I tried:
A = [1,2,3]
B = A
B[0] = A[0]*3
should give:
B == [3,2,3]
True
A == [1,2,3]
True
but what it really does:
B == [3,2,3]
A == [3,2,3]
how do I copy A over in B without keeping A linked to B? Thanks for your help.
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hpaulj
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`B = A[:]` instead of `B = A`? This makes a shallow copy. – Willem Van Onsem Mar 24 '17 at 12:06
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you can do B = A.copy() returns a shallow copy – bvmcode Mar 24 '17 at 12:15
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you can use this too B = [x for x in A] – Ashish Mar 24 '17 at 12:16
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Why did you tag this NumPy? You only talk about lists in this question. If you *are* using NumPy, keep in mind that NumPy arrays are **not lists** and do not behave like lists. `A[:]` won't copy a NumPy array. – user2357112 Mar 24 '17 at 17:32
1 Answers
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This is because you are only pointing the reference to B, without making a copy at all. Instead do the following to actually create a copy.
A = [1,2,3]
B = A[:]
This will work if there are no referenced variables. If you don't want that behaviour, then use a deep_copy method as below
B = copy.deepcopy(A)
Then if you change A, it won't change B
A. N. Other
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No, the OP does not make a copy at all: `B` and `A` refer to the **same** list. One can solve it by making a shallow copy. – Willem Van Onsem Mar 24 '17 at 12:07
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