functions inside of a loop:
functions = []
for i in range(3):
def f():
return i
# alternatively: f = lambda: i
f
The issue here is that the value of i
is not saved when the function f
is created. Rather, f
looks up the value of i
when it is called.
If you think about it, this behavior makes perfect sense. In fact, it's the only reasonable way functions can work. Imagine you have a function that accesses a global variable, like this:
global_var = 'foo'
def my_function():
print(global_var)
global_var = 'bar'
my_function()
When you read this code, you would - of course - expect it to print "bar", not "foo", because the value of global_var
has changed after the function was declared. The same thing is happening in your own code: By the time you call f
, the value of i
has changed and been set to 2
.
There are actually many ways to solve this problem. Here are a few options:
i
by using it as a default argumentUnlike closure variables (like i
), default arguments are evaluated immediately when the function is defined:
for i in range(3):
def f(i=i): # <- right here is the important bit
return i
functions.append(f)
To give a little bit of insight into how/why this works: A function's default arguments are stored as an attribute of the function; thus the current value of i
is snapshotted and saved.
>>> i = 0
>>> def f(i=i):
... pass
>>> f.__defaults__ # this is where the current value of i is stored
(0,)
>>> # assigning a new value to i has no effect on the function's default arguments
>>> i = 5
>>> f.__defaults__
(0,)
i
in a closureThe root of your problem is that i
is a variable that can change. We can work around this problem by creating another variable that is guaranteed to never change - and the easiest way to do this is a closure:
def f_factory(i):
def f():
return i # i is now a *local* variable of f_factory and can't ever change
return f
for i in range(3):
f = f_factory(i)
functions.append(f)
functools.partial
to bind the current value of i
to f
functools.partial lets you attach arguments to an existing function. In a way, it too is a kind of function factory.
import functools
def f(i):
return i
for i in range(3):
f_with_i = functools.partial(f, i) # important: use a different variable than "f"
functions.append(f_with_i)
Caveat: These solutions only work if you assign a new value to the variable. If you modify the object stored in the variable, you'll experience the same problem again:
>>> i = [] # instead of an int, i is now a *mutable* object
>>> def f(i=i):
... print('i =', i)
...
>>> i.append(5) # instead of *assigning* a new value to i, we're *mutating* it
>>> f()
i = [5]
Notice how i
still changed even though we turned it into a default argument! If your code mutates i
, then you must bind a copy of i
to your function, like so:
def f(i=i.copy()):
f = f_factory(i.copy())
f_with_i = functools.partial(f, i.copy())
You're running into a problem with late binding -- each function looks up i
as late as possible (thus, when called after the end of the loop, i
will be set to 2
).
Easily fixed by forcing early binding: change def f():
to def f(i=i):
like this:
def f(i=i):
return i
Default values (the right-hand i
in i=i
is a default value for argument name i
, which is the left-hand i
in i=i
) are looked up at def
time, not at call
time, so essentially they're a way to specifically looking for early binding.
If you're worried about f
getting an extra argument (and thus potentially being called erroneously), there's a more sophisticated way which involved using a closure as a "function factory":
def make_f(i):
def f():
return i
return f
and in your loop use f = make_f(i)
instead of the def
statement.