Python Pandas:计算每行数据帧中特定值的频率?

问题描述

我有一个数据框 df:

I have a dataframe df:

domain               country     out1 out2 out3
oranjeslag.nl           NL          1    0   NaN    
pietervaartjes.nl       NL          1    1    0
andreaputting.com.au    AU          NaN  1    0 
michaelcardillo.com     US          0    0    NaN

我想定义两列 sum_0 和 sum_1 并计算每行列 (out1,out2,out3) 中 0 和 1 的数量.所以预期的结果是:

I would like to define two columns sum_0 and sum_1 and count the number of 0s and 1s in columns (out1,out2,out3),per row. So expected results would be:

domain               country     out1 out2 out3   sum_0  sum_1
oranjeslag.nl           NL          1    0   NaN    1      1
pietervaartjes.nl       NL          1    1    0     1      2
andreaputting.com.au    AU          NaN  1    0     1      1
michaelcardillo.com     US          0    0    NaN   2      0

我有这个计算1个数的代码，但我不知道如何计算0个数.

I have this code for counting the number of 1s, but I do not know how to count the number of 0s.

df['sum_1'] = df[['out_1','out_2','out_3']].sum(axis=1)

有人可以帮忙吗?

推荐答案

你可以为每个条件调用sum，1条件很简单，只是一个直接的axis=1 上的 sum，第二次您可以将 df 与 0 值进行比较，然后像以前一样调用 sum:

You can call sum for each condition, the 1 condition is simple just a straight sum on axis=1, for the second you can compare the df against 0 value and then call sum as before:

In [102]:
df['sum_1'] = df[['out1','out2','out3']].sum(axis=1)
df['sum_0'] = (df[['out1','out2','out3']] == 0).sum(axis=1)
df

Out[102]:
                 domain country  out1  out2  out3  sum_0  sum_1
0         oranjeslag.nl      NL     1     0   NaN      1      1
1     pietervaartjes.nl      NL     1     1     0      1      2
2  andreaputting.com.au      AU   NaN     1     0      1      1
3   michaelcardillo.com      US     0     0   NaN      2      0

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