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python - Pandas:根据特定列值重置增量

转载 作者:行者123 更新时间:2023-12-01 01:13:37 26 4
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我想创建一个列,对于差异中不是 NaT 的每一行,该列递增 1。如果值为 NaT,我希望重置增量

下面是一个示例数据框:

              x        y      min      z        o     diffs
0 0 0 0 1 1 NaT
1 0 0 0 2 1 00:00:01
2 0 0 0 6 1 00:00:04
3 0 0 0 11 1 00:00:05
4 0 0 0 14 0 NaT
5 0 0 2 18 0 NaT
6 0 0 2 41 1 NaT
7 0 0 2 42 0 NaT
8 0 0 8 13 1 00:00:54
9 0 0 8 16 1 00:00:03
10 0 0 8 17 1 00:00:01
11 0 0 8 20 0 NaT
12 0 0 8 32 1 NaT

这是我的预期输出:

              x        y      min      z        o     diffs   increment
0 0 0 0 1 1 NaT 0
1 0 0 0 2 1 00:00:01 1
2 0 0 0 6 1 00:00:04 2
3 0 0 0 11 1 00:00:05 3
4 0 0 0 14 0 NaT 0
5 0 0 2 18 0 NaT 0
6 0 0 2 41 1 NaT 0
7 0 0 2 42 0 NaT 0
8 0 0 8 13 1 00:00:54 1
9 0 0 8 16 1 00:00:03 2
10 0 0 8 17 1 00:00:01 3
11 0 0 8 20 0 NaT 0
12 0 0 8 32 1 NaT 0

最佳答案

使用numpy.where设置不缺失值来计数器 cumcount具有连续的非缺失组:

m = df['diffs'].notnull()
df['increment'] = np.where(m, df.groupby(m.ne(m.shift()).cumsum()).cumcount()+1, 0)
print (df)
x y min z o diffs increment
0 0 0 0 1 1 NaT 0
1 0 0 0 2 1 00:00:01 1
2 0 0 0 6 1 00:00:04 2
3 0 0 0 11 1 00:00:05 3
4 0 0 0 14 0 NaT 0
5 0 0 2 18 0 NaT 0
6 0 0 2 41 1 NaT 0
7 0 0 2 42 0 NaT 0
8 0 0 8 13 1 00:00:54 1
9 0 0 8 16 1 00:00:03 2
10 0 0 8 17 1 00:00:01 3
11 0 0 8 20 0 NaT 0
12 0 0 8 32 1 NaT 0

如果性能很重要,替代解决方案:

b = m.cumsum()
df['increment'] = b-b.mask(m).ffill().fillna(0).astype(int)

关于python - Pandas:根据特定列值重置增量,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54595479/

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