Question

我有一个看起来像这样的数据框：

boat_type   boat_type_2
Not Known   Not Known
Not Known   kayak
ship        Not Known
Not Known   Not Known
ship        Not Known

我想创建第三列boat_type_final，其内容应如下所示：

boat_type   boat_type_2  boat_type_final
Not Known   Not Known    cruise
Not Known   kayak        kayak
ship        Not Known    ship  
Not Known   Not Known    cruise
ship        Not Known    ship

因此，基本上，如果boat_type和boat_type_2中都存在“未知”，则该值应为“巡航”。但是，如果在前两列中有一个不是“未知”的字符串，则boat_type_final应该用该字符串填充，即“皮划艇”或“船”。

最优雅的方法是什么？我已经看到了多个选项，例如where，创建函数和/或逻辑，而且我想知道真正的pythonista会做什么。

到目前为止，这是我的代码：

import pandas as pd
import numpy as np
data = [{'boat_type': 'Not Known', 'boat_type_2': 'Not Known'},
    {'boat_type': 'Not Known',  'boat_type_2': 'kayak'},
    {'boat_type': 'ship',  'boat_type_2': 'Not Known'},
    {'boat_type': 'Not Known',  'boat_type_2': 'Not Known'},
    {'boat_type': 'ship',  'boat_type_2': 'Not Known'}]
df = pd.DataFrame(data
df['phone_type_final'] = np.where(df.phone_type.str.contains('Not'))...

Answer 1

使用：

df['boat_type_final'] = (df.replace('Not Known',np.nan)
                           .ffill(axis=1)
                           .iloc[:, -1]
                           .fillna('cruise'))
print (df)
   boat_type boat_type_2 boat_type_final
0  Not Known   Not Known          cruise
1  Not Known       kayak           kayak
2       ship   Not Known            ship
3  Not Known   Not Known          cruise
4       ship   Not Known            ship

说明：

第一个replace Not Known缺失值：

print (df.replace('Not Known',np.nan))
  boat_type boat_type_2
0       NaN         NaN
1       NaN       kayak
2      ship         NaN
3       NaN         NaN
4      ship         NaN

然后通过向前填充每行来替换NaN：

print (df.replace('Not Known',np.nan).ffill(axis=1))
  boat_type boat_type_2
0       NaN         NaN
1       NaN       kayak
2      ship        ship
3       NaN         NaN
4      ship        ship

按iloc按位置选择最后一列：

print (df.replace('Not Known',np.nan).ffill(axis=1).iloc[:, -1])
0      NaN
1    kayak
2     ship
3      NaN
4     ship
Name: boat_type_2, dtype: object

如果可能的话，NaN添加fillna：

print (df.replace('Not Known',np.nan).ffill(axis=1).iloc[:, -1].fillna('cruise'))
0    cruise
1     kayak
2      ship
3    cruise
4      ship
Name: boat_type_2, dtype: object

如果仅使用几列，则使用另一种解决方案：numpy.select：

m1 = df['boat_type'] == 'ship'
m2 = df['boat_type_2'] == 'kayak'

df['boat_type_final'] = np.select([m1, m2], ['ship','kayak'], default='cruise')
print (df)
   boat_type boat_type_2 boat_type_final
0  Not Known   Not Known          cruise
1  Not Known       kayak           kayak
2       ship   Not Known            ship
3  Not Known   Not Known          cruise
4       ship   Not Known            ship

Answer 2

另一种解决方案是在具有映射的位置定义函数：

def my_func(row):
    if row['boat_type']!='Not Known':
        return row['boat_type']
    elif row['boat_type_2']!='Not Known':
        return row['boat_type_2']
    else: 
        return 'cruise'

[注意：您没有提到当两列都不为'Unknown'时会发生什么。]

然后只需应用以下功能：

df.loc[:,'boat_type_final'] = df.apply(my_func, axis=1)

print(df)

输出：

   boat_type boat_type_2 boat_type_final
0  Not Known   Not Known          cruise
1  Not Known       kayak           kayak
2       ship   Not Known            ship
3  Not Known   Not Known          cruise
4       ship   Not Known            ship

根据其他两列中的字符串创建pandas数据框列

2 个答案: