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Home/ Questions/Q 593625
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T15:51:57+00:00 2026-05-13T15:51:57+00:00

I’m looking for a nice way to process a list where some elements need

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I’m looking for a “nice” way to process a list where some elements need to be expanded into more elements (only once, no expansion on the results).

Standard iterative way would be to do:

i=0
while i < len(l):
   if needs_expanding(l[i]):
      new_is = expand(l[i])
      l[i:i] = new_is
      i += len(new_is)
   else:
      i += 1

which is pretty ugly. I could rewrite the contents into a new list with:

nl = []
for x in l:
   if needs_expanding(x):
      nl += expand(x)
   else:
      nl.append(x)

But they both seem too long. Or I could simply do 2 passes and flatten the list later:

flatten(expand(x) if needs_expanding(x) else x for x in l)
# or
def try_expanding(x)....
flatten(try_expanding(x) for x in l)

but this doesn’t feel “right” either.

Are there any other clear ways of doing this?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T15:51:58+00:00Added an answer on May 13, 2026 at 3:51 pm

    If you do not need random access in the list you are generating, you could also use write a generator.

    def iter_new_list(old_list):    
        for x in old_list:
           if needs_expanding(x):
               for y in expand(x):
                   yield y
           else:
               yield x
    
    new_list = list(iter_new_list(old_list))
    

    This is functionally equivalent to your second example, but it might be more readable in your real-world situation.

    Also, Python coding standards forbid the use of lowercase-L as a variable name, as it is nearly indistinguishable from the numeral one.

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