How do I extract a certain substring using recursion?

Question

I'm trying to write a method that takes a substring consisting of two parts:

name: idnumber

Yes the name and the id are separated by a colon (:), for example:

A:1313
B:4141

The method should return A for the first example and B for the second but it also should accept special cases, for example:

A$%#@$@#!__231:2 
A:::$@$@:::!$@:4

The above code should return A$%#@$@#!__231 for the first line and A:::$@$@:::!$@ for the second line.

I tried to do this recursively and here is my code:

j=0
def name(input):
  global j
  item = ''
  item = input[j:]
  if ':' in item:
    return name(input[j+1:])
  else:
    return input[:j - 1]

This code returns nothing and I'm not sure why it's not working.

Any help is appreciated.

Split won't work in a special case where the input is A$@$@:::::$@$!:4, here I have multiple ":", splitting by ":" results in a faulty output. — Tino Mosso
– Tino Mosso, Commented Apr 24, 2019 at 15:18

brunns · Accepted Answer · 2019-04-24 15:23:49Z

1

Sounds like a job for rpartition():

input = ["A:1313",
         "B:4141",
         "A$%#@$@#!__231:2",
         "A:::$@$@:::!$@:4",
        ]

for item in input:
    print(item.rpartition(":")[0])

Gives:

A
B
A$%#@$@#!__231
A:::$@$@:::!$@

answered Apr 24, 2019 at 15:23

brunns

2,7741 gold badge15 silver badges25 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Daweo · Accepted Answer · 2019-04-24 15:24:25Z

0

I would do it via split join combination, i.e. as follows:

example1 = 'A$%#@$@#!__231:2'
example2 = 'A:::$@$@:::!$@:4'
def getname(x):
    return ':'.join(x.split(':')[:-1])
print(getname(example1)) # A$%#@$@#!__231
print(getname(example2)) # A:::$@$@:::!$@

Function that I created gets string, split it at :, discards last element (idnumber in your case) and then join it together with :. Note that it will return empty str if there is not : inside input.

answered Apr 24, 2019 at 15:24

Daweo

38.2k3 gold badges17 silver badges32 bronze badges

Comments

Wang Liang · Accepted Answer · 2019-04-24 15:26:16Z

0

From your code, 'j' is not neccessory

This is works with your request

r = 'A:::$@$@:::!$@:4'
pos = r.rfind(':') #https://stackoverflow.com/questions/9572490
print(r[:pos])

edited Apr 24, 2019 at 15:26

answered Apr 24, 2019 at 15:16

Wang Liang

4,4527 gold badges26 silver badges50 bronze badges

Comments

Malekai · Accepted Answer · 2019-04-24 15:32:54Z

0

You could simply ignore values with multiple colons, then use the split() method as you normally would:

def splitter(id_strings = []):
  return [id if "::" in id else id.split(":")[0] for id in id_strings]

You could also convert it to a lambda method:

splitter = lambda id_strings = []: [id if "::" in id else id.split(":")[0] for id in id_strings]

Usage for splitter() method:

print(splitter(["A:1313", "B:1414", "A$%#@$@#!__231", "A:::$@$@:::!$@"]))

Expected output for both methods:

["A", "B", "A$%#@$@#!__231", "A:::$@$@:::!$@"]

Good luck.

edited Apr 24, 2019 at 15:32

answered Apr 24, 2019 at 15:27

Malekai

5,0706 gold badges34 silver badges65 bronze badges

Comments

loretoparisi · Accepted Answer · 2019-04-24 16:09:42Z

0

Using a regex with search

import re
samples=["A$%#@$@#!__231:2","A:::$@$@:::!$@:4","A:1313","B:4141"]
res=[ re.search("((.*( by )?.*):(.*))",s).group(2,4) for s in samples]
print(res)

and you get (key, value) dict

[('A$%#@$@#!__231', '2'), ('A:::$@$@:::!$@', '4'), ('A', '1313'), ('B', '4141')]

Repl: https://repl.it/@loretoparisi/How-do-I-extract-a-certain-substring-using-recursion

edited Apr 24, 2019 at 16:09

answered Apr 24, 2019 at 16:01

loretoparisi

16.3k12 gold badges113 silver badges159 bronze badges

Collectives™ on Stack Overflow

How do I extract a certain substring using recursion?

5 Answers 5

Comments

Comments

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related