如何根据列值在数据框单元格内附加字符串

Question

Given a dataframe:

import pandas as pd

df = pd.DataFrame(data= {'Col1': ['No', 'Yes', 'No', 'Maybe'], 'Col2': ['Yes', 'No', 'No', 'No'], 'Result': ''})

I want to populate Result with a list that may need to be appended based upon a column value. In this case, the parameters would be:

If the value is 'Yes' keep the current value of Result , if the value is 'Maybe' append 'Attention needed (insert column name)', if the value is 'No' append 'Failure (insert column name)'

Desired result:

Answer 1

Not very pretty, but you could create a dict , then use stack , map and groupby with join aggregation:

d = {'No': 'Failure', 'Maybe': 'Attention needed'}
s = df[['Col1', 'Col2']].stack().map(d).dropna()

df['Result'] = (s + ' ' + s.index.get_level_values(1)).groupby(level=0).agg(', '.join)

[out]

    Col1 Col2                               Result
0     No  Yes                         Failure Col1
1    Yes   No                         Failure Col2
2     No   No           Failure Col1, Failure Col2
3  Maybe   No  Attention needed Col1, Failure Col2

Answer 2

Try this one liner code using lambda function:

df['Result'] = df[['Col1','Col2']].apply(lambda x: 'Failure Col1' if (x[0]=='No' and x[1]=='Yes') else ('Failure Col2' if (x[1]=='No' and x[0]=='Yes') else ('Failure Col1, Failure Col2' if (x[0]=='No' and x[1]=='No') else("Attention needed Col1, Failure Col2" if (x[0]=='Maybe' and x[1]=='No') else None))), axis=1)

Output:

   Col1     Col2    Result
0   No      Yes     Failure Col1
1   Yes     No      Failure Col2
2   No      No      Failure Col1, Failure Col2
3   Maybe   No      Attention needed Col1, Failure Col2

Answer 3

您可以首先将结果列构造为一个 numpy 数组，同时遍历数据框列并检查值，然后您可以添加新的结果列并删除旧的结果列。

Answer 4

Construct a dictionary to replace values in df and Using * and + to construct a series of appropriate message strings and finally join them and assign to df.Result

d = {'Yes': '', 'No': 'Failure ', 'Maybe': 'Attention needed '}
df1 = df[['Col1', 'Col2']]
df['Result'] = ((df1.replace(d) 
                + df1.ne('Yes').values * df1.columns.values).agg(','.join, axis=1)
                                                            .str.strip(','))

Or

df['Result'] = ((df1.replace(d) 
                + df1.ne('Yes').values * (df1.columns+',').values).sum(1)
                                                                  .str.strip(','))

Out[267]:
    Col1 Col2                              Result
0     No  Yes                        Failure Col1
1    Yes   No                        Failure Col2
2     No   No           Failure Col1,Failure Col2
3  Maybe   No  Attention needed Col1,Failure Col2

---

Here the detail

df1.replace(d) + df1.ne('Yes').values * df1.columns.values

Out[268]:
                    Col1          Col2
0           Failure Col1
1                         Failure Col2
2           Failure Col1  Failure Col2
3  Attention needed Col1  Failure Col2

((df1.replace(d) + df1.ne('Yes').values * df1.columns.values).agg(','.join, axis=1)
                                                             .str.strip(','))

Out[269]:
0                          Failure Col1
1                          Failure Col2
2             Failure Col1,Failure Col2
3    Attention needed Col1,Failure Col2
dtype: object