Python Pandas：计算每行数据帧中特定值的频率？
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Python Pandas: Counting the frequency of a specific value in each row of dataframe?
提问by UserYmY
I have a dataframe df:
我有一个数据框 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
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:
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：
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
I have this code for counting the number of 1s, but I do not know how to count the number of 0s.
我有这个用于计算 1 数量的代码，但我不知道如何计算 0 的数量。
df['sum_1'] = df[['out_1','out_2','out_3']].sum(axis=1)
Can anybody help?
有人可以帮忙吗？
回答by EdChum
You can call sumfor each condition, the 1condition is simple just a straight sumon axis=1, for the second you can compare the df against 0value and then call sumas before:
您可以调用sum每个条件，1条件很简单，只是直接sum在axis=1，第二个您可以将 df 与0值进行比较，然后sum像以前一样调用：
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
回答by AntonyBrd
I would do :
我会做 ：
df["sum_0"] = df.apply(lambda row: sum(row[0:3]==0) ,axis=1)
