pandas 引入条件时不能使用 fillna
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Cannot use fillna when a condition is introduced
提问by mlee_jordan
I am very new to python. Trying to do some imputation in my data. however, I could not manage. Here is the simple code:
我对python很陌生。试图对我的数据进行一些插补。然而，我无法应付。这是简单的代码：
df['a'] = ""
df.loc[(df['c'] >= 0) & (df['c'] <= 43), 'a'] = 1
df.loc[(df['c'] >= 44) & (df['c'] <= 96), 'a'] = 2
df.loc[(df['c'] >= 97) & (df['c'] <= 151), 'a'] = 3
df.loc[(df['c'] >= 152) & (df['c'] <= 273), 'a'] = 4
print(df[df['a'] == 1]['b'].median())
print(df[df['a'] == 2]['b'].median())
print(df[df['a'] == 3]['b'].median())
print(df[df['a'] == 4]['b'].median())
print(df[df['a'] == 1]['b'].median())
df[df['a'] == 1]['b'].fillna(df[df['a'] == 1]['b'].median(), inplace=True)
When I tried this it threw a warning:
当我尝试这样做时，它发出了警告：
A value is trying to be set on a copy of a slice from a DataFrame
How can I apply fillna properly?
如何正确使用fillna？
回答by jezrael
Use loc:
使用loc：
df = pd.DataFrame({'c':[10,50,100,200] * 3,
'b':[1,3,8,np.nan,5,8,np.nan,7, np.nan, 4,1,0]})
#print (df)
m1 = (df['c'] >= 0) & (df['c'] <= 43)
m2 = (df['c'] >= 44) & (df['c'] <= 96)
m3 = (df['c'] >= 97) & (df['c'] <= 151)
m4 = (df['c'] >= 152) & (df['c'] <= 273)
df.loc[m1,'b'] = df.loc[m1,'b'].fillna(df.loc[m1,'b'].median())
df.loc[m2,'b'] = df.loc[m2,'b'].fillna(df.loc[m2,'b'].median())
df.loc[m3,'b'] = df.loc[m3,'b'].fillna(df.loc[m3,'b'].median())
df.loc[m4,'b'] = df.loc[m4,'b'].fillna(df.loc[m4,'b'].median())
print (df)
b c
0 1.0 10
1 3.0 50
2 8.0 100
3 3.5 200
4 5.0 10
5 8.0 50
6 4.5 100
7 7.0 200
8 3.0 10
9 4.0 50
10 1.0 100
11 0.0 200
But better is use cutfor category column and then groupbywith custom function with fillnaand median:
但更好的是cut用于类别列，然后groupby使用带有fillna和的自定义函数median：
bins = [0,43,96,151,273]
labels=[1,2, 3, 4]
df['a'] = pd.cut(df['c'], bins=bins, labels=labels, include_lowest=True)
df['b'] = df.groupby('a')['b'].apply(lambda x: x.fillna(x.median()))
print (df)
b c a
0 1.0 10 1
1 3.0 50 2
2 8.0 100 3
3 3.5 200 4
4 5.0 10 1
5 8.0 50 2
6 4.5 100 3
7 7.0 200 4
8 3.0 10 1
9 4.0 50 2
10 1.0 100 3
11 0.0 200 4
回答by Allen
#Use.loc when you try to change df values.
df.loc[df.a==1,'b'] = df.loc[df.a==1,'b'].fillna(df[df['a'] == 1]['b'].median())
