I have a 2D array in python, like the following:
2d_array_1 =[[0,1],[1,0]]
And want to create a second array of the same size, with all values initialized to zero.
How can I do that in python, if the size of the initial array can vary?
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Read stackoverflow.com/questions/2582138/… for a 2d list you can use a nested list comprehension.Kasravnd– Kasravnd2017-07-15 09:24:54 +00:00Commented Jul 15, 2017 at 9:24
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Possible duplicate of finding and replacing elements in a list (python)Oblomov– Oblomov2017-07-17 09:40:02 +00:00Commented Jul 17, 2017 at 9:40
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2 Answers
I think above answer should be fixed.
first_array = [[0,1],[1,0]]
second_array = [ [0] * len(first_array[0]) ] * len( first_array )
second_array[0][0] = 1
print(second_array) # [[1, 0], [1, 0]]
To prevent this, you can use below code
second_array = [ [0 for _ in range(len(first_array[0]))] for _ in range(len( first_array )) ]
second_array[0][0] = 1
print(second_array) # [[1, 0], [0, 0]]
Comments
You can get the size of a list by len(list) and you can initialise a list of length n with a given value val like this: list = [val] * n. Putting these two together you get:
first_array = [[0,1],[1,0]]
second_array = [ [0] * len(first_array[0]) ] * len( first_array )
assuming of course that first_array is of the form you're expecting.
1 Comment
CoffeeSyntax
you are simply duplicating the array len(first_array) times. This would lead to side effects. Sangwon has a better answer below that prevents that.