python replace specific elements in a string

Question

I have a string '1472_1 2014-6-19' and I want to replace whatever number is after the underscore(in this case number one) with the word 'Normal', what I did was to find the index of the element that I want replaced:

print line.replace(line[line.find('_') + 1:line.find(' ')],'Normal', 1)

But instead of getting '1472_Normal 2014-6-19', I got 'Normal472_1 2014-6-19' It seems like my code replace the first 1 instead of the one after the underscore.

I have read this post: Replacing specific words in a string (Python) but still wondering is there a way to specify which element to be replaced instead of using regex?

And to be clear the number after underscore could be any number from 0 to 237

are you okay with using regex or you want to keep it as string operations? — taesu, Aug 25 '15 at 19:18
You have to use regex. Anywhere where a value is not static requires wildcard. Which entails regex — FirebladeDan, Aug 25 '15 at 19:18
Actually I want to avoid regex so I'm wondering is there any other way to do it? — Enkri_, Aug 25 '15 at 19:19
@Enkri_ - It's okay if you don't understand it. I still don't understand some of them that get quite complicated. However, so you have some background this tool should become your best friend. https://regex101.com/ — FirebladeDan, Aug 25 '15 at 19:20
Might the number (in this case `1`) be two digits, or is it always only one digit? — TigerhawkT3, Aug 25 '15 at 19:28
`b = (a.split('_')[0]+'_Normal '+a.split(' ')[1])` don't work with `temp` values ! — dsgdfg, Aug 25 '15 at 19:29

TigerhawkT3 · Accepted Answer · 2015-08-25T19:32:18.510

3

You could use str.partition():

first, delimiter, second = line.partition('_')
a, s, b = second.partition(' ')
print first + delimiter + 'Normal' + s + b

edited Aug 25 '15 at 19:32

answered Aug 25 '15 at 19:22

TigerhawkT3

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Edited to handle multiple digits after underscore (with another `partition()`). – TigerhawkT3 Aug 25 '15 at 19:32
@TigerhawkT3 still have one question tho, what was the problem with the method I used in the original post? – Enkri_ Aug 25 '15 at 19:48
`line[line.find('_') + 1:line.find(' ')]` evaluated to just `'1'`, so it replaced the first `'1'` that it found with `'Normal'`. – TigerhawkT3 Aug 25 '15 at 19:51
@TigerhawkT3 Thanks so much – Enkri_ Aug 25 '15 at 19:53
You could fix it by removing the `+ 1` and adding an underscore before `Normal`: `line.replace(line[line.find('_') :line.find(' ')], '_Normal', 1)`. – TigerhawkT3 Aug 25 '15 at 19:53

Padraic Cunningham · Answer 2 · 2015-08-25T19:59:45.760

2

without regex if there is always a space after the digit which based on line.find(' ') there is, split once to divide the string and rstrip any digits:

s = '1472_1 2014-6-19'

a,b = s.split(None,1)

print("{}Normal {}".format(a.rstrip("0123456789"),b))) 
1472_Normal 2014-6-19

stripping will work for any amount of digits after the underscore.

With regex:

 import  re

s = '1472_1 2014-6-19'

print(re.sub("(?<=_)\d+","Normal",s))
1472_Normal 2014-6-19

Why your own code fails is because you have a 1 at the start of your string so you replace that not the one after the underscore

edited Aug 25 '15 at 19:59

answered Aug 25 '15 at 19:18

Padraic Cunningham

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OP is specifically asking for a non-regex solution. – TigerhawkT3 Aug 25 '15 at 19:19

score 2 · Answer 3 · edited Jun 20 '20 at 09:12

2

no regex

string = '1472_1 2014-6-19'
to_replace = string[string.find('_'):string.find(' ')]
string = string.replace(to_replace,'_Normal')
print string

edited Jun 20 '20 at 09:12

Community

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answered Aug 25 '15 at 19:20

taesu

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How do you know there will always be a space after? You don't sir – FirebladeDan Aug 25 '15 at 19:21
Beceause the OP is expecting that? – taesu Aug 25 '15 at 19:22
Ha hopefully! Then yes it can be this easy – FirebladeDan Aug 25 '15 at 19:22
the string he provided is obviously 'whatever date'. How do you know if there will be space, or no space? You don't sir. – taesu Aug 25 '15 at 19:23
1

Thanks guys, the data come with a space after – Enkri_ Aug 25 '15 at 19:28
This removes the underscore entirely. – TigerhawkT3 Aug 25 '15 at 19:29
@TigerhwakT3 try running the code in your head, or in your computa – taesu Aug 25 '15 at 19:33
Weird - the first time I ran it it removed the underscore, but this time it didn't. ...Oh I see, I didn't include the `_` before the `Normal` in the `replace()`. :) – TigerhawkT3 Aug 25 '15 at 19:34
your eyes did, maybe :) – taesu Aug 25 '15 at 19:37

score 1 · Answer 4 · answered Aug 25 '15 at 19:26

If you don't want to use RegEx, you could use slicing to isolate the section of the string to replace in. So for instance:

def replace_normal(s)
    ui = s.index('_')  # index of first underscore
    si = s.index(' ')  # index of first space
    return s[:ui+1] + 'Normal' + s[si:]

s1 = '1472_1 2014-6-19'
s2 = '1571_44 2014-7-24'

print replace_normal(s1)  # 1472_Normal 2014-6-19
print replace_normal(s2)  # 1571_Normal 2014-7-24

python replace specific elements in a string

4 Answers4

no regex