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?
解决方案
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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