Python的 - 每个日期和月份大熊猫的所有非零列计数

Question

我有这样的数据帧，并且我希望所有非零值的计数每月，日期和电子邮件互动

    DATE    LOC      EMAIL         INTERATION
    1/11    INDIA    [email protected]     0
    1/11    INDIA    [email protected]     11
    1/11    LONDON   [email protected]     2
    2/11    INDIA    [email protected]     5
    2/11    INDIA    [email protected]     5
    2/11    LONDON   [email protected]     0
    3/11    LONDON   [email protected]     1

所以我得到的数据帧应该是这样的：

    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

Answer 1

使用groupby与agg和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

另一种解决方案是创建布尔面具由不COMPRE通过ne相等，转换为整数和骨料sum：

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

如果列EMAIL需要组太：

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

Answer 2

一个不一定有效的解决方法是将转换为bool然后sum。这用事实0 / 1相当于在计算分别False / True：

res = df.groupby(['DATE', 'LOC'])['INTERATION']\
        .apply(lambda x: x.astype(bool).sum()).reset_index()

print(res)

   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