Enumerate columns with same prefix

The idea is to group columns with the same prefix, and establish a cumcount for them.

Since we need to handle column without a prefix separately, we will need to do this in two steps using GroupBy.cumcount and np.where:

cols = df.columns.str.split('_').str[0].to_series()

df.columns = np.where(
    cols.groupby(level=0).transform('count') > 1, 
    cols.groupby(level=0).cumcount().add(1).astype(str).radd(df.columns), 
    cols
)

df
   A  B  Data_mean1  Data_std2  Data_corr3 Text_one1 Text_two2 Text_three3
0  a  e           1          5           9       foo       bar         bar
1  b  f           2          6          10       bar       foo         bar
2  c  g           3          7          11    foobar    barfoo      barbar
3  d  h           4          8          12    barfoo    foobar      foofoo

---

A simpler solution would be to set columns you don't want to add a suffix to as the index. Then you can simply do

df.set_index(['A', 'B'], inplace=True)
df.columns = (
    df.columns.str.split('_')
      .str[0]
      .to_series()
      .groupby(level=0)
      .cumcount()
      .add(1)
      .astype(str)
      .radd(df.columns))

df
     Data_mean1  Data_std2  Data_corr3 Text_one1 Text_two2 Text_three3
A B                                                                   
a e           1          5           9       foo       bar         bar
b f           2          6          10       bar       foo         bar
c g           3          7          11    foobar    barfoo      barbar
d h           4          8          12    barfoo    foobar      foofoo

You could also use a defaultdict to create a counter for each prefix.

from collections import defaultdict

prefix_starting_location = 2
columns = df.columns[prefix_starting_location:]
prefixes = set(col.split('_')[0] for col in columns)

new_cols = []
dd = defaultdict(int)
for col in columns:
    prefix = col.split('_')[0]
    dd[prefix] += 1
    new_cols.append(col + str(dd[prefix]))
df.columns = df.columns[:prefix_starting_location].tolist() + new_cols
>>> df
   A  B  Data_mean1  Data_std2  Data_corr3 Text_one1 Text_two2 Text_three3
0  a  e           1          5           9       foo       bar         bar
1  b  f           2          6          10       bar       foo         bar
2  c  g           3          7          11    foobar    barfoo      barbar
3  d  h           4          8          12    barfoo    foobar      foofoo
​

If the prefixes are known:

prefixes = ['Data', 'Text']
new_cols = []
dd = defaultdict(int)
for col in df.columns:
    prefix = col.split('_')[0]
    if prefix in prefixes:
        dd[prefix] += 1
        new_cols.append(col + str(dd[prefix]))
    else:
        new_cols.append(col)

If your split character _ is not in any of your data fields:

new_cols = []
dd = defaultdict(int)
for col in df.columns:
    if '_' in col:
        prefix = col.split('_')[0]
        dd[prefix] += 1
        new_cols.append(col + str(dd[prefix]))
    else:
        new_cols.append(col)

df.columns = new_cols

you can use rename such as:

l_word = ['Data','Text']
df = df.rename(columns={ col:col+str(i+1) 
                         for word in l_word 
                         for i, col in enumerate(df.filter(like=word))})