remove elements from list of strings while traversing

Question

how to remove elements from a list of strings while traversing through it. I have a list

list1 = ['', '$', '32,324', '$', '32', '$', '(35', ')', '$', '32,321']

i want to remove $ fro the list and if a ) or )% or % comes add that to the previous elemt of the list.
expected output is :

['', '32,324', '32', '(35)', '32,321']

what i have tried is

for j,element in enumerate(list1):
   if element == '%' or element == ")%" or element ==')':
      list1[j-1] = list1[j-1] + element
      list1.pop(j)
   elif element == '$':
      list1.pop(j)

but the output i am getting is

['', '32,324', '32', '(35)', '$', '32,321']

whis is not the expected output. Please help

This question is different from the suggested reference is, here I have to do a concatenation with the previous element if the current element is ),)% or %.

It will probably be easier for you to copy the elements you want to a new list instead of trying to filter it AND traverse it at the same time. — Calvin Godfrey, May 30 '19 at 18:20
@LanteDellarovere, that also should be concatenated to the first element and the last two should be removed, output: `["(3$)"] — Shijith, May 30 '19 at 18:29
if `["(3$")]` is the output I think you should re-formulate the rules, because there are incongruences — Lante Dellarovere, May 30 '19 at 18:34
Question have been put on hold few seconds before I've posted my answer. [Here](https://pastebin.com/Aah7HG1m) is code which works without clonning list, check it. — Olvin Roght, May 30 '19 at 18:36
@OlvinRoght If you believe you have a solution which is not already covered by any answer on the duplicates, you should post it there. Duplicate are meant to gather all answer at the same place, not to prevent new good answers. — Olivier Melançon, May 30 '19 at 19:06
the question isn't different. The problem is that you cannot use `pop` when iterating on a list. — Jean-François Fabre, May 30 '19 at 19:34

score 3 · Answer 1 · answered May 30 '19 at 18:24

Instead of attempting to remove and merge elements dynamically while iterating on the list, it will be much easier to make a new list based on the conditions here.

list1 = ['', '$', '32,324', '$', '32', '$', '(35', ')', '$', '32,321']

out = []
for element in list1:
    if element == "$":
        continue #skip if $ present
    elif element in ("%", ")", ")%"):
        out[-1] = out[-1] + element #merge with last element of out so far.
    else:
        out.append(element)

print(out)
#Output:
['', '32,324', '32', '(35)', '32,321']

`if element in ("%", ")", ")%"): ... elif element != "$": ...` should be enough. — Olvin Roght, May 30 '19 at 18:50

Ryan Drew · Accepted Answer · 2019-05-30T18:40:34.070

What Green Cloak Guy said is mostly correct. Editing the size of the list (by calling .pop()) is causing you to have an unexpected j value. To me, the easiest way to fix this problem while keeping your existing code is to simply not mutate your list, and build up a new one instead:

new_list = []
for j,element in enumerate(list1):
   if element == '%' or element == ")%" or element ==')':
      ret[len(ret) - 1] += element  # add at the end of the previous element
   elif element != '$':
      new_list.push(element)

However, I would encourage you to think about your edge cases here. What happens when a ')' is followed by another ')' in your list? This may be a special case in your if statement. Hope this helped!

Nice call about edge cases. – Scott Hunter May 30 '19 at 18:34 — Scott Hunter, May 30 '19 at 18:34

score 1 · Answer 3 · answered May 30 '19 at 18:32

I think this list comprehension works (haven't seen an example of how % is handled):

[ (a+b if b in (')',')%','%') else a) for a,b in zip(list1,list1[1:]+['']) if a not in ('$',')',')%','%')]

The idea is to:

make a list of pairings of elements and their successors
filter out elements that should be removed
add the successor as appropriate to those that we keep

remove elements from list of strings while traversing

3 Answers3