这是我的熊猫数据框的简化示例： User Binary0 UserA 01 UserA 02 UserA 03 UserA 14 UserA 05 UserA 16 UserA 07 UserA 08 UserB 09 UserB 010 UserB 011 UserB 012 UserB 013 UserB 114 UserB 115 UserB 016 UserC 017 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 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
2 回答
胡说叔叔
TA贡献1804条经验 获得超8个赞
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 User
g = df.Binary.eq(1).groupby(df.User)
# transform to either idxmax or the last index depending
# on whether there are any Trues or not
m = g.transform(lambda x: x.idxmax() if x.any() else x.index[-1])
# index the dataframe where the index is smaler or eq m
out = df[df.index <= m]
print(out)
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
SMILET
TA贡献1796条经验 获得超4个赞
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 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
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