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
– taesu, Commented Aug 25, 2015 at 19:18
You have to use regex. Anywhere where a value is not static requires wildcard. Which entails regex — FirebladeDan
– FirebladeDan, Commented Aug 25, 2015 at 19:18
Actually I want to avoid regex so I'm wondering is there any other way to do it? — Enkri_
– Enkri_, Commented Aug 25, 2015 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. regex101.com — FirebladeDan
– FirebladeDan, Commented Aug 25, 2015 at 19:20

TigerhawkT3 · Accepted Answer · 2015-08-25 19:32:18Z

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, 2015 at 19:32

answered Aug 25, 2015 at 19:22

TigerhawkT3

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5 Comments

TigerhawkT3 Over a year ago

Edited to handle multiple digits after underscore (with another partition()).

Enkri_ Over a year ago

@TigerhawkT3 still have one question tho, what was the problem with the method I used in the original post?

TigerhawkT3 Over a year ago

line[line.find('_') + 1:line.find(' ')] evaluated to just '1', so it replaced the first '1' that it found with 'Normal'.

Enkri_ Over a year ago

@TigerhawkT3 Thanks so much

TigerhawkT3 Over a year ago

You could fix it by removing the + 1 and adding an underscore before Normal: line.replace(line[line.find('_') :line.find(' ')], '_Normal', 1).

Community · Accepted Answer · 2020-06-20 09:12:55Z

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, 2020 at 9:12

CommunityBot

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answered Aug 25, 2015 at 19:20

taesu

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9 Comments

FirebladeDan Over a year ago

How do you know there will always be a space after? You don't sir

taesu Over a year ago

Beceause the OP is expecting that?

FirebladeDan Over a year ago

Ha hopefully! Then yes it can be this easy

taesu Over a year ago

the string he provided is obviously 'whatever date'. How do you know if there will be space, or no space? You don't sir.

Enkri_ Over a year ago

Thanks guys, the data come with a space after

|

Padraic Cunningham · Accepted Answer · 2015-08-25 19:59:45Z

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, 2015 at 19:59

answered Aug 25, 2015 at 19:18

Padraic Cunningham

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1 Comment

TigerhawkT3 Over a year ago

OP is specifically asking for a non-regex solution.

Bob Dylan · Accepted Answer · 2015-08-25 19:26:40Z

1

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

answered Aug 25, 2015 at 19:26

Bob Dylan

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