Why lambda in loop requires default?
Antoon Pardon
antoon.pardon at rece.vub.ac.be
Mon Mar 28 16:26:48 EDT 2016
Op 27-03-16 om 03:46 schreef gvim:
> Given that Python, like Ruby, is an object-oriented language why doesn't this:
It has nothing to do with being object-oriented but by how scopes are used
> def m():
> a = []
> for i in range(3): a.append(lambda: i)
> return a
Python doesn't create a new scope for the suite of the for loop. If you want an
intermediate scope, you have to provide it your self. Like the following.
def m():
a = []
for i in range(3):
a.append((lambda i: (lambda : i))(i))
return a
> b = m()
> for n in range(3): print(b[n]()) # => 2 2 2
>
> ... work the same as this in Ruby:
>
> def m
> a = []
> (0..2).each {|i| a << ->(){i}}
> a
> end
I don't know ruby but I guess the block creates a new scope and
thus running the block is like calling an anonymous function.
So the i in each run of the block is a new instantiation of the
variable instead of being the same variable.
--
Antoon Pardon
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