問題描述

我有一個(gè)數(shù)據(jù)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

我想定義兩列 sum_0 和 sum_1 并計(jì)算每行列 (out1,out2,out3) 中 0 和 1 的數(shù)量.所以預(yù)期的結(jié)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

我有這個(gè)計(jì)算1個(gè)數(shù)的代碼，但我不知道如何計(jì)算0個(gè)數(shù).

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)

有人可以幫忙嗎?

推薦答案

你可以為每個(gè)條件調(diào)用sum，1條件很簡單，只是一個(gè)直接的axis=1 上的 sum，第二次您可以將 df 與 0 值進(jìn)行比較，然后像以前一樣調(diào)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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