从其他列中的匹配单词创建熊猫列

Question

我有一个包含多个物质列的数据帧，如下所示：

import random
values = ['oil', 'gas', 'water']
modifier = ['dirty', 'crude', 'fuel']
wordzip = [modifier, values]
data = [[wordzip[0][random.randint(0,2)] + ' ' + wordzip[1][random.randint(0,2)] for c in wordzip[0]] for i in range(7)]
pd.DataFrame(data = data, columns = ['A', 'B', 'C'])
A            B            C   
0    dirty gas    crude oil  dirty water 
1  dirty water     fuel gas    dirty gas  
2  dirty water     fuel gas    dirty oil  
3     fuel oil  crude water    crude gas  
4  dirty water     fuel oil  dirty water  
5    crude oil   fuel water    dirty oil
6   fuel water    crude gas  crude water 

我想创建一个新列，其中包含包含单词"oil"的列中的值。所以最终的df应该看起来像这样：

A            B            C          D
0    dirty gas    crude oil  dirty water  crude oil
1  dirty water     fuel gas    dirty gas  NaN
2  dirty water     fuel gas    dirty oil  dirty oil
3     fuel oil  crude water    crude gas  fuel oil
4  dirty water     fuel oil  dirty water  fuel oil
5    crude oil   fuel water    dirty oil  crude oil
6   fuel water    crude gas  crude water  NaN

我尝试了df[['A', 'B', 'C']].apply(lambda x: x.str.contains('oil'))但它返回了一个布尔数据帧，而不是值本身。

Answer 1

让我们使用stack+extract：

df['D'] = df.stack().str.extract(r'(.* oil)').groupby(level=0).first()[0]
df
A            B            C          D
0    dirty gas    crude oil  dirty water  crude oil
1  dirty water     fuel gas    dirty gas        NaN
2  dirty water     fuel gas    dirty oil  dirty oil
3     fuel oil  crude water    crude gas   fuel oil
4  dirty water     fuel oil  dirty water   fuel oil
5    crude oil   fuel water    dirty oil  crude oil
6   fuel water    crude gas  crude water        NaN

Answer 2

将applymap与bfill一起使用

df[df.applymap(lambda x : 'oil' in x)].bfill(1).loc[:,'A']
Out[80]: 
0          NaN
1          NaN
2     fuel oil
3    crude oil
4    crude oil
5     fuel oil
6          NaN
Name: A, dtype: object

Answer 3

像这样：

import pandas as pd
import random
values = ['oil', 'gas', 'water']
modifier = ['dirty', 'crude', 'fuel']
wordzip = [modifier, values]
data = [[wordzip[0][random.randint(0,2)] + ' ' + wordzip[1][random.randint(0,2)] for c in wordzip[0]] for i in range(7)]
df=pd.DataFrame(data = data, columns = ['A', 'B', 'C'])
temp=df[df[['A', 'B', 'C']].apply(lambda x: x.str.contains('oil'))]
df['D'] = temp.A.combine_first(temp.B).combine_first(temp.C)

Answer 4

这个答案对字符串水平求和，然后使用正则表达式提取来获得所需的输出：

# insert temporary columns containing spaces for this regex implementation to work
df.insert(1,'a',' ')
df.insert(3,'b',' ')
# this regex contains a capture group which will get 'oil' instances and the preceding word
df['D'] = df.sum(axis=1).str.extract('([a-z]+ oil)')
# remove the temporary columns
df.drop(['a', 'b'], axis=1, inplace=True)