Remove element of list until getting to the first empty one in Python

Question

I have this list of strings

list = ['1', '2', '3', '4', '', '    5', '    ', '    6', '', '']

and I want to get every item after the first empty string to get this result

list = ['    5', '    ', '    6', '', '']

note that I want to leave the empty strings that comes after

I wrote this function:

def get_message_text(list):
    for i, item in enumerate(list):
        del list[i]
        if item == '':
            break
    return list

but I am getting this wrong results for no reason I know:

['2', '4', '    5', '    ', '    6', '', '']

any help ?

Jean-François Fabre, I can see now what I did wrong, thanks — Jean Khechfe
– Jean Khechfe, Commented Nov 6, 2020 at 22:30

ti7 · Accepted Answer · 2020-11-06 22:15:16Z

5

Just find the index of the first empty string and slice on it:

def get_message_text(lst):
    try:
        return lst[lst.index("") + 1:]
    except ValueError:  # '' is not in list
        return [] # if there's no empty string then don't return anything

edited Nov 6, 2020 at 22:15

ti7

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answered Nov 6, 2020 at 22:07

Aplet123

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Comments

rioV8 · Accepted Answer · 2020-11-06 22:24:29Z

0

Construct a generator that drops while empty string not yet found.

Use list(get_message_text(lst)) if you need a list,

otherwise use generator in for expression

for e in get_message_text(lst):

def get_message_text(lst):
    output = False
    for e in lst:
        if not output:
            if e == '': output = True
            continue
        yield e

answered Nov 6, 2020 at 22:24

rioV8

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