Count number of empty array occurrences within a 2D array

Question

I'm looking to count the number of occurrences of an empty array within a 2D array in the most efficient manner. For example :

array = [[a,b],[a,b,c],[a],[],[],[],[]]

The answer I would like to get should be 4.

How do we get around doing that without using a lengthy for loop process? It shouldn't also use Numpy, just plain simple Python. I tried .count, but I doesn't quite work with empty arrays.

check https://stackoverflow.com/questions/53513/how-do-i-check-if-a-list-is-empty — deadvoid, Oct 05 '18 at 15:18
@cryptonome Doesn't exactly quite address my problem being it a 2D array and counting each empty array in a easy manner. — Lotus, Oct 05 '18 at 15:20

score 2 · Accepted Answer · answered Oct 05 '18 at 15:27

2

Here is my short solution using the list.count() method:

array = [[a,b],[a,b,c],[a],[],[],[],[]]

print(array.count([])) # 4

answered Oct 05 '18 at 15:27

Laurent H.

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damn, this is better – deadvoid Oct 05 '18 at 15:30

score 1 · Answer 2 · answered Oct 05 '18 at 15:29

1

array = [[a,b],[a,b,c],[a],[],[],[],[]]
answer = len([a for a in array if not a])

>>> answer
4

answered Oct 05 '18 at 15:29

deadvoid

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score 0 · Answer 3 · answered Oct 05 '18 at 15:31

0

array = [[1,2],[1,2,3],[1],[],[],[],[]]

print(sum(1 for arr in array if not arr)) # 4

answered Oct 05 '18 at 15:31

Tanmay jain

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score 0 · Answer 4 · answered Oct 05 '18 at 15:43

0

I would use the array.count([]) method but just introducing an option that has not been displayed

print(len(list(filter(lambda x: x == [], array))))
# 4

answered Oct 05 '18 at 15:43

vash_the_stampede

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Count number of empty array occurrences within a 2D array

4 Answers4