pandas 计算pandas/python中df列中非零数字的数量
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/34968223/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
counting the number of non-zero numbers in a column of a df in pandas/python
提问by user5826447
I have a df that looks something like:
我有一个 df 看起来像：
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
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. Any help would be really appreciated!
我想输出 c 列中不为零的数字数量。任何帮助将非常感激！
回答by jezrael
Use double sum:
使用双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 cor d:
如果您只需要计算列c或d：
print (df['c'] != 0).sum()
2
print (df['d'] != 0).sum()
3
EDIT: Solution with numpy.sum:
编辑：解决方案numpy.sum：
print ((df != 0).values.sum())
14
回答by Stig Johan B.
Numpy's count_nonzerofunction is efficient for this.
Numpy 的count_nonzero功能对此非常有效。
np.count_nonzero(df["c"])
np.count_nonzero(df["c"])
