Changing the value of a copy also changes the value of the original

Question

Changing the value of a variable b, which is a copy of a, also changes the value of a.

a = [[0]]
b = a.copy()

print("a before", a)

b[0][0] = 1

print("a after ", a)

prints:

a before [[0]]
a after  [[1]]

Although this works:

a = [0]
b = a.copy()

print("a before", a)

b[0] = 1

print("a after ", a)

prints:

a before [[0]]
a after  [[0]]

a copy is only a shallow copy, the "outermost" container was created from scratch, but the items inside are still references. use `from copy import deepcopy` — Paritosh Singh, Dec 04 '19 at 15:55
Does this answer your question? [Copying nested lists in Python](https://stackoverflow.com/questions/2541865/copying-nested-lists-in-python) — abhilb, Dec 04 '19 at 15:55

score 1 · Answer 1 · answered Dec 04 '19 at 16:05

1

Turns out copy.deepcopy works.

import copy

a = [[0]]
b = copy.deepcopy(a)

print("a before", a)

b[0][0] = 1

print("a after ", a)

prints:

a before [[0]]
a after  [[0]]

answered Dec 04 '19 at 16:05

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