Clash Royale CLAN TAG#URR8PPP

I would expect the output to be like [[0,0,0,0,],[0,1,0,0],[0,2,0,0],[0,3,0,0]]

time_count = [[0, 0, 0, 0]] * 4
j = 0
for i in range(len(time_count)):
time_count[i][1] = j
j += 1
print(time_count)


Output:

[[0, 3, 0, 0], [0, 3, 0, 0], [0, 3, 0, 0], [0, 3, 0, 0]]


I would expect the output to be like:

[[0,0,0,0],[0,1,0,0],[0,2,0,0],[0,3,0,0]]


can someone explain why every index[1] is 3 ?

index[1]

3

*

3 Answers
3

Easy fix:

time_count = [[0, 0, 0, 0] for _ in range(4)]


As Klaus D. has alluded, using the * operator on nested lists is usually a bad idea, as it duplicates the contents by copying the references. You won't notice this when multiplying sequences of immutable types, e.g. a list of integers, but when the elements are mutable, you get duplicates.

*

The * operator just creates a new reference to the original list (here [0, 0, 0, 0]).

[0, 0, 0, 0]

If you change the original list as time_count[i][1] = j, all the references reflect the same changes.

time_count[i][1] = j

If you want to create a new list, you can use the extend method with a loop on it :

time_count =
time_count.extend([0,0,0,0] for _ in range(4))


Now, all the element list in time_count store separate memory and can have different values.

Do:

time_count = [[0, 0, 0, 0]] * 4
print([[v if x!=1 else i for x,v in enumerate(a)] for i,a in enumerate(time_count)])


Output:

[[0, 0, 0, 0], [0, 1, 0, 0], [0, 2, 0, 0], [0, 3, 0, 0]]


Explanation:

Use enumerate to iterate trough both the indexes and values of the list time_count.

enumerate

time_count

Use enumerate again to iterate trough both the indexes and values of the list a. which is the iterator for time_count.

enumerate

a

time_count

Go trough the indexes of a and say i (the index iterator for time_count) if x (the index iterator for a) is 1, otherwise say v (the iterator for a)

a

i

time_count

x

a

v

a

Note: This is all in a list comprehension

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