问题
I have a titanic Dataset. It has attributes and i was working manly on 1.Age 2.Embark ( from which port passengers embarked..There are total 3 ports..S,Q and C) 3.Survived ( 0 for did not survived,1 for survived)
I was filtering the useless data. Then i needed to fill Null values present in Age. So i counted how many passengers survived and didn't survived in each Embark i.e. S,Q and C
I find out the mean age of Passengers who survived and who did not survived after embarking from each S,Q and C port. But now i have no idea how to fill these 6 values ( 3 for survived from each S,Q and C and 3 for who did not survived from each S,Q and C...So total 6) in the original titanic Age column. If i do simply titanic.Age.fillna('With one of the six values') it will fill All the Null values of Age with that one value which i don't want.
After giving some time,i tried this.
titanic[titanic.Survived==1][titanic.Embarked=='S'].Age.fillna(SurvivedS.Age.mean(),inplace=True)
titanic[titanic.Survived==1][titanic.Embarked=='Q'].Age.fillna(SurvivedQ.Age.mean(),inplace=True)
titanic[titanic.Survived==1][titanic.Embarked=='C'].Age.fillna(SurvivedC.Age.mean(),inplace=True)
titanic[titanic.Survived==0][titanic.Embarked=='S'].Age.fillna(DidntSurvivedS.Age.mean(),inplace=True)
titanic[titanic.Survived==0][titanic.Embarked=='Q'].Age.fillna(DidntSurvivedQ.Age.mean(),inplace=True)
titanic[titanic.Survived==0][titanic.Embarked=='C'].Age.fillna(DidntSurvivedC.Age.mean(),inplace=True)
This showed no error but still it doesn't work. Any idea what should i do?
回答1:
I think you need groupby with apply with fillna by mean:
titanic['age'] = titanic.groupby(['survived','embarked'])['age']
.apply(lambda x: x.fillna(x.mean()))
import seaborn as sns
titanic = sns.load_dataset('titanic')
#check NaN rows in age
print (titanic[titanic['age'].isnull()].head(10))
survived pclass sex age sibsp parch fare embarked class \
5 0 3 male NaN 0 0 8.4583 Q Third
17 1 2 male NaN 0 0 13.0000 S Second
19 1 3 female NaN 0 0 7.2250 C Third
26 0 3 male NaN 0 0 7.2250 C Third
28 1 3 female NaN 0 0 7.8792 Q Third
29 0 3 male NaN 0 0 7.8958 S Third
31 1 1 female NaN 1 0 146.5208 C First
32 1 3 female NaN 0 0 7.7500 Q Third
36 1 3 male NaN 0 0 7.2292 C Third
42 0 3 male NaN 0 0 7.8958 C Third
who adult_male deck embark_town alive alone
5 man True NaN Queenstown no True
17 man True NaN Southampton yes True
19 woman False NaN Cherbourg yes True
26 man True NaN Cherbourg no True
28 woman False NaN Queenstown yes True
29 man True NaN Southampton no True
31 woman False B Cherbourg yes False
32 woman False NaN Queenstown yes True
36 man True NaN Cherbourg yes True
42 man True NaN Cherbourg no True
idx = titanic[titanic['age'].isnull()].index
titanic['age'] = titanic.groupby(['survived','embarked'])['age']
.apply(lambda x: x.fillna(x.mean()))
#check if values was replaced
print (titanic.loc[idx].head(10))
survived pclass sex age sibsp parch fare embarked \
5 0 3 male 30.325000 0 0 8.4583 Q
17 1 2 male 28.113184 0 0 13.0000 S
19 1 3 female 28.973671 0 0 7.2250 C
26 0 3 male 33.666667 0 0 7.2250 C
28 1 3 female 22.500000 0 0 7.8792 Q
29 0 3 male 30.203966 0 0 7.8958 S
31 1 1 female 28.973671 1 0 146.5208 C
32 1 3 female 22.500000 0 0 7.7500 Q
36 1 3 male 28.973671 0 0 7.2292 C
42 0 3 male 33.666667 0 0 7.8958 C
class who adult_male deck embark_town alive alone
5 Third man True NaN Queenstown no True
17 Second man True NaN Southampton yes True
19 Third woman False NaN Cherbourg yes True
26 Third man True NaN Cherbourg no True
28 Third woman False NaN Queenstown yes True
29 Third man True NaN Southampton no True
31 First woman False B Cherbourg yes False
32 Third woman False NaN Queenstown yes True
36 Third man True NaN Cherbourg yes True
42 Third man True NaN Cherbourg no True
#check mean values
print (titanic.groupby(['survived','embarked'])['age'].mean())
survived embarked
0 C 33.666667
Q 30.325000
S 30.203966
1 C 28.973671
Q 22.500000
S 28.113184
Name: age, dtype: float64
来源：https://stackoverflow.com/questions/44586346/how-to-fill-null-values-in-a-dataset-using-python-that-matches-with-two-other-co