Python/Pandas dataframe - return column name

Question

Is there a way to return the name/header of a column into a string in a pandas dataframe? I want to work with a row of data which has the same prefix. The dataframe header looks like this:

col_00 | col_01 | ... | col_51 | bc_00 | cd_00 | cd_01 | ... | cd_90

I'd like to apply a function to each row, but only from col_00 to col_51 and to cd_00 to cd_90 separately. To do this, I thought I'd collect the column names into a list, fe. to_work_with would be the list of columns starting with the prefix 'col', apply the function to df[to_work_with]. Then I'd change the to_work_with and it would contain the list of columns starting with the 'cd' prefix et cetera. But I don't know how to iterate through the column names.

So basically, the thing I'm looking for is this function:

to_work_with = column names in the df that start with "thisstring"

How can I do that? Thank you!

Answer 1

You can use boolean indexing with str.startswith:

cols = df.columns[df.columns.str.startswith('cd')]
print (cols)
Index(['cd_00', 'cd_01', 'cd_02', 'cd_90'], dtype='object')

Sample:

print (df)
   col_00  col_01  col_02  col_51  bc_00  cd_00  cd_01  cd_02  cd_90
0       1       2       3       4      5      6      7      8      9

cols = df.columns[df.columns.str.startswith('cd')]
print (cols)
Index(['cd_00', 'cd_01', 'cd_02', 'cd_90'], dtype='object')

#if want apply some function for filtered columns only
def f(x):
    return x + 1

df[cols] = df[cols].apply(f)    
print (df)
   col_00  col_01  col_02  col_51  bc_00  cd_00  cd_01  cd_02  cd_90
0       1       2       3       4      5      7      8      9     10

Another solution with list comprehension:

cols = [col for col in df.columns if col.startswith("cd")]
print (cols)
['cd_00', 'cd_01', 'cd_02', 'cd_90']