Backward Concatenation of Column values(Strings) in Python pandas

I trying to concatenate strings by decremented columns and assuming the table is as below. I'm creating fixed number (5) of NEW columns & want every column to have string concatenated by non-empty strings. Let say for the row-1, first concatenation will be STR_1;STR_2;STR_3 and the second concatenation should be STR_1;STR_2 and third concatenation should be STR_1 & the rest of the cells should remain empty.

row-1

STR_1;STR_2;STR_3

STR_1;STR_2

STR_1

MAJOR_BRAND | SUB_SEGMENT | PACKAGE_TYPE | PACKAGE | SIZE | FLAVOR | VOLUME ------------------------------------------------------------------------------------- STR_1 | STR_2 | STR_3 | | | | STR_1 | STR_2 | STR_3 | | | | STR_7 STR_1 | | | STR_4 | STR_5 | STR_6 | STR_7 STR_1 | | STR_3 | | | STR_6 | STR_7 STR_1 | | STR_3 | | | STR_6 |

Output should looks like

Kind help will be greatly appreciated !!!

So the concatenation operation should go from left to right over all the columns of the dataframe, yielding each additional concatenation of one more string, and stopping the process at the first empty string found in a column? Also, regarding the second row, when you say that the second concatenation should be STR_1;STR_2, why is it not STR_1;STR_2;STR_3? EDIT: I got it, by "first concatenation" you mean the last concatenation, you got the operations in reverse order. You will need to recursively .apply() a function on the dataframe, and this function should check for empty strings first.

– Daniel R.
Sep 11 '18 at 8:32

the second concatenation should be STR_1;STR_2

2 Answers
2

Quick and dirty. Note f-strings require Python 3.6+.

res = pd.DataFrame(columns=[f'STR_JOIN_i' for i in range(1, 6)]) res['STR_JOIN_1'] = df.apply(lambda x: ';'.join(filter(None, x)), axis=1) for i in range(2, 6): res[f'STR_JOIN_i'] = res[f'STR_JOIN_i-1'].str.split(';').str[:-1].str.join(';')

Result:

print(res) STR_JOIN_1 STR_JOIN_2 STR_JOIN_3 0 STR_1;STR_2;STR_3 STR_1;STR_2 STR_1 1 STR_1;STR_2;STR_3;STR_7 STR_1;STR_2;STR_3 STR_1;STR_2 2 STR_1;STR_4;STR_5;STR_6;STR_7 STR_1;STR_4;STR_5;STR_6 STR_1;STR_4;STR_5 3 STR_1;STR_3;STR_6;STR_7 STR_1;STR_3;STR_6 STR_1;STR_3 4 STR_1;STR_3;STR_6 STR_1;STR_3 STR_1 STR_JOIN_4 STR_JOIN_5 0 1 STR_1 2 STR_1;STR_4 STR_1 3 STR_1 4

Another Approach..i guess

new_data = pd.DataFrame(columns=['STR_JOIN_'.format(i) for i in range(1, len(initial_data.columns))]) loc = 0 for i in range(len(initial_data)):#initial_data being the input temp_list = [ x for x in list(initial_data.loc[i].values) if x ] new_data.loc[loc] = ['']*(len(initial_data.columns)-1) for j in range(len(temp_list)): new_data['STR_JOIN_'.format(j+1)][loc] = ';'.join(data[:len(data)-j]) loc+=1

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