In the following Python code value of lis is not changed, then why it is printing two different values?
lis = [[]]
x = lis[0]
print(lis)
x.append(1)
print(lis)
output:
[[]]
[[1]]
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1"value of lis is not changed" Why do you think so? – Karl Knechtel Dec 12 '21 at 09:48
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Another good post https://stackoverflow.com/questions/24292174/are-python-lists-mutable – nikeros Dec 12 '21 at 09:57
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after `x.append` python adds `1` in the list `x`. – Faraaz Kurawle Dec 12 '21 at 11:49
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and the list `x` is a sub-list of `lis`, and `x=lis[0]`, so this will change the value of list `lis`. Hope you got it. – Faraaz Kurawle Dec 12 '21 at 11:51
2 Answers
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Because in python when you assign to var a list from another var, you won't get it's value directly. You only get a link to its address. To get a list from lis directly you need to write:
lis = [[]]
x = lis[0].copy()
This will create a new list for x and empty list would not change.
Даниял
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As lis[0] and x are referenced to the same memory location, changes in x are causing changes in lis.
Instead use:
lis = [[]]
x = lis[0].copy()
print(lis)
x.append(1)
print(x)
Kedar U Shet
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