如果多列中的值固定在一个值上，请使用Python使用NaN进行更改

我应该在发帖时问以下问题。如下所示，my

df

包含一些相同的连续值，即1、2和3

Date                London  Paris   Dubai   Tokyo
18/07/2017 22:35    1       2406    4348    70715
18/07/2017 22:40    1       4756    3744    3
18/07/2017 22:45    1       3988    2915    3
18/07/2017 22:50    2280    3058    2120    3
18/07/2017 22:55    2       1       1939    3
18/07/2017 23:00    2       1       2256    3
18/07/2017 23:05    2121    1       2640    2025
18/07/2017 23:10    3367    2       2202    1916
18/07/2017 23:15    3247    3       1       2
18/07/2017 23:20    2491    3       1       2
18/07/2017 23:25    2010    3       1       1560
18/07/2017 23:30    1899    3       1366    1355
18/07/2017 23:35    1992    2265    1236    1
18/07/2017 23:40    2196    4407    2       1
18/07/2017 23:45    1961    3848    2       1
18/07/2017 23:50    3       2880    2809    4565
18/07/2017 23:55    3       2143    2397    3725
19/07/2017 00:00    3       1981    3       2921
19/07/2017 00:05    3       2227    3       2131
19/07/2017 00:10    1366    2526    3       1990

我想在这些“死值”至少停留在3行时检测它们，然后用NaN将它们全部更改，因为我想稍后消除它们。有了来自的代码，我可以为伦敦专栏做这件事

g = df.London.diff().fillna(0).ne(0).cumsum()
m = df.groupby(g).London.transform('size').ge(3)
df.loc[m,'London'] = np.nan
df.assign(grouper=g, mask=m, result=df.London)

但是，我想对其他人（大约250列）也这样做

以下是预期的输出，其中所有1和3都被转换为NaN，因为它们的值至少被固定在3个连续行中

Date                London  Paris   Dubai   Tokyo
18/07/2017 22:35    NaN     2406    4348    70715
18/07/2017 22:40    NaN     4756    3744    NaN
18/07/2017 22:45    NaN     3988    2915    NaN
18/07/2017 22:50    2280    3058    2120    NaN
18/07/2017 22:55    2       NaN     1939    NaN
18/07/2017 23:00    2       NaN     2256    NaN
18/07/2017 23:05    2121    NaN     2640    2025
18/07/2017 23:10    3367    2       2202    1916
18/07/2017 23:15    3247    NaN     NaN     2
18/07/2017 23:20    2491    NaN     NaN     2
18/07/2017 23:25    2010    NaN     NaN     1560
18/07/2017 23:30    1899    NaN     1366    1355
18/07/2017 23:35    1992    2265    1236    NaN
18/07/2017 23:40    2196    4407    2       NaN
18/07/2017 23:45    1961    3848    2       NaN
18/07/2017 23:50    NaN     2880    2809    4565
18/07/2017 23:55    NaN     2143    2397    3725
19/07/2017 00:00    NaN     1981    NaN     2921
19/07/2017 00:05    NaN     2227    NaN     2131
19/07/2017 00:10    1366    2526    NaN     1990

---

如果您的代码运行良好且快速-只需跨列迭代即可：

for col in df.columns:
    g = df.[col].diff().fillna(0).ne(0).cumsum()
    # and so on...

---

shift

使用

np.logical\u和.reduce

和

np.logical\u或.reduce

创建掩码（或双精度）

输出：

df

---

1号和3号仍然有效there@k.koen把它分配回来<例如，code>df=df.where（df.where（m.to_numpy（））.bfill（limit=Nmin-1）.isnull（））Ah通过将其分配回df，它现在可以工作了。谢谢

import numpy as np
import pandas as pd

Nmin = 3  # At least 2

m = pd.DataFrame(np.logical_and.reduce([(df == df.shift(i)).to_numpy() for i in range(1, Nmin)]))

df = df.where(df.where(m.to_numpy()).bfill(limit=Nmin-1).isnull())
#df = df.mask(np.logical_or.reduce([m.shift(-i).fillna(False).to_numpy() for i in range(Nmin)]))

                Date  London   Paris   Dubai    Tokyo
0   18/07/2017-22:35     NaN  2406.0  4348.0  70715.0
1   18/07/2017-22:40     NaN  4756.0  3744.0      NaN
2   18/07/2017-22:45     NaN  3988.0  2915.0      NaN
3   18/07/2017-22:50  2280.0  3058.0  2120.0      NaN
4   18/07/2017-22:55     2.0     NaN  1939.0      NaN
5   18/07/2017-23:00     2.0     NaN  2256.0      NaN
6   18/07/2017-23:05  2121.0     NaN  2640.0   2025.0
7   18/07/2017-23:10  3367.0     2.0  2202.0   1916.0
8   18/07/2017-23:15  3247.0     NaN     NaN      2.0
9   18/07/2017-23:20  2491.0     NaN     NaN      2.0
10  18/07/2017-23:25  2010.0     NaN     NaN   1560.0
11  18/07/2017-23:30  1899.0     NaN  1366.0   1355.0
12  18/07/2017-23:35  1992.0  2265.0  1236.0      NaN
13  18/07/2017-23:40  2196.0  4407.0     2.0      NaN
14  18/07/2017-23:45  1961.0  3848.0     2.0      NaN
15  18/07/2017-23:50     NaN  2880.0  2809.0   4565.0
16  18/07/2017-23:55     NaN  2143.0  2397.0   3725.0
17  19/07/2017-00:00     NaN  1981.0     NaN   2921.0
18  19/07/2017-00:05     NaN  2227.0     NaN   2131.0
19  19/07/2017-00:10  1366.0  2526.0     NaN   1990.0