Convert a Pandas DataFrame into a single row DataFrame

Question

I\'ve seen similar questions but mine is more direct and abstract.

I have a dataframe with \"n\" rows, being \"n\" a small number.We can assume the index is just the

Answer 1

Another way using list comprehension -

ndf = pd.DataFrame(df.values.reshape(1, -1)[0]).T
ndf.columns = [j + '_' + str(i) for i in range(1, 4) for j in df.columns]

Answer 2

Unstack and map i.e

ndf = df.unstack().to_frame().T

ndf.columns = ndf.columns.map('{0[0]}_{0[1]}'.format) 

    A_0  A_1  A_2  B_0  B_1  B_2  C_0  C_1  C_2  D_0  D_1  D_2  E_0  E_1  E_2
0    1    6   11    2    7   12    3    8   13    4    9   14    5   10    5

In case you want the sorted columns then you can do

ndf = df.unstack().to_frame().T.sort_index(1,1)

Answer 3

Let's try this, using stack, to_frame, and T:

df.index = df.index + 1
df_out = df.stack()
df_out.index = df_out.index.map('{0[1]}_{0[0]}'.format)
df_out.to_frame().T

Output:

   A_1  B_1  C_1  D_1  E_1  A_2  B_2  C_2  D_2  E_2  A_3  B_3  C_3  D_3  E_3
0    1    2    3    4    5    6    7    8    9   10   11   12   13   14    5

Answer 4

We need stack and swaplevel

df1=df.stack().swaplevel()
df1.index=df1.index.map('{0[0]}_{0[1]}'.format) 
df1.to_frame().T
Out[527]: 
   A_0  B_0  C_0  D_0  E_0  A_1  B_1  C_1  D_1  E_1  A_2  B_2  C_2  D_2  E_2
0    1    2    3    4    5    6    7    8    9   10   11   12   13   14    5

Or you can using numpy

pd.DataFrame(data=np.concatenate(df.values),index=[m+'_'+str(n) for m,n in zip(df.columns.tolist()*3,np.repeat([1,2,3],df.shape[1]))]).T
Out[551]: 
   A_1  B_1  C_1  D_1  E_1  A_2  B_2  C_2  D_2  E_2  A_3  B_3  C_3  D_3  E_3
0    1    2    3    4    5    6    7    8    9   10   11   12   13   14    5