Convert data to new columns

Question

columns = ['a', "b","cin",'cout', 'din', 'dout']
rows = [[1 , 1, 3, 4, 2, 3], [2,3,3,1, 8 , 4], [3,1,3,1, 2, 1]]
dff1 = pd.DataFrame(rows, columns=columns)

result

How to make a conclusion as 6 lines, where Cin, Cout and din, dout are taken as 2 lines with a and b first? Also with added in / out column.

In general, I need to manipulate the above data framework to bring the below

Answer 1

You can use lreshape:

df = pd.lreshape(dff, {'c':['c1','c2'], 'd':['d1','d2']})
print (df)
   a  b  c  d
0  1  1  3  2
1  2  3  3  8
2  3  1  3  2
3  1  1  4  3
4  2  3  1  4
5  3  1  1  1

Or wide_to_long:

dff = dff.reset_index()
a = (pd.wide_to_long(dff, stubnames=['c', 'd'], i='index', j='B')
       .reset_index(drop=True)
       .reindex(columns=['a','b','c', 'd']))
print (a)
   a  b  c  d
0  1  1  3  2
1  2  3  3  8
2  3  1  3  2
3  1  1  4  3
4  2  3  1  4
5  3  1  1  1

EDIT: If you want to use the columns melt, extract, assignand the last : sort_values

a = dff1.melt(id_vars=['a','b'],value_vars=['cin','cout'],value_name = 'c',var_name='in/out')
b = dff1.melt(id_vars=['a','b'],value_vars=['din','dout'],value_name = 'd',var_name='in/out')
a['in/out'] = a['in/out'].str.extract('(in|out)', expand=False)
b['in/out'] = b['in/out'].str.extract('(in|out)', expand=False)
print (a)
   a  b in/out  c
0  1  1     in  3
1  2  3     in  3
2  3  1     in  3
3  1  1    out  4
4  2  3    out  1
5  3  1    out  1

print (b)
   a  b in/out  d
0  1  1     in  2
1  2  3     in  8
2  3  1     in  2
3  1  1    out  3
4  2  3    out  4
5  3  1    out  1

c = a.assign(d=b['d']).sort_values(['a','b'])
#same as
#c = pd.merge(a,b).sort_values(['a','b'])

print (c)
   a  b in/out  c  d
0  1  1     in  3  2
3  1  1    out  4  3
1  2  3     in  3  8
4  2  3    out  1  4
2  3  1     in  3  2
5  3  1    out  1  1

Solution rewritten for pandas 0.15.0:

a=pd.melt(dff1,id_vars=['a','b'],value_vars=['cin','cout'],value_name='c',var_name='in/out') 
b=pd.melt(dff1,id_vars=['a','b'],value_vars=['din','dout'],value_name='d',var_name='in/out')
a['in/out'] = a['in/out'].str.extract('(in|out)')
b['in/out'] = b['in/out'].str.extract('(in|out)')
c = pd.merge(a,b).sort_values(['a','b'])

Another solution from wen's remote answer is to replacestring to numeric and then use the wide_to_longlast one mapback:

#define columns
L = ['in','out']
d = dict(enumerate(L))
d1 = {v: str(k) for k, v in d.items()}
print (d)
{0: 'in', 1: 'out'}

print (d1)
{'out': '1', 'in': '0'}

dff1.columns = dff1.columns.to_series().replace(d1,regex=True)
a = pd.wide_to_long(dff1, stubnames=['c', 'd'], j='in/out', i=['a','b']).reset_index()
a['in/out'] = a['in/out'].astype(int).map(d)
a = a[['a','b','c','d','in/out']]
print (a)
   a  b  c  d in/out
0  1  1  3  2     in
1  1  1  4  3    out
2  2  3  3  8     in
3  2  3  1  4    out
4  3  1  3  2     in
5  3  1  1  1    out

EDIT:

To return the process, use:

df = df.set_index(['a', 'b', 'in/out']).unstack()
df.columns = df.columns.map(''.join)
df = df.reset_index()
print (df)
   a  b  cin  cout  din  dout
0  1  1    3     4    2     3
1  2  3    3     1    8     4
2  3  1    3     1    2     1