Remove last 3 characters of a string
Question:
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).
Answers:
It doesn’t work as you expect because strip is character based. You need to do this instead:
foo = foo.replace(' ', '')[:-3].upper()
What’s wrong with this?
foo.replace(" ", "")[:-3].upper()
>>> foo = 'BS1 1AB'
>>> foo.replace(" ", "").rstrip()[:-3].upper()
'BS1'
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)
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()
Aren’t you performing the operations in the wrong order? You requirement seems to be foo[:-3].replace(" ", "").upper()
I try to avoid regular expressions, but this appears to work:
string = re.sub("s","",(string.lower()))[:-3]
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")
>>> foo = "Bs12 3ab"
>>> foo[:-3]
'Bs12 '
>>> foo[:-3].strip()
'Bs12'
>>> foo[:-3].strip().replace(" ","")
'Bs12'
>>> foo[:-3].strip().replace(" ","").upper()
'BS12'
split sliceconcentrate
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])
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).
It doesn’t work as you expect because strip is character based. You need to do this instead:
foo = foo.replace(' ', '')[:-3].upper()
What’s wrong with this?
foo.replace(" ", "")[:-3].upper()
>>> foo = 'BS1 1AB'
>>> foo.replace(" ", "").rstrip()[:-3].upper()
'BS1'
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)
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()
Aren’t you performing the operations in the wrong order? You requirement seems to be foo[:-3].replace(" ", "").upper()
I try to avoid regular expressions, but this appears to work:
string = re.sub("s","",(string.lower()))[:-3]
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")
>>> foo = "Bs12 3ab"
>>> foo[:-3]
'Bs12 '
>>> foo[:-3].strip()
'Bs12'
>>> foo[:-3].strip().replace(" ","")
'Bs12'
>>> foo[:-3].strip().replace(" ","").upper()
'BS12'
splitsliceconcentrate
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])