Unpivot multiple columns with same name in pandas dataframe

Question

I have the following dataframe:

pp  b          pp   b
5   0.001464    6   0.001853
5   0.001459    6   0.001843

Is there a way to unpivot columns with the same name into multiple rows?

This is the required output:

pp  b         
5   0.001464    
5   0.001459    
6   0.001853
6   0.001843

Answer 1

Try groupby with axis=1

df.groupby(df.columns.values, axis=1).agg(lambda x: x.values.tolist()).sum().apply(pd.Series).T.sort_values('pp')
Out[320]: 
          b   pp
0  0.001464  5.0
2  0.001459  5.0
1  0.001853  6.0
3  0.001843  6.0

A fun way with wide_to_long

s=pd.Series(df.columns)
df.columns=df.columns+s.groupby(s).cumcount().astype(str)

pd.wide_to_long(df.reset_index(),stubnames=['pp','b'],i='index',j='drop',suffix='\d+')
Out[342]: 
            pp         b
index drop              
0     0      5  0.001464
1     0      5  0.001459
0     1      6  0.001853
1     1      6  0.001843

Answer 2

This is possible using numpy:

res = pd.DataFrame({'pp': df['pp'].values.T.ravel(),
                    'b': df['b'].values.T.ravel()})

print(res)

          b  pp
0  0.001464   5
1  0.001459   5
2  0.001853   6
3  0.001843   6

---

Or without referencing specific columns explicitly:

res = pd.DataFrame({i: df[i].values.T.ravel() for i in set(df.columns)})

Answer 3

Let's use melt, cumcount and unstack:

dm = df.melt()
dm.set_index(['variable',dm.groupby('variable').cumcount()])\
  .sort_index()['value'].unstack(0)

Output:

variable         b   pp
0         0.001464  5.0
1         0.001459  5.0
2         0.001853  6.0
3         0.001843  6.0

Answer 4

I'm a little bit surprise that nobody has mentioned so far the use of pd.concat... Take a look below:

df1 = pd.DataFrame({'Col1':[1,2,3,4], 'Col2':[5,6,7,8]})
df1
      Col1  Col2
   0     1     5
   1     2     6
   2     3     7
   3     4     8 

Now if you make:

   df2 = pd.concat([df1,df1])

you get:

   Col1  Col2
0     1     5
1     2     6
2     3     7
3     4     8
0     1     5
1     2     6
2     3     7
3     4     8

This is what you wanted, isn't?