Creating a list consisting of elements popped from another list (Python)

Question

Here's my code:

txt = '''food: bacon, eggs, toast

drinks: orange juice, coffee'''

groups = txt.split('\n\n')
for group in groups:
    nameslist = group.split(': ').pop(0)

print(groups)
print(nameslist)

The output for print(groups) is as expected: ['food: bacon, eggs, toast', 'drinks: orange juice, coffee']

I'm trying to get a list ['food', 'drinks'] as the output for print(nameslist), but Python gives me 'drinks' as the output, as if it only grabs from the second element rather than iterating.

Answer 1

The problem with your approach is that nameslist is never initialized, and is reset at each iteration, you probably wanted to do:

txt = '''food: bacon, eggs, toast

drinks: orange juice, coffee'''

nameslist = [] # initialize namelist
groups = txt.split('\n\n')
for group in groups:
    nameslist.append(group.split(': ').pop(0)) # append instead of overwriting

print(groups)
print(nameslist)  # ['food', 'drinks']

Additionally, you can throw away the loop and use a list comprehension to achieve that:

namelist = [x.split(':')[0] for x in groups]  # ['food', 'drinks']

Answer 2

The way you currently have it set up, you are not adding anything to a list. You are simply setting the variable nameslist as the last pop. Try this:

nameslist = []  # Define an empty list
groups = txt.split('\n\n')
for group in groups:
    nameslist.append(group.split(': ').pop(0)) #Add popped element to the list