How do i change the last 3 characters of a string in python?

text = "ABCD"

size = len(text)     # text length
replacement = "XYZ"  # replace with this

text = text.replace(text[size - 3:], replacement)

print(text)  # AXYZ

text = "ABCD"

print("String before:", text)

size = len(text)     # text length
replacement = "XYZ"  # replace with this

text = text.replace(text[size - 3:], replacement)

print("String after: ", text)  # AXYZ

String before: ABCD
String after:  AXYZ

I'm trying to remove the last 3 characters from a string in Python, I don't know what these characters are so I can't use rstrip, I also need to remove any white space and convert to upper-case.

An example would be:

foo = "Bs12 3ab"
foo.replace(" ", "").rstrip(foo[-3:]).upper()

This works and gives me "BS12" which is what I want, however if the last 4th & 3rd characters are the same I lose both, e.g. if foo = "BS11 1AA" I just get "BS".

Examples of foo could be:

BS1 1AB
bs11ab
BS111ab

The string could be 6 or 7 characters and I need to drop the last 3 (assuming no white space).

mkrieger1

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asked Nov 25, 2009 at 17:14

2

Removing any and all whitespace:

foo = ''.join(foo.split())

Removing last three characters:

foo = foo[:-3]

Converting to capital letters:

foo = foo.upper()

All of that code in one line:

foo = ''.join(foo.split())[:-3].upper()

answered Nov 25, 2009 at 17:23

Noctis SkytowerNoctis Skytower

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1

It doesn't work as you expect because strip is character based. You need to do this instead:

foo = foo.replace(' ', '')[:-3].upper()

answered Nov 25, 2009 at 17:17

Nadia AlramliNadia Alramli

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1

>>> foo = "Bs12 3ab"
>>> foo[:-3]
'Bs12 '
>>> foo[:-3].strip()
'Bs12'
>>> foo[:-3].strip().replace(" ","")
'Bs12'
>>> foo[:-3].strip().replace(" ","").upper()
'BS12'

answered Nov 26, 2009 at 1:15

ghostdog74ghostdog74

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2

You might have misunderstood rstrip slightly, it strips not a string but any character in the string you specify.

Like this:

>>> text = "xxxxcbaabc"
>>> text.rstrip("abc")
'xxxx'

So instead, just use

text = text[:-3] 

(after replacing whitespace with nothing)

answered Nov 25, 2009 at 17:22

Mattias NilssonMattias Nilsson

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>>> foo = 'BS1 1AB'
>>> foo.replace(" ", "").rstrip()[:-3].upper()
'BS1'

answered Nov 25, 2009 at 17:18

SilentGhostSilentGhost

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I try to avoid regular expressions, but this appears to work:

string = re.sub("\s","",(string.lower()))[:-3]

answered Nov 25, 2009 at 17:29

krs1krs1

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2

1. split
2. slice
3. concentrate

This is a good workout for beginners and it's easy to achieve.

Another advanced method is a function like this:

def trim(s):
    return trim(s[slice])

And for this question, you just want to remove the last characters, so you can write like this:

def trim(s):
    return s[ : -3] 

I think you are over to care about what those three characters are, so you lost. You just want to remove last three, nevertheless who they are!

If you want to remove some specific characters, you can add some if judgements:

def trim(s):
    if [conditions]:   ### for some cases, I recommend using isinstance().
        return trim(s[slice])

answered Nov 28, 2018 at 1:50

What's wrong with this?

foo.replace(" ", "")[:-3].upper()

answered Nov 25, 2009 at 17:18

abyxabyx

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Aren't you performing the operations in the wrong order? You requirement seems to be foo[:-3].replace(" ", "").upper()

answered Nov 25, 2009 at 17:26

AndreaGAndreaG

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1

It some what depends on your definition of whitespace. I would generally call whitespace to be spaces, tabs, line breaks and carriage returns. If this is your definition you want to use a regex with \s to replace all whitespace charactors:

import re

def myCleaner(foo):
    print 'dirty: ', foo
    foo = re.sub(r'\s', '', foo)
    foo = foo[:-3]
    foo = foo.upper()
    print 'clean:', foo
    print

myCleaner("BS1 1AB")
myCleaner("bs11ab")
myCleaner("BS111ab")

answered Nov 25, 2009 at 17:33

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