CODEWITHSUNDEEP
pythonpandas
Trace R.
Trace R.

Reputation: 347

How to append strings inside dataframe cells based on column values

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: enter image description here

Upvotes: 1

Views: 454

Answers (4)

Andy L.
Andy L.

Reputation: 25239

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

Upvotes: 0

Shadab Hussain
Shadab Hussain

Reputation: 814

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

Upvotes: 1

Chris Adams
Chris Adams

Reputation: 18647

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

Upvotes: 1

Meto
Meto

Reputation: 658

You may first construct the result column as a numpy array while traversing the data frame columns and checking the values then you can add the new result column and drop the old one.

Upvotes: 0

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