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Jul 8th, 2009

Using List in Python

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What is the difference between these two methods;

python Syntax (Toggle Plain Text)
# Solution 1
list1 = ['a', 'b', 'c', 'd']
temp_list1 = list1
for i in range(len(list1)):
    temp_list1[i] = "1"
print list1;              # ['1', '1', '1', '1']
print temp_list1;         # ['1', '1', '1', '1']
 
# Solution 2
list2 = ['a', 'b', 'c', 'd']
temp_list2 = list2
temp_list2 = ['1', '1', '1', '1']
print list2;              # ['a', 'b', 'c', 'd']
print temp_list2;         # ['1', '1', '1', '1']
# Solution 1
list1 = ['a', 'b', 'c', 'd']
temp_list1 = list1
for i in range(len(list1)):
    temp_list1[i] = "1"
print list1;              # ['1', '1', '1', '1']
print temp_list1;         # ['1', '1', '1', '1']

# Solution 2
list2 = ['a', 'b', 'c', 'd']
temp_list2 = list2
temp_list2 = ['1', '1', '1', '1']
print list2;              # ['a', 'b', 'c', 'd']
print temp_list2;         # ['1', '1', '1', '1']

By using solution1, want to update the temp_list1 with some other value at the same time want to retain the old value in list1. But this updated the both lists.

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kes_ee

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Jul 8th, 2009

Re: Using List in Python

Because your temporary list is the same thing as your other list. You said temp_list1 = list1 . This means that the value for temp_list1 will be the same address in memory as the value for list1, i.e. they reference the same thing. Modifying one modifies the other because they are just 2 different names for the same data.

shadwickman

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Jul 8th, 2009

Re: Using List in Python

I forgot to mention that in solution #2, you assign temp_list2 to the same address of memory as list2, but then that's undone as you create a new list (and new address in memory for it) consisting of four ones. Therefore, temp_list2 no longer refers to the same data as list2.

If you want to copy the list over without making the new name reference the same data, how about

python Syntax (Toggle Plain Text)
list1 = ['a', 'b', 'c', 'd']
list2 = [x for x in list1]
list1 = ['a', 'b', 'c', 'd']
list2 = [x for x in list1]

That'll copy all the indices of list1 into a new address for list2, and thus, they are independent of each other.

Last edited by shadwickman; Jul 8th, 2009 at 6:17 am.

shadwickman

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Jul 8th, 2009

Re: Using List in Python

Simply use list():

python Syntax (Toggle Plain Text)
list1 = ['a', 'b', 'c', 'd']
# this will create a new list with a different mem ref
temp_list1 = list(list1)
for i in range(len(list1)):
    temp_list1[i] = "1"
print list1;              # ['a', 'b', 'c', 'd']
print temp_list1;         # ['1', '1', '1', '1']
list1 = ['a', 'b', 'c', 'd']
# this will create a new list with a different mem ref
temp_list1 = list(list1)
for i in range(len(list1)):
    temp_list1[i] = "1"
print list1;              # ['a', 'b', 'c', 'd']
print temp_list1;         # ['1', '1', '1', '1']

For nested lists you need to use deepcopy() from module copy.

Last edited by Ene Uran; Jul 8th, 2009 at 11:07 am.

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Jul 8th, 2009

Re: Using List in Python

Here is an example for the use of deepcopy with a nested list. First without ...

python Syntax (Toggle Plain Text)
# a nested list
list1 = ['a', 'b', 'c', 'd', [1, 2, 3]]
list2 = [x for x in list1]
 
# now replace 2 with 77 in list1
list1[4][1] = 77
 
print(list1)  # ['a', 'b', 'c', 'd', [1, 77, 3]]
print(list2)  # ['a', 'b', 'c', 'd', [1, 77, 3]] oops!!!
# a nested list
list1 = ['a', 'b', 'c', 'd', [1, 2, 3]]
list2 = [x for x in list1]

# now replace 2 with 77 in list1
list1[4][1] = 77

print(list1)  # ['a', 'b', 'c', 'd', [1, 77, 3]]
print(list2)  # ['a', 'b', 'c', 'd', [1, 77, 3]] oops!!!

now with ...

python Syntax (Toggle Plain Text)
import copy
 
# a nested list
list1 = ['a', 'b', 'c', 'd', [1, 2, 3]]
list2 = copy.deepcopy(list1)
 
# now replace 2 with 77 in list1
list1[4][1] = 77
 
print(list1)  # ['a', 'b', 'c', 'd', [1, 77, 3]]
print(list2)  # ['a', 'b', 'c', 'd', [1, 2, 3]]   okay!!!
import copy

# a nested list
list1 = ['a', 'b', 'c', 'd', [1, 2, 3]]
list2 = copy.deepcopy(list1)

# now replace 2 with 77 in list1
list1[4][1] = 77

print(list1)  # ['a', 'b', 'c', 'd', [1, 77, 3]]
print(list2)  # ['a', 'b', 'c', 'd', [1, 2, 3]]   okay!!!

In the first example the list comprehension simply copied the inner list as an alias list which shares the memory reference.

Last edited by vegaseat; Jul 8th, 2009 at 7:35 pm.

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