I tried the following code in Python 3.5.1:
>>> f = {x: (lambda y: x) for x in range(10)}
>>> f[5](3)
9
It's obvious that this should return 5. I don't understand where the other value comes from, and I wasn't able to find anything.
It seems like it's something related to reference - it always returns the answer of f[9], which is the last function assigned.
What's the error here, and how should this be done so that it works properly?
Python scoping is lexical. A closure will refer to the name and scope of the variable, not the actual object/value of the variable.
What happens is that each lambda is capturing the variable x not the value of x.
At the end of the loop the variable x is bound to 9, therefore every lambda will refer to this x whose value is 9.
Why @ChrisP's answer works:
make_funcforces the value ofxto be evaluated (as it is passed into a function). Thus, the lambda is made with value ofxcurrently and we avoid the above scoping issue.
def make_func(x):
return lambda y: x
f = {x: make_func(x) for x in range(10)}
f = {x: (lambda y, x=x: x) for x in range(10)}, because function default arguments are bound at definition time, so they'd bind the value of x, not the name x.
Lazy binding paragraph in Common Gotchas In Python.The following should work:
def make_func(x):
return lambda y: x
f = {x: make_func(x) for x in range(10)}
The x in your code ends up referring to the last x value, which is 9, but in mine it refers to the x in the function scope.
