Question

对于数据框df，如果列b中的值为空2017-01-01或a中的值，我试图用值NaNs填充列Others df = pd.DataFrame({'a':['Coffee','Muffin','Donut','Others',pd.np.nan, pd.np.nan]}) a 0 Coffee 1 Muffin 2 Donut 3 Others 4 NaN 5 NaN：

        a           b
0  Coffee  2017-01-01
1  Muffin  2017-01-01
2   Donut  2017-01-01
3  Others         NaN
4     NaN         NaN
5     NaN         NaN

预期结果如下：

NaNs

我尝试过的不排除df.loc[~df['a'].isin(['nan', 'Others']), 'b'] = '2017-01-01' a b 0 Coffee 2017-01-01 1 Muffin 2017-01-01 2 Donut 2017-01-01 3 Others NaN 4 NaN 2017-01-01 5 NaN 2017-01-01的内容：

SELECT
  company_name,
  warehouse_city,
  COUNT(*)         AS total_orders,
  SUM(CASE WHEN arrival_status = 'cancelled' THEN 1 ELSE 0 END)   AS cancelled_orders
FROM
  your_table
GROUP BY
  company_name,
  warehouse_city
HAVING
  COUNT(*) < 5 * SUM(CASE WHEN arrival_status = 'cancelled' THEN 1 ELSE 0 END)

谢谢！

Answer 1

使用np.nan代替nan：

df.loc[~df['a'].isin([np.nan, 'Others']), 'b'] = '2017-01-01'

或者在比较之前，用Others替换缺失值：

df.loc[~df['a'].fillna('Others').eq('Others'), 'b'] = '2017-01-01'
print (df)
        a           b
0  Coffee  2017-01-01
1  Muffin  2017-01-01
2   Donut  2017-01-01
3  Others         NaN
4     NaN         NaN
5     NaN         NaN

Answer 2

检查一下：

import numpy as np
import pandas as pd

df = pd.DataFrame({'a': ['Coffee', 'Muffin', 'Donut', 'Others', pd.np.nan, pd.np.nan]})
conditions = [
    (df['a'] == 'Others'),
    (df['a'].isnull())
]
choices = [np.nan, np.nan]
df['color'] = np.select(conditions, choices, default='2017-01-01')

print(df)

Answer 3

import pandas as pd
import numpy as np

df = pd.DataFrame({'a':['Coffee','Muffin','Donut','Others',pd.np.nan, pd.np.nan]})


df.loc[df['a'].replace('Others',np.nan).notnull(),'b'] = '2017-01-01'

print(df)

通过排除熊猫中另一列中的特定值来填充一列

3 个答案: