Return only index zero in a list comprehension

Question

To create a list with the values that I retrieve from a spreadsheet, I do it like this:

result = {'range': 'Winning_Margin!G2:O1000', 'majorDimension': 'ROWS', 
          'values': [['10148734', 'CD Olimpia', 'CD Vida', '1', 
                      '71', '85', '14.00', '1.40', '2-0'], 
                     ['', '', '', '', '', '', '', '', ''], 
                     ['10024627', 'Sporting Kansas City', 'FC Dallas', 
                      '2', '35', '1', '-34.00', '2.88']]}

for a in result["values"]:
    a.extend([''] * (max(map(len, result["values"])) - len(a)))

spreadsheets_match_id = [item for sublist in result["values"] for item in (sublist or ['']) if sublist[-1] != '']

This way returns the entire index:

['10148734', 'CD Olimpia', 'CD Vida', '1', '71', '85', '14.00', '1.40', '2-0']

But my need is just the first index of each list of list, in this case the result is:

['10148734']

Tried to make each item become item[0] but when trying to do that, it either says that item doesn't exist or returns the first letter of each of the list's indexes within the list like ['1', 'C', 'C', '1', '7', '8', '1', '1', '2'].

Answer 1

Edit: You can check if the last element of the sublist is '' or not and keep the first elements of the ones that aren't:

out = [sublist[0] for sublist in result["values"] if sublist[-1] !='']

Old answer:

IIUC, you want to get the first element of a sublist that is not '', right? Then you can create a single element list in a list comprehension and iterate over it in a nested loop to check if it is '' or not:

out = [item for sublist in result["values"] for item in [sublist[0]] if item !='']

Output:

['10148734', '10024627']