如何从字符串中删除空格/哈希并将其替换为特定长度的哈希？

Question

I would like to change the string of numbers in a way that all extra spaces and hashes will be removed and the output will have a proper string with hashes at the right length. It must be after every 3rd digit unless the remaining length of the string is 2 or 4.

The code I tried with is able to give me half of the result, I am not able to write a code where the remaining length of the string is 4 and then divide them into 2 with a hash in between. It divides remaining 4 also into 3 and 1. eg - 1234 must be 12-34 not 123-1.

def solution(s):
    # this will remove all extra spaces and hashes
    number = "".join(s.split()).replace('-','')
    return '-'.join(number[i:i+3] for i in range(0, len(number), 3))


print(solution("00-44  48 5555 8361")) # Output is 004-448-555-583-61
print(solution("0 - 22 1985--324")) # Output is 022-198-532-4 (not expected)

Output must be like below:

print(solution("00-44  48 5555 8361")) # Expected O/P - 004-448-555-583-61
print(solution("0 - 22 1985--324")) # Expected O/P -  022-198-53-24

Answer 1

Here you go:

def solution(s):
    number = "".join(s.split()).replace('-','')
    return formatnumber(number)

def formatnumber(number):
    if len(number) == 0:
        return ""
    if len(number) == 4:
        return number [:2] + "-" + number[2:]
    if len(number) == 2 or len(number) == 3:
        return number
    return number[:3]+ "-" +solution(number[3:])


print(solution("00-44  4 8 5555 8361")) # Output is 004-448-555-583-61
print(solution("0 - 22 1985--324")) # Output is 022-198-53-24 
print(solution("00-44  48 5555 83613")) # Output is 004-448-555-583-613

Answer 2

The reason your code does not work as intended I think is obvious--placing a hyphen between every 3rd character in a greedy fashion like you are doing will leave len(number)%3 (either 0, 1, or 2) characters at the end of the string. Based on your requirements, all you need to worry about is the case when there is exactly 1 remaining character:

1. "123-456-789" has zero remaining characters, and it's fine.
2. "123-456-789-0" has one remaining character, but should be "123-456-78-90 "
3. "123-456-789-01" has two remaining characters, and it's fine.
4. "123-456-789-012" has three remaining characters, but it's the same pattern as (1.), above, so it's fine.

This means all you have to do is deal specially with the case when there is one remaining character. Even better, all you have to do to correct it is swap the third-to-last and second-to-last characters, like so:

"123-456-789-0"  # this is wrong
#          ^^
#          ||    # swap these two, however, and you get...
"123-456-78-90"  # the correct output

So you can keep your original code and add a simple check at the end to make the swap. Unfortunately, strings are immutable in python, so you have to reconstruct a new string with the swapped characters. Doing this the long way:

def solution(s):
    # this will remove all extra spaces and hashes
    number = "".join(s.split()).replace('-','')
    dashed = '-'.join(number[i:i+3] for i in range(0, len(number), 3))
    if len(number)%3 == 1:
        dashed = dashed[:-3] + "-" + dashed[-3::2]
    return dashed

Answer 3

def solution(s):
    s.strip()
    number = "".join(s.split()).replace('-','')
    newN=''
    while(len(number)>4):
        newN=newN+number[0:3]+'-'
        number=number[3:]
    if len(number)==4:
        newN=newN+number[0:2]+'-'+number[2:]
    elif len(number)==2 or len(number)==3:
        newN=newN+number
    print(newN)

solution("00-44  48 5555 8361") # Output is 004-448-555-583-61
solution("0 - 22 1985--324") # 022-198-53-24

Answer 4

You can check condition for 2 and 4 manually.

import re
def solution(s):
    number = re.sub('[- ]','',s)
    ans = ''
    if len(number)%3 == 2:
        for i in range(0, len(number)-2,3):
            ans += number[i:i+3]
            ans += '-'
        ans += number[-2:]
    elif len(number)%3 == 1:
        for i in range(0, len(number)-4,3):
            ans += number[i:i+3]
            ans += '-'
        ans += (number[-4:-2] + '-' + number[-2:])
    else:
        for i in range(0, len(number)-4,3):
            ans += number[i:i+3]
            ans += '-'
        ans = ans[:-1] # this is required because we are appending `'-'` after each slice and last slice will also have dash at the end.
    return ans

print(solution("00-44  48 5555 8361")) # Expected O/P - 004-448-555-583-61
print(solution("0 - 22 1985--324")) # Expected O/P -  022-198-53-24   

Output:

004-448-555-583-61
022-198-53-24

UPD: This is oneliner solution for you.

def solution(s):
    ss = re.sub('[- ]','',s)
    l = len(ss)
    r = l - (4 if l%3==1 else l%3)
    return '-'.join(ss[i:i+3] for i in range(0,r,3)) + (('-' + ss[-4:-2]) if l%3 == 1 else '') + (('-' + ss[-2:]) if l%3!=0 else '' )

PS: this code can be reduced by few lines. But the main idea is same that you need to handle both 2 and 4 case.

Answer 5

Another way of solving this is to work with the length of the string. If the length - 4 is modulo 3, then format the last 4 characters specifically into -nn-nn format, otherwise treat the overall string as 'chunks of 3 with whatever remains at end of string'.

from textwrap import wrap

def solution(s):
    output_string = ""
    s = s.replace(' ', '').replace('-', '')
    if (len(s)-4) % 3 == 0:
        end_text = s[-4:][0:2] + '-' + s[-4:][2:]
        output_string = '-'.join(wrap(s[0:-4], 3)) + '-' + end_text
    else:
        output_string = '-'.join(wrap(s, 3)) 
    return output_string

print(solution("00-44  48 5555 8361"))
print(solution("0 - 22 1985--324"))