Inserting string to string regularly ( 1234567891234 -> 1,2345,6789,1234 )

Question

How to insert ' # ' for each n index from backward?
ex) n=4
evil = '01234567891234oooooooooooooooo321'
to
stan = '0#1234#5678#9123#4ooo#oooo#oooo#oooo#o321'
i tried using list with for,if statement, got stuck. something shameful like this

a = 1234567891234
b = [ a[-i] for i in range(1,len(a)+1)]
for i in range(len(b)):
    c += b[i]
    if i%4==0: #stuck
        c += ',' 
c.reverse()

What is the optimum way?

Welcome to Stack Overflow. Given https://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-sized-chunks and https://stackoverflow.com/questions/1876191/what-exactly-does-the-join-method-do, can you see how to solve the problem? — Karl Knechtel, Apr 28 '22 at 15:32
Do you actually want to insert `#`, or `,`? *Why* do you want to do this? If it's to format a number for some locale, does https://stackoverflow.com/questions/1823058/how-to-print-number-with-commas-as-thousands-separators help? — Karl Knechtel, Apr 28 '22 at 15:35
Thank you, i was just curious about how f'{value:,}' works like — bbiot426, Apr 28 '22 at 15:44

score 2 · Answer 1 · answered Apr 28 '22 at 15:09

You might use a pattern asserting optional repetitions of 4 characters to the right, and replace that position with #

import re

pattern = r"(?=(?:.{4})*$)"
s = "01234567891234oooooooooooooooo321"
print(re.sub(pattern, "#", s))

Output

0#1234#5678#9123#4ooo#oooo#oooo#oooo#o321#

Python demo

Z Li · Accepted Answer · 2022-04-28T15:10:12.383

1

cut the string into chunks (backwards) and then concat them using the seperator

evil = '01234567891234oooooooooooooooo321'
l = 4
sep = '#'
sep.join([evil[max(i-l,0):i] for i in range(len(evil), 0, -l)][::-1])

'0#1234#5678#9123#4ooo#oooo#oooo#oooo#o321'

edited Apr 28 '22 at 15:10

answered Apr 28 '22 at 15:04

Z Li

4,133
1
4
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This isn't the desired result? – Nin17 Apr 28 '22 at 15:05
the request is to do it backward. A slight modification on the indexes and it will work :) – Apr 28 '22 at 15:05
did not see the reverse thing. updated. Thanks! – Z Li Apr 28 '22 at 15:10

Nin17 · Answer 3 · 2022-04-28T15:27:54.470

0

You can do it like this:

evil = '01234567891234oooooooooooooooo321'
''.join(j if i%4  else f'#{j}' for i, j in enumerate(evil[::-1]))[::-1][:-1]

Output:

'0#1234#5678#9123#4ooo#oooo#oooo#oooo#o321'

edited Apr 28 '22 at 15:27

answered Apr 28 '22 at 15:06

Nin17

2,821
2
4
14

score 0 · Answer 4 · answered Apr 28 '22 at 15:07

chunks function as in this answer

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in range(0, len(lst), n):
        yield lst[i:i + n]

evil = '01234567891234oooooooooooooooo321'
n = 4
stan = "#".join(chunks(evil[::-1], n))[::-1]
print(stan) # Output: 0#1234#5678#9123#4ooo#oooo#oooo#oooo#o321

Input string is reversed ([::-1]), split into chunks, joined by "#" and then reversed back again. (It's possible to skip reverses if you calculate how many characters there will be in the first set of characters)

score 0 · Answer 5 · answered Apr 28 '22 at 15:08

A naive solution would be using parts of evil string:

evil = '01234567891234oooooooooooooooo321'
n = 4
start = len(evil) % n
insert = '#'

stan = evil[:start] + insert
for i in range(start, len(evil) - n, n):
    stan += evil[i:i+n] + insert
stan += evil[-n:]

score 0 · Answer 6 · answered Apr 28 '22 at 15:09

For this, I would go backwards through your string evil by reversing the string and iterating through it in a for loop. Then I set a count variable to keep track of how many loops it's done, and reset to 0 when it equals 4. All of this looks like the below:

count = 0
for char in evil[::-1]:
    if count == 4:
        count = 0
    count += 1

You can then establish a new empty string (new_str), and append each character of evil to, each time checking if count is 4, and adding a # to the string as well before resetting the count. Full code:

count = 0
new_str = ''
for char in evil[::-1]:
    if count == 4:
        new_str += '#'
        count = 0
    count += 1
    new_str += char

This will produce the new string reversed, so you need to reverse it again to get the desired result:

new_str = new_str[::-1]

Output:

'123o#oooo#oooo#oooo#ooo4#3219#8765#4321#0'

cards · Answer 7 · 2022-04-28T17:57:01.390

An exact method: use divmod to get the reminder and quotient of the string when divided in "blocks" of size 4 then slice.

evil = '01234567891234oooooooooooooooo321'

size = 4
q, r = divmod(len(evil), size)
sep = '#'
stan = f"{evil[:r]}{sep}{sep.join(evil[r+i*size: r+(i+1)*size] for i in range(q))}"

print(stan)

Remark: if the length of the string is a multiple of the block's size the new string will start with sep. Assumed as default behavior since lake of explanation

Inserting string to string regularly ( 1234567891234 -> 1,2345,6789,1234 )

7 Answers7