CODEWITHSUNDEEP
pythonpython-3.xlist
ddm_ingram
ddm_ingram

Reputation: 111

New list created by slicing old list; modifying 'new' changes 'old' as well?

Summary: I created a new list variable by taking a slice of an original one. When I tweak the new list, the previous one also changes, but I'm fairly certain they are two distinct lists! Here's my general method:

1.Create initial list all_info by calling a function. Each list item is itself a list with 5 elements.

all_info = CrossReference(start_info, RBS_info, promoter_info)

2.Create a new list all_info_mod by slicing the initial list to remove the first item (I need to do this for later code). Then confirm thet are distinct.

all_info_mod = all_info[1:]

all_info #Shows original list
all_info_mod #Shows sliced list

3.Increment the first 3 elements of each outer list by 1. Then check.

for i in all_info_mod:
    for j in range(0, 3):
        i[j] += 1

all_info_mod #Successfully altered
all_info #Also altered?!

The initial list is also altered when it shouldn't be. I have tested with simpler examples and get the same result. Perhaps it's a result of my previous code that comes before this? (Can provide more details if needed).

Thanks in advance!

Upvotes: 0

Views: 135

Answers (3)

Laurent H.
Laurent H.

Reputation: 6526

When you do this: 

all_info = [[1,2,3], [2,3,4], [5,6,7]]
all_info_mod = all_info[1:]

You perform slicing on all_info to create a new list all_info_mod. The problem is that the elements of all_info are (references to) lists, so all_info_mod will contain the copy of the references to lists. The references are copied, so at the end both lists contain references to the same lists.
This explains why when you modifies the lists referenced by the elements of all_info_mod, you also modify the lists referenced by the elements of all_info.

Note that if you modify directly an element of all_info_mod by assignment, i.e. if you replace the reference to a list with another one, then all_info is not changed: 

all_info_mod[1] = [9,9,9]
print(all_info)      # [[1, 2, 3], [2, 3, 4], [5, 6, 7]]
print(all_info_mod)  # [[2, 3, 4], [9, 9, 9]]

If you really want to duplicate the data in another list, you have to use the method copy.deepcopy(), as follows: 

import copy
all_info = ...
all_info_mod = copy.deepcopy(all_info[1:])

Upvotes: 1

Patrick Haugh
Patrick Haugh

Reputation: 60944

This is happening because you're modifying mutable objects that both lists contain references to. Modifying the list you sliced won't change the slice, but modifying a mutable object in that list will also be visible from the other list, because both lists contain the same object.

list_ = [[1, 2, 3], [4, 5, 6]]
slice_ = list_[:]
list_[0] = [7, 8, 9]
print(slice_)
# [[1, 2, 3], [4, 5, 6]]

list_[1].append(0)
print(slice_[1])
# [4, 5, 6, 0]

print(list_[1] is slice_[1])  # The two lists both contain the same list object
# True

Upvotes: 1

Oli
Oli

Reputation: 1393

You should import copy module and make a deepcopy if you don't want changes in original list.

import copy

all_info_mod = copy.deepcopy(all_info[1:])

Upvotes: 3

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