specific characters printing with Python

Question

given a string as shown below,

"[xyx],[abc].[cfd],[abc].[dgr],[abc]"

how to print it like shown below ?

1.[xyz]
2.[cfd]
3.[dgr]

The original string will always maintain the above-mentioned format.

Answer 1

I did not realize you had periods and commas... that adds a bit of trickery. You have to split on the periods too

I would use something like this...

list_to_parse = "[xyx],[abc].[cfd],[abc].[dgr],[abc]"

count = 0
for  i in list_to_parse.split('.'):
    for j in i.split(','):
        string = str(count + 1) + "." + j
        if string:
            count += 1
            print(string)
        string = None

Another option is split on the left bracket, and then just re-add it with enumerate - then strip commas and periods - this method is also probably a tiny bit faster, as it's not a loop inside a loop

list_to_parse = "[xyx],[abc].[cfd],[abc].[dgr],[abc]"

for index, i in enumerate(list.split('[')):
    if i:
        print(str(index) + ".[" + i.rstrip(',.'))

also strip is really "what characters to remove" not a specific pattern. so you can add any characters you want removed from the right, and it will work through the list until it hits a character it can't remove. there is also lstrip() and strip()

string manipulation can always get tricky, so pay attention. as this will output a blank first object, so index zero isn't printed etc... always practice and learn your needs :D

Answer 2

You can use split() function:

a = "[xyx],[abc].[cfd],[abc].[dgr],[abc]"

desired_strings = [i.split(',')[0] for i in a.split('.')]

for i,string in enumerate(desired_strings):
    print(f"{i+1}.{string}")

Answer 3

You can use RegEx:

import regex as re
pattern=r"(\[[a-zA-Z]*\])\,\[[a-zA-Z]*\]\.?"
results=re.findall(pattern, '[xyx],[abc].[cfd],[abc].[dgr],[abc]')
print(results)

Answer 4

This is just a fun way to solve it:

lst = "[xyx],[abc].[cfd],[abc].[dgr],[abc]"

count = 1
var = 1
for char in range(0, len(lst), 6):
    if var % 2:
        print(f"{count}.{lst[char:char + 5]}")
        count += 1
    var += 1

output:

1.[xyx]
2.[cfd]
3.[dgr]

explanation : "[" appears in these indexes: 0, 6, 12, etc. var is for skipping the next pair. count is the counting variable.

---

Here we can squeeze the above code using list comprehension and slicing instead of those flag variables. It's now more Pythonic:

lst = "[xyx],[abc].[cfd],[abc].[dgr],[abc]"

lst = [lst[i:i+5] for i in range(0, len(lst), 6)][::2]

res = (f"{i}.{item}" for i, item in enumerate(lst, 1))

print("\n".join(res))

Answer 5

Using re.findall:

import re

s = "[xyx],[abc].[cfd],[abc].[dgr],[abc]"

print('\n'.join(f'{i+1}.{x}' for i,x in
                enumerate(re.findall(r'(\[[^]]+\])(?=,)', s))))

Output:

1.[xyx]
2.[cfd]
3.[dgr]