Is this a valid python behavior? I would think that the end result should be [0,0,0] and the id() function should return identical values each iteration. How to make it pythonic, and not use enumerate or range(len(bar))?
bar = [1,2,3]
print bar
for foo in bar:
print id (foo)
foo=0
print id(foo)
print bar
output:
[1, 2, 3] 5169664 5169676 5169652 5169676 5169640 5169676 [1, 2, 3]
Answers:
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Method 1
First of all, you cannot reassign a loop variable—well, you can, but that won’t change the list you are iterating over. So setting foo = 0 will not change the list, but only the local variable foo (which happens to contain the value for the iteration at the begin of each iteration).
Next thing, small numbers, like 0 and 1 are internally kept in a pool of small integer objects (This is a CPython implementation detail, doesn’t have to be the case!) That’s why the ID is the same for foo after you assign 0 to it. The id is basically the id of that integer object 0 in the pool.
If you want to change your list while iterating over it, you will unfortunately have to access the elements by index. So if you want to keep the output the same, but have [0, 0, 0] at the end, you will have to iterate over the indexes:
for i in range(len(bar)):
print id(bar[i])
bar[i] = 0
print id(bar[i])
print bar
Otherwise, it’s not really possible, because as soon as you store a list’s element in a variable, you have a separate reference to it that is unlinked to the one stored in the list. And as most of those objects are immutable and you create a new object when assigning a new value to a variable, you won’t get the list’s reference to update.
Method 2
Yes, the output you got is the ordinary Python behavior. Assigning a new value to foo will change foo’s id, and not change the values stored in bar.
If you just want a list of zeroes, you can do:
bar = [0] * len(bar)
If you want to do some more complicated logic, where the new assignment depends on the old value, you can use a list comprehension:
bar = [x * 2 for x in bar]
Or you can use map:
def double(x):
return x * 2
bar = map(double, bar)
Method 3
you actually didnt change the list at all.
the first thing for loop did was to assign bar[0] to foo(equivalent to foo = bar[0]). foo is just an reference to 1. Then you assign another onject 0 to foo. This changed the reference of foo to 0. But you didnt change bar[0]. Remember, foo as a variable, references bar[0], but assign another value/object to foo doesn’t affect bar[0] at all.
Method 4
bar = [0 for x in bar]
Long answer : foo is just a local name, rebinding does not impact the list. Python variables are really just key:value pairs, not symbolic names for memory locations.
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0