如何将df列中的多个值更改为np.nan公司，基于其他列中的条件？

Question

我没有太多的编码经验，这是我的第一个问题，所以请耐心等待我。我需要找到一种方法将一列的多个值更改为np.nan公司，基于另一列中的条件。因此，我创建了所需列“Vorgabe”和“Temp”的副本。你知道吗

每当“Grad”中的值不是0时，我想将“Vorgabe”和“Temp”中定义区域中的值更改为np.nan公司. 你知道吗

print(df)  

    OptOpTemp  OpTemp  BSP  Grad  Vorgabe  Temp
0        22.0    20.0    5   0.0     22.0  20.0
1        22.0    20.5    7   0.0     22.0  20.5
2        22.0    21.0    8   1.0     22.0  21.0
3        22.0    21.0    6   0.0     22.0  21.0
4        22.0    23.5    7   0.0     22.0  20.0
5        23.0    21.5    1   0.0     23.0  21.5
6        24.0    22.5    3   1.0     24.0  22.5
7        24.0    23.0    4   0.0     24.0  23.0
8        24.0    25.5    9   0.0     24.0  25.5

所以我想实现这样的目标：

    OptOpTemp  OpTemp  BSP  Grad  Vorgabe  Temp
0        22.0    20.0    5   0.0     22.0  20.0
1        22.0    20.5    7   0.0     nan   nan      <-one row above
2        22.0    21.0    8   1.0     nan   nan
3        22.0    21.0    6   0.0     nan   nan      <-one row among
4        22.0    23.5    7   0.0     22.0  20.0
5        23.0    21.5    1   0.0     nan   nan
6        24.0    22.5    3   1.0     nan   nan
7        24.0    23.0    4   0.0     nan   nan
8        24.0    25.5    9   0.0     24.0  25.5

有人能解决我的问题吗？你知道吗

编辑：我可能不清楚。目标是将定义区域中“Vorgabe”和“Temp”中的每个值都更改为nan。在我的示例中，区域将是上面的一行，其中包含1.0的行，以及中间的一行。所以不仅是1.0所在的行，还有上面和下面的行。你知道吗

Answer 1

带|的链3条件对于bitwise OR，对于1上方和下方的行，使用带^{}的掩码：

mask1 = df['Grad'] == 1
mask2 = df['Grad'].shift() == 1
mask3 = df['Grad'].shift(-1) == 1

---

mask1 = df['Grad'] != 0
mask2 = df['Grad'].shift() != 0
mask3 = df['Grad'].shift(-1) != 0

mask = mask1 | mask2 | mask3

df.loc[mask, ['Vorgabe', 'Temp']] = np.nan
print (df)
   OptOpTemp  OpTemp  BSP  Grad  Vorgabe  Temp
0       22.0    20.0    5   0.0     22.0  20.0
1       22.0    20.5    7   0.0      NaN   NaN
2       22.0    21.0    8   1.0      NaN   NaN
3       22.0    21.0    6   0.0      NaN   NaN
4       22.0    23.5    7   0.0     22.0  20.0
5       23.0    21.5    1   0.0      NaN   NaN
6       24.0    22.5    3   1.0      NaN   NaN
7       24.0    23.0    4   0.0      NaN   NaN
8       24.0    25.5    9   0.0     24.0  25.5

多行通用解决方案：

N = 1
#create range for test value betwen -N to N
r = np.concatenate([np.arange(0, N+1), np.arange(-1, -N-1, -1)])
#create boolean mask by comparing with shift and join together by reduce 
mask = np.logical_or.reduce([df['Grad'].shift(x) == 1 for x in r])

df.loc[mask, ['Vorgabe', 'Temp']] = np.nan

编辑：

可以将两个遮罩连接在一起：

N = 1
r1 = np.concatenate([np.arange(0, N+1), np.arange(-1, -N-1, -1)])
mask1 = np.logical_or.reduce([df['Grad'].shift(x) == 1 for x in r1])

N = 2
r2 = np.concatenate([np.arange(0, N+1), np.arange(-1, -N-1, -1)])
mask2 = np.logical_or.reduce([df['Grad'].shift(x) == 1.5 for x in r2])
#if not working ==1.5 because precision of floats
#mask2 = np.logical_or.reduce([np.isclose(df['Grad'].shift(x), 1.5) for x in r2])

mask = mask1 | mask2
df.loc[mask, ['Vorgabe', 'Temp']] = np.nan
print (df)
   OptOpTemp  OpTemp  BSP  Grad  Vorgabe  Temp
0       22.0    20.0    5   0.0     22.0  20.0
1       22.0    20.5    7   0.0      NaN   NaN
2       22.0    21.0    8   1.0      NaN   NaN
3       22.0    21.0    6   0.0      NaN   NaN
4       22.0    23.5    7   0.0      NaN   NaN
5       23.0    21.5    1   0.0      NaN   NaN
6       24.0    22.5    3   1.5      NaN   NaN <- changed value to 1.5
7       24.0    23.0    4   0.0      NaN   NaN
8       24.0    25.5    9   0.0      NaN   NaN

Answer 2

你可以用numpy.where。你知道吗

import numpy as np

df['Vorbage']=np.where(df['Grad']!=0, df['OptOpTemp'], np.nan)
df['Temp']=np.where(df['Grad']!=0, df['OpTemp'], np.nan)

Answer 3

使用loc：

df.loc[df.Grad != 0.0, ['Vorgabe', 'Temp']] = np.nan
print(df)

输出

   OptOpTemp  OpTemp  BSP  Grad  Vorgabe  Temp
0       22.0    20.0    5   0.0     22.0  20.0
1       22.0    20.5    7   0.0     22.0  20.5
2       22.0    21.0    8   1.0      NaN   NaN
3       22.0    21.0    6   0.0     22.0  21.0
4       22.0    23.5    7   0.0     22.0  20.0
5       23.0    21.5    1   0.0     23.0  21.5
6       24.0    22.5    3   1.0      NaN   NaN
7       24.0    23.0    4   0.0     24.0  23.0
8       24.0    25.5    9   0.0     24.0  25.5