如何使用Python Pandas在特定切片中制作一片DataFrame和“fillna”？

The problem: let us take Titanic dataset from Kaggle. I have dataframe with columns "Pclass", "Sex" and "Age". I need to fill NaN in column "Age" with a median for certain group. If it is a woman from 1st class, I would like to fill her age with the median for 1st class women, not with the median for whole Age column.

问题是:让我们从Kaggle中获取Titanic数据集。我的数据框有“Pclass”,“Sex”和“Age”列。我需要在“年龄”栏填写NaN,其中某个组的中位数。如果是一等女性,我想在她的年龄中填写一等女性的中位数,而不是整个年龄段的中位数。

The question is how to make this change in a certain slice?

问题是如何在某个切片中进行此更改?

I tried:

data['Age'][(data['Sex'] == 'female')&(data['Pclass'] == 1)&(data['Age'].isnull())].fillna(median)

where the "median" is my value, but nothing changes "inplace=True" didn't help.

其中“中位数”是我的值,但没有任何变化“inplace = True”没有帮助。

Thanks alot!

2 个解决方案

#1

I believe you need filter by masks and assign back:

我相信你需要通过面具过滤并分配回来:

data = pd.DataFrame({'a':list('aaaddd'),
                     'Sex':['female','female','male','female','female','male'],
                     'Pclass':[1,2,1,2,1,1],
                     'Age':[40,20,30,20,np.nan,np.nan]})

print (data)
    Age  Pclass     Sex  a
0  40.0       1  female  a
1  20.0       2  female  a
2  30.0       1    male  a
3  20.0       2  female  d
4   NaN       1  female  d
5   NaN       1    male  d

#boolean mask
mask1 = (data['Sex'] == 'female')&(data['Pclass'] == 1)

#get median by mask without NaNs
med = data.loc[mask1, 'Age'].median()
print (med)
40.0

#repalce NaNs
data.loc[mask1, 'Age'] = data.loc[mask1, 'Age'].fillna(med)
print (data)
    Age  Pclass     Sex  a
0  40.0       1  female  a
1  20.0       2  female  a
2  30.0       1    male  a
3  20.0       2  female  d
4  40.0       1  female  d
5   NaN       1    male  d

What is same as:

同样如下:

mask2 = mask1 &(data['Age'].isnull())

data.loc[mask2, 'Age'] = med
print (data)
    Age  Pclass     Sex  a
0  40.0       1  female  a
1  20.0       2  female  a
2  30.0       1    male  a
3  20.0       2  female  d
4  40.0       1  female  d
5   NaN       1    male  d

EDIT:

If need replace all groups NaNs by median:

如果需要用中位数替换所有组NaN:

data['Age'] = data.groupby(["Sex","Pclass"])["Age"].apply(lambda x: x.fillna(x.median()))
print (data)

    Age  Pclass     Sex  a
0  40.0       1  female  a
1  20.0       2  female  a
2  30.0       1    male  a
3  20.0       2  female  d
4  40.0       1  female  d
5  30.0       1    male  d

#2

In case you want to do the same for every groups you can use this trick

如果你想为每个组做同样的事情,你可以使用这个技巧

data = pd.DataFrame({'a':list('aaaddd'),
                    'Sex':['female','female','male','female','female','male'],
                    'Pclass':[1,2,1,2,1,1],
                    'Age':[40,20,30,20, np.nan, np.nan]})
df = data.groupby(["Sex","Pclass"])["Age"].median().to_frame().reset_index()
df.rename(columns={"Age":"Med"}, inplace=True)
data = pd.merge(left=data,right=df, how='left', on=["Sex", "Pclass"])
data["Age"] = np.where(data["Age"].isnull(), data["Med"], data["Age"])

UPDATE:

# dummy dataframe
n = int(1e7)
data = pd.DataFrame({"Age":np.random.choice([10,20,20,30,30,40,np.nan], n),
                     "Pclass":np.random.choice([1,2,3], n),
                     "Sex":np.random.choice(["male","female"], n),
                     "a":np.random.choice(["a","b","c","d"], n)})

In my machine running this (is as the previous without renaming)

在我的机器上运行它(如前所述没有重命名)

df = data.groupby(["Sex","Pclass"])["Age"].agg(['median']).reset_index()
data = pd.merge(left=data,right=df, how='left', on=["Sex", "Pclass"])
data["Age"] = np.where(data["Age"].isnull(), data["median"], data["Age"])

CPU times: user 1.98 s, sys: 216 ms, total: 2.2 s
Wall time: 2.2 s

While the mask solution took:

虽然掩模解决方案采取:

for sex in ["male", "female"]:
    for pclass in range(1,4):
        mask1 =(data['Sex'] == sex)&(data['Pclass'] == pclass)
        med = data.loc[mask1, 'Age'].median()
        data.loc[mask1, 'Age'] = data.loc[mask1, 'Age'].fillna(med)

CPU times: user 5.13 s, sys: 60 ms, total: 5.19 s
Wall time: 5.19 s

@jezrael solution is even faster

@jezrael解决方案更快

data['Age'] = data.groupby(["Sex","Pclass"])["Age"].apply(lambda x: x.fillna(x.median()))

CPU times: user 1.34 s, sys: 92 ms, total: 1.44 s
Wall time: 1.44 s

#1