Python variable assignment changes value

Question

I am a Network Engineer trying to learn Python programming as a job requirement.

I wrote this code below to

# Funtion to chop the first and last item in the list

def chop(t):
    t.pop(0) and t.pop(len(t)-1)
    return t

When I run the function on the list t and assign it to a variable a. a gets the remainder of the list after the function was executed and a becomes a new list.This works perfect.

>>> t = ['a', 'b', 'c', 'd', 'e' ,'f','g','h','i','j','k','l']
>>> a=chop(t)
>>> a
['b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k']
>>>

Later when i try it works well but the value of a also changes to the output of print chop(t) whereas I did not run the variable a through the function chop(t). Can someone explain why does this happen?

>>> print chop(t)
['c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']
>>> a
['c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']

Regards Umesh

You're not copying your list, so a and t refer to the same list. If you alter one, they are both being altered. — khelwood
– khelwood, Commented Mar 28, 2017 at 10:28
You may find this article helpful: Facts and myths about Python names and values, which was written by SO veteran Ned Batchelder. Also see Other languages have "variables", Python has "names" for a brief summary with nice diagrams. — PM 2Ring
– PM 2Ring, Commented Mar 28, 2017 at 10:29
To add to what khelwood has said, the chop function mutates the list object that you pass to it, and returns that same list object. So the assignment a = chop(t) makes a another name for the same list object that is also named t, just as if you'd done a = t. — PM 2Ring
– PM 2Ring, Commented Mar 28, 2017 at 10:32
If you want to create a new list object instead of mutating the existing one, you can just do t[1:-1] — PM 2Ring
– PM 2Ring, Commented Mar 28, 2017 at 10:34

darvark · Accepted Answer · 2017-03-28 10:42:44Z

1

To save unmodified list t, create copy of it by a = t[:], and test chop on list a

>>> t = ['a', 'b', 'c', 'd', 'e' ,'f','g','h','i','j','k','l']
>>> a=t[:]
>>> a.pop(0)
'a'
>>> a
['b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l']
>>> t
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l']
>>>

answered Mar 28, 2017 at 10:42

darvark

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1 Comment

PM 2Ring Over a year ago

Rather than doing a=t[:]; a.pop(0) it's more efficient to do a=t[1:]. When you pop from the start of a list all the subsequent list items have to be moved down. That happens at C speed, so it's relatively fast, but it's better to avoid it if you can.

codemonkey · Accepted Answer · 2017-03-28 10:40:03Z

0

When you call chop(t) , t is being passed by reference. So when you do the operation "t.pop(0) and t.pop(len(t)-1)" , it is being done on the original object and not a copy.

Because of the above , the below is true as well

>>> t = ['a', 'b', 'c', 'd', 'e' ,'f','g','h','i','j','k','l']
>>> chop(t)
>>> t
['b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k']
>>>

answered Mar 28, 2017 at 10:40

codemonkey

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Comments

Abhishake Gupta · Accepted Answer · 2017-03-28 10:47:07Z

0

t = ['a', 'b', 'c', 'd', 'e' ,'f','g','h','i','j','k','l']
def chop(t):
    t.pop(0) and t.pop(len(t)-1)
    return t
a=chop(t)

In python, Assignment never copies data. By doing a=chop(t) is like Simply storing the reference to the function in a variable. Hence, every time we call a we get chop() function executed.

answered Mar 28, 2017 at 10:47

Abhishake Gupta

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