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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.

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  • check stackoverflow.com/questions/53513/… Commented Oct 5, 2018 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. Commented Oct 5, 2018 at 15:20
  • This is a list not an array Commented Oct 5, 2018 at 15:25

4 Answers 4

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

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

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

damn, this is better
1 
array = [[a,b],[a,b,c],[a],[],[],[],[]]
answer = len([a for a in array if not a])

>>> answer
4
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0 
array = [[1,2],[1,2,3],[1],[],[],[],[]]

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