How can I add a if statement that goes back in string if there is a number, and to stop if there is-CodePudding

So let's say I have a string, like this:

string = '12345 67890'

I want to go back to search for that plus, and add a letter 'H', so the end result would be like this:

string = '12345 h67890

Also, what if I want to get the latest , so if there is two 's, I want to get the last ?

Any help would be appreciated!

CodePudding user response：

You could use reverse split to split on the last instance of , then join with H:

>>> ' H'.join('123 456 789'.rsplit(' ',1))
'123 456 H789'

CodePudding user response：

Convert into list, iterate through list, when you find the plus add an 'h' into the next index.

string = '12345 67890'
stringList = list(string)
i = 0
newString = ''

for i in range(len(stringList)):
    if stringList[i] == ' ':
          stringList.insert(i 1,'h')
for letter in stringList:
    newString  = letter
print(newString)

CodePudding user response：

Since you asked how to do it with if statements:

i = -1
for i in range(len(my_str)):
    if my_str[i] == ' ':
        plus = i 1 # <- will update every time you see a   symbol
if i != -1:
    my_str = my_str[:i]   'h'   my_str[i:]

Alternatively, you can search backwards:

i = -1
for i in reversed(range(len(my_str))):
    if my_str[i] == ' ':
        plus = i 1 # <- will update every time you see a   symbol
        break
if i != -1:
    my_str = my_str[:i]   'h'   my_str[i:]

As others have mentioned you can use built-in functions like find and rfind. My personal choice is to refer to regular expressions for this type of thing:

import re
my_str = re.sub('(.*\ )(.*)',r"\1h\2", my_str))