按熊猫中唯一的一对列计算

Question

我试图找出如何根据每对唯一列(ip,useragent)的行数来计算,例如

d = pd.DataFrame({'ip': ['192.168.0.1', '192.168.0.1', '192.168.0.1', '192.168.0.2'], 'useragent': ['a', 'a', 'b', 'b']})

     ip              useragent
0    192.168.0.1     a
1    192.168.0.1     a
2    192.168.0.1     b
3    192.168.0.2     b

生产:

ip           useragent  
192.168.0.1  a           2
192.168.0.1  b           1
192.168.0.2  b           1

想法？

Answer 1

如果你使用groupby,你会得到你想要的.

d.groupby(['ip', 'useragent']).size()

生产:

ip          useragent               
192.168.0.1 a           2
            b           1
192.168.0.2 b           1

Answer 2

print(d.groupby(['ip', 'useragent']).size().reset_index().rename(columns={0:''}))

给出：

            ip useragent   
0  192.168.0.1         a  2
1  192.168.0.1         b  1
2  192.168.0.2         b  1

另一个不错的选择可能是pandas.crosstab：

print(pd.crosstab(d.ip, d.useragent) )
print('\nsome cosmetics:')
print(pd.crosstab(d.ip, d.useragent).reset_index().rename_axis('',axis='columns') )

给出：

useragent    a  b
ip               
192.168.0.1  2  1
192.168.0.2  0  1

some cosmetics:
            ip  a  b
0  192.168.0.1  2  1
1  192.168.0.2  0  1