我已经定义了一个熊猫数据框：
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我想创建一个新列C，该列将为每个值df.A = i计算B相应元素的过滤器scipy.signal.savgol_filter，即df.loc [df.A == i]的过滤器.B代表i = 1,2,3 ...
我使用以下代码：
df = pd.DataFrame( {
'A': [1,1,1,1,2,2,2,3,3],
'B': [5,4,8,6,5,6,6,7,7])
df
A B
0 1 5
1 1 4
2 1 8
3 1 6
4 2 5
5 2 6
6 2 6
7 3 7
8 3 7
这不会创建列“ C”并给我消息：
for i in df.A.unique() :
df.loc[df.A==i]['C']=scipy.signal.savgol_filter(df.loc[df.A==i].B, 3, 1)
我已经阅读了文档，但是没有找到定义新列的正确方法。我应该用什么方法来做？
谢谢您的帮助。
说明：
此问题与scipy.signal.savgol_filter函数无关，并且与使用df.B的N个元素创建要放入df.C的N个其他元素（例如执行fft的其他任何函数）相同。 df.loc [df.A == i] .B for i = 1,2,3 ...
答案 0 :(得分：3)
它叫做chaining indexing，更好的是：
for i in df.A.unique() :
df.loc[df.A==i, 'C']=scipy.signal.savgol_filter(df.loc[df.A==i, 'B'], 3, 1)
但是最好使用GroupBy.transform：
import scipy.signal
#added last row to sample for avoid error
df = pd.DataFrame( {
'A': [1,1,1,1,2,2,2,3,3,3],
'B': [5,4,8,6,5,6,6,7,7,5]})
#print (df)
df['C'] = df.groupby('A')['B'].transform(lambda x: scipy.signal.savgol_filter(x, 3, 1))
print (df)
A B C
0 1 5 4.166667
1 1 4 5.666667
2 1 8 6.000000
3 1 6 7.000000
4 2 5 5.166667
5 2 6 5.666667
6 2 6 6.166667
7 3 7 7.333333
8 3 7 6.333333
9 3 5 5.333333
答案 1 :(得分：1)
代替
df.loc[df.A==i]['C']
使用
df.loc[df.A==i, 'C']
通过df.loc[df.A==i]['C']，您实际上是在更改df的副本，而不是原定的原始副本