python - Python - 每个日期和月份 PANDAS 的所有非零列的计数

Question

我有这个数据框，我想要每个月、日期和电子邮件的交互的所有非零值的计数

    DATE    LOC      EMAIL         INTERATION
    1/11    INDIA    [e-mail]     0
    1/11    INDIA    [e-mail]     11
    1/11    LONDON   [e-mail]     2
    2/11    INDIA    [e-mail]     5
    2/11    INDIA    [e-mail]     5
    2/11    LONDON   [e-mail]     0
    3/11    LONDON   [e-mail]     1

所以我得到的数据框应该是这样的：

    DATE    LOC        INTERATION
    1/11    INDIA         1
    1/11    LONDON        1
    2/11    INDIA         2
    2/11    LONDON        0
    3/11    LONDON        1

提前致谢

标签： pythonpandasdataframe

Answer 1

groupby与agg和一起使用numpy.count_nonzero：

df1 = df.groupby(['DATE','LOC'], as_index=False)['INTERATION'].agg(np.count_nonzero)
print (df1)
   DATE     LOC  INTERATION
0  1/11   INDIA           1
1  1/11  LONDON           1
2  2/11   INDIA           2
3  2/11  LONDON           0
4  3/11  LONDON           1

另一种解决方案是通过 compre by not equal by 创建布尔掩码ne，转换为整数和聚合sum：

df1 = (df.assign(INTERATION = df['INTERATION'].ne(0).astype(int))
       .groupby(['DATE','LOC'], as_index=False)['INTERATION']
       .sum())

如果也需要按列分组EMAIL：

df2 = df.groupby(['DATE','LOC','EMAIL'], as_index=False)['INTERATION'].agg(np.count_nonzero)
print (df2)
   DATE     LOC        EMAIL  INTERATION
0  1/11   INDIA  [e-mail]           1
1  1/11   INDIA  [e-mail]           0
2  1/11  LONDON  [e-mail]           1
3  2/11   INDIA  [e-mail]           1
4  2/11   INDIA  [e-mail]           1
5  2/11  LONDON  [e-mail]           0
6  3/11  LONDON  [e-mail]           1