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根据一列将缺失值填充到另一列

[英]fill missing value based on one column to another

我有两列这样的:

在此处输入图像描述

我想要做的是假设'age'列值在30-39之间,我想填充age_band = 30的缺失值。就像假设'age'列值在80-89之间,我想填充缺失的age_band 的值 = 80。我如何在 pandas dataframe 中做到这一点?

我试过这样,但循环一直在运行

for ages in data['age']:
if 0<=ages<=9:
    data['age_band']= data['age_band'].fillna(0)
elif 10<=ages<=19:
    data['age_band']= data['age_band'].fillna(10)
elif 20<=ages<=29:
    data['age_band']= data['age_band'].fillna(20)
elif 30<=ages<=39:
    data['age_band']= data['age_band'].fillna(30)
elif 40<=ages<=49:
    data['age_band']= data['age_band'].fillna(40)
elif 50<=ages<=59:
    data['age_band']= data['age_band'].fillna(50)
elif 60<=ages<=69:
    data['age_band']= data['age_band'].fillna(60)
elif 70<=ages<=79:
    data['age_band']= data['age_band'].fillna(70)
elif 80<=ages<=89:
    data['age_band']= data['age_band'].fillna(80)
elif 90<=ages<=99:
    data['age_band']= data['age_band'].fillna(90)
elif 100<=ages<=109:
    data['age_band']= data['age_band'].fillna(100)

请帮我

试试这个快捷方式:

data['age_band'] = data['age_band'].fillna(data['age'] // 10 * 10).astype(int)
print(data)

# Output
   age  age_band
0   93        90
1   46        40
2   50        50
3   56        50
4   89        80
5   19        10
6   25        20
7   17        10
8   54        50
9   42        40

设置:

import pandas as pd
import numpy as np

np.random.seed(2022)
data = pd.DataFrame({'age': np.random.randint(1, 111, 10), 'age_band': np.nan})
print(data)

# Output
   age  age_band
0   93       NaN
1   46       NaN
2   50       NaN
3   56       NaN
4   89       NaN
5   19       NaN
6   25       NaN
7   17       NaN
8   54       NaN
9   42       NaN

上述答案仅在年龄箱相等时才有效,您可以尝试 pd.cut 在所有情况下都可以使用。

您也可以对 pd.cut() 使用标签。 以下示例包含 0-9 范围内的年龄。 我们正在添加一个名为“age alband”的新列来对年龄进行分类

bins代表区间:0-9为1个区间,10-19为1个区间,以此类推对应的标签为“0-9”等

bins = [0, 9,19,29,39,49,59,69,79,89,99,109]
labels = ["0-9","10-19","20-29","30-39","40-49","50-59","60-69","70-79","80-89","90-99","100-109",">109"]
data['age_band']= pd.cut(data['age'], bins=bins, labels=labels)

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