保持到每组值最后一次出现
这是我的熊猫数据框的简化示例：
User Binary
0 UserA 0
1 UserA 0
2 UserA 0
3 UserA 1
4 UserA 0
5 UserA 1
6 UserA 0
7 UserA 0
8 UserB 0
9 UserB 0
10 UserB 0
11 UserB 0
12 UserB 0
13 UserB 1
14 UserB 1
15 UserB 0
16 UserC 0
17 UserC 0
对于每个用户，我想在第一次出现 Binary=1 之后删除所有行。注意，会有一些用户没有 Binary=1 的情况，例如本例中的 UserC。
输出如下所示：
User Binary
0 UserA 0
1 UserA 0
2 UserA 0
3 UserA 1
8 UserB 0
9 UserB 0
10 UserB 0
11 UserB 0
12 UserB 0
13 UserB 1
16 UserC 0
17 UserC 0
青春有我浏览 76回答 2
2回答
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胡说叔叔
这是使用groupby自定义函数和进行转换的一种方法：# check which Binary values are 1 and group the series by Userg = df.Binary.eq(1).groupby(df.User)# transform to either idxmax or the last index depending# on whether there are any Trues or notm = g.transform(lambda x: x.idxmax() if x.any() else x.index[-1])# index the dataframe where the index is smaler or eq mout = df[df.index <= m]print(out) User Binary0 UserA 01 UserA 02 UserA 03 UserA 18 UserB 09 UserB 010 UserB 011 UserB 012 UserB 013 UserB 116 UserC 017 UserC 0 -
SMILET
想法是按交换顺序测试连续值的最大值DataFrame.iloc，如果仅0或仅1正确分组值，什么也有效：def f(x): s = x.cumsum() return s.eq(s.max())df = df[df.iloc[::-1].groupby('User')['Binary'].transform(f).sort_index()]print (df) User Binary0 UserA 01 UserA 02 UserA 03 UserA 18 UserB 09 UserB 010 UserB 011 UserB 012 UserB 013 UserB 116 UserC 017 UserC 0
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Python