I have a df that looks something like:
a b c d e 0 1 2 3 5 1 4 0 5 2 5 8 9 6 0 4 5 0 0 0
I would like to output the number of numbers in column c that are not zero.
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Sunderam Dubey
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user5826447
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I saw that, thanks. Unfortunately their question is different because they want it for each row, and to be divided by the sum, so none of the code that was used there is applicable to my question. – user5826447 Jan 23 '16 at 20:17
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1The first line of the top answer there reads "To count nonzero values, just do `(column!=0).sum()`, where `column` is the data you want to do it for." That seems to be exactly what you're asking ;-) – Alex Riley Jan 23 '16 at 20:22
2 Answers
13
Use double sum:
print df
a b c d e
0 0 1 2 3 5
1 1 4 0 5 2
2 5 8 9 6 0
3 4 5 0 0 0
print (df != 0).sum(1)
0 4
1 4
2 4
3 2
dtype: int64
print (df != 0).sum(1).sum()
14
If you need count only column c or d:
print (df['c'] != 0).sum()
2
print (df['d'] != 0).sum()
3
EDIT: Solution with numpy.sum:
print ((df != 0).values.sum())
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jezrael
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That makes sense, but how do I do it just for column c? I don't want the total number. – user5826447 Jan 23 '16 at 20:13
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Numpy's count_nonzero function is efficient for this.
np.count_nonzero(df["c"])
Stig Johan B.
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