Python string is out of index

Question

I’m trying to delete ‘X’ and the char before X and here is my code

s = input()
for i in range(len(s)):
    if s[i] == 'X':
        s = s.replace("X", "")
        s = s.replace(s[i], "")

i got out of index error

    if s[i] == 'X':
IndexError: string index out of range

What’s the problem here and how to fix it?

Asked By: user12505209

||

Source

Answer 1

Solution

f= []
x = input()
for i in range(len(x)-1):
    if x[i+1]=='X' or x[i]=='X':
        continue
    else:
        f.append(x[i])

final = ''.join(f)

Answered By: Lovleen Kaur

Answer 2

You can use regular expressions:

import re
s = "aXbcXd" #example string
s = re.sub(".?X", "", s)
print(s)
#output: bd

Answered By: Amin Guermazi

Answer 3

Solution:

s = input()
ls = len(s)
x_pos = []
for i in range(ls):
    if s[i] == 'X':
        if i != 0:
            x_pos.append(i - 1)
            x_pos.append(i)
        else:
            x_pos.append(i)
new_s = ''
for j in range(ls):
    if j not in x_pos:
        new_s += s[j]

print(new_s)

Answered By: maghsood026

Answer 4

In my opinion, best way to address this problem would be using regular expressions. An alternative method could be this.

Solution:

s = input()
def removeX(s):
    pos = s.find('X')
    if pos == -1:
        print(s)
        return
    elif pos == 0:
        s = s[pos+1:]
    else:
        s = s[:pos-1]+s[pos+1:]
    removeX(s)

removeX(s)

Answered By: Onion_Hash

Answer 5

You can use while loop

s = input()
while True:
    a = s.find('X')
    if a==-1:
        break
    if a==0:
        s = s[1:]
    else:
        s = s[:a-1]+s[a+1:]

Answered By: Sakibulislam

Python string is out of index

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