I initiate a list as below -
arr = [1,2,3,4,5]
and assign two lists to it as below -
arrA = arr
arrB = arr
When I do arrA.pop(0), why does it also pops the 0th element from arrB and arr? What is the logic behind this lists behavior in Python ?
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``print(id(arrA))`` & ``print(id(arrB))`` you will see both objects pointed to same ref. – sushanth Jun 28 '20 at 17:32
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Thanks, this clears the logic. any way to avoid it ? – Apoorv Mehta Jun 28 '20 at 17:34
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Yes.. see the below answer, using deepcopy. – sushanth Jun 28 '20 at 17:35
1 Answers
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because you just copy the address of the arr value (like pointers in C).
to independently copy the value to another list, you can use
arrA = arr.copy()
or
arrA = arr[:]
when you pop the arr, the value in arrA doesn't change.
Vincent Andreas
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