Convert column headers to row values

Question

I have a dataframe which looks like this

       Wash_Month  Wash_Day

0           3         2
1           4         3

I need to convert the same to this

          Input              Value

0     Wash_Month/Wash_Day     3,2
1     Wash_Month/Wash_Day     4,3

I tried merging the two columns but wasn't able to convert the column headers to row values

Thanks.

Answer 1

Here is cute way of doing it

pd.DataFrame(dict(
    Input='/'.join(df),
    Value=[*map(','.join, zip(*map(df.astype(str).get, df)))]
))

                 Input Value
0  Wash_Month/Wash_Day   3,2
1  Wash_Month/Wash_Day   4,3

Answer 2

This is a more efficient solution. I break down the steps:

# Compute the Value column with `agg`.
v = df.astype(str).agg(','.join)
# Compute the Input column with `df.columns.str.cat`
v.index = [df.columns.str.cat(sep='/')] * len(v)
# Reset the index.
v.rename_axis('Input').to_frame('Value').reset_index()

                 Input Value
0  Wash_Month/Wash_Day   3,2
1  Wash_Month/Wash_Day   4,3

---

Alternative (slower). Reshape your data a bit with stack :

v = df.stack().astype(str).reset_index(level=1)
v.columns = ['Input', 'Value']

print(v)
        Input Value
0  Wash_Month     3
0    Wash_Day     2
1  Wash_Month     4
1    Wash_Day     3

Look at the index(!). Now, call groupby and agg :

v.groupby(level=0).agg({'Input': '/'.join, 'Value':','.join})

                 Input Value
0  Wash_Month/Wash_Day   3,2
1  Wash_Month/Wash_Day   4,3

Answer 3

Using groupby with dict

d={'Wash_Month':'Wash_Month/Wash_Day','Wash_Day':'Wash_Month/Wash_Day'} 
df.T.astype(str).groupby(d).agg(','.join).stack()
    Out[319]: 
    Wash_Month/Wash_Day  0    3,2
                         1    4,3