根据来自其他两列的字符串创建熊猫数据框列答案

【问题标题】：Create pandas data frame column based on strings from two other columns根据来自其他两列的字符串创建熊猫数据框列
【发布时间】：2018-07-25 09:42:18
【问题描述】：

我有一个如下所示的数据框：

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，应该如下所示：

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

所以基本上如果boat_type 和boat_type_2 中都存在“未知”，那么该值应该是“巡航”。但如果前两列中存在“Not Known”以外的字符串，则应使用该字符串填充boat_type_final，“kayak”或“ship”。

最优雅的方法是什么？我见过几个选项，例如where、创建函数和/或逻辑，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'))...

【问题讨论】：

标签： python python-3.x pandas numpy dataframe

【解决方案1】：

用途：

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

解释：

首先replaceNot Known 到缺失值：

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

然后通过每行前向填充替换NaNs：

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的位置选择最后一列：

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NaNs 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

如果只有几列的另一种解决方案是使用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

【讨论】：

你能解释一下它是如何/为什么起作用的吗？特别是这部分：.ffill(axis=1).iloc[:, -1]
@bzier - 当然，等一下。
@bzier - 答案已修改。

【解决方案2】：

另一种解决方案是在您拥有映射的地方定义您的函数：

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是“未知”时应该发生什么。]

然后简单地应用函数：

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

【讨论】：