根据条件获取数据帧的行数 [英] get dataframe row count based on conditions

问题描述

我想基于条件选择获取数据帧行数.我尝试了以下代码.

I want to get the count of dataframe rows based on conditional selection. I tried the following code.

print df[(df.IP == head.idxmax()) & (df.Method == 'HEAD') & (df.Referrer == '"-"')].count()

输出:

IP          57
Time        57
Method      57
Resource    57
Status      57
Bytes       57
Referrer    57
Agent       57
dtype: int64

输出显示数据帧中每一列的计数.相反，我需要获得满足以上所有条件的单个计数?这该怎么做?如果您需要有关我的数据框的更多说明，请告诉我.

The output shows the count for each an every column in the dataframe. Instead I need to get a single count where all of the above conditions satisfied? How to do this? If you need more explanation about my dataframe please let me know.

推荐答案

您要查询所有条件都为真的条件， 因此，除非我误解了您的要求，否则答案就是答案

You are asking for the condition where all the conditions are true, so len of the frame is the answer, unless I misunderstand what you are asking

In [17]: df = DataFrame(randn(20,4),columns=list('ABCD'))

In [18]: df[(df['A']>0) & (df['B']>0) & (df['C']>0)]
Out[18]: 
           A         B         C         D
12  0.491683  0.137766  0.859753 -1.041487
13  0.376200  0.575667  1.534179  1.247358
14  0.428739  1.539973  1.057848 -1.254489

In [19]: df[(df['A']>0) & (df['B']>0) & (df['C']>0)].count()
Out[19]: 
A    3
B    3
C    3
D    3
dtype: int64

In [20]: len(df[(df['A']>0) & (df['B']>0) & (df['C']>0)])
Out[20]: 3

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