How to replace some characters from the end of a string?
Question:
I want to replace characters at the end of a python string. I have this string:
s = "123123"
I want to replace the last 2 with x. Suppose there is a method called replace_last:
>>> replace_last(s, '2', 'x')
'1231x3'
Is there any built-in or easy method to do this?
It’s similar to python’s str.replace():
>>> s.replace('2', 'x', 1)
'1x3123'
But it’s from the end to beginning.
Answers:
Here is a solution based on a simplistic interpretation of your question. A better answer will require more information.
>>> s = "aaa bbb aaa bbb"
>>> separator = " "
>>> parts = s.split(separator)
>>> separator.join(parts[:-1] + ["xxx"])
'aaa bbb aaa xxx'
Update
(After seeing edited question) another very specific answer.
>>> s = "123123"
>>> separator = "2"
>>> parts = s.split(separator)
>>> separator.join(parts[:-1]) + "x" + parts[-1]
'1231x3'
Update 2
There is far better way to do this. Courtesy @mizipzor.
>>> s = "aaa bbb aaa bbb"
>>> s[::-1].replace('bbb','xxx',1)[::-1]
'aaa bbb aaa xxx'
For your second example
>>> s = "123123"
>>> s[::-1].replace('2','x',1)[::-1]
'1231x3'
When the wanted match is at the end of string, re.sub comes to the rescue.
>>> import re
>>> s = "aaa bbb aaa bbb"
>>> s
'aaa bbb aaa bbb'
>>> re.sub('bbb$', 'xxx', s)
'aaa bbb aaa xxx'
>>>
Using regular expression function re.sub to replace words at end of string
import re
s = "123123"
s = re.sub('23$', 'penguins', s)
print s
Prints:
1231penguins
or
import re
s = "123123"
s = re.sub('^12', 'penguins', s)
print s
Prints:
penguins3123
This is exactly what the rpartition function is used for:
S.rpartition(sep) -> (head, sep, tail)
Search for the separator sep in S, starting at the end of S, and return
the part before it, the separator itself, and the part after it. If the
separator is not found, return two empty strings and S.
I wrote this function showing how to use rpartition in your use case:
def replace_last(source_string, replace_what, replace_with):
head, _sep, tail = source_string.rpartition(replace_what)
return head + replace_with + tail
s = "123123"
r = replace_last(s, '2', 'x')
print r
Output:
1231x3
This is one of the few string functions that doesn’t have a left and right version, but we can mimic the behaviour using some of the string functions that do.
>>> s = '123123'
>>> t = s.rsplit('2', 1)
>>> u = 'x'.join(t)
>>> u
'1231x3'
or
>>> 'x'.join('123123'.rsplit('2', 1))
'1231x3'
I got a tricky answer, but it is not efficient enough
fname = '12345.png.pngasdfg.png'
suffix = '.png'
fname_rev = fname[::-1]
suffix_rev = suffix[::-1]
fullname_rev = fname_rev.replace(suffix_rev, '', 1)
fullname = fullname_rev[::-1]
fullname '12345.png.pngasdfg'
Built-in function replace() takes three arguments
str.replace(old, new, max_time)
So you can delete the last mached string from the origional string
For the second example, I would recommend rsplit, as it is very simple to use and and directly achieves the goal. Here is a little function with it:
def replace_ending(sentence, old, new):
if sentence.endswith(old):
i = sentence.rsplit(old,1)
new_sentence =new.join(i)
return new_sentence
return sentence
print(replace_ending("aaa bbb aaa bbb", "bbb", "xxx"))
Output:
aaa bbb aaa xxx
There is a way to handle replace to make it work backwards.
The idea is to reverse the string, text to replace and new text using [::-1]. The logic stays the same.
For example:
>>> a = "123 456 123 456 123 456"
and we want to replace the last one, (or two, or n) occurrences of "123" with "abc"
>>> a[::-1].replace("123"[::-1], "abc"[::-1], 1)[::-1]
'123 456 123 456 abc 456'
This could be put as a function that behaves exactly like replace.
def replace_right(text, old, new, count=-1):
"""
Return a copy of text with all occurrences of substring old
replaced by new, starting from the right.
If count is given and is not -1, only the first count
occurrences are replaced.
"""
return text[::-1].replace(old[::-1], new[::-1], count)[::-1]
Execution:
>>> replace_right(a, "123", "abc", 1)
'123 456 123 456 abc 456'
>>> replace_right(a, "123", "abc", 2)
'123 456 abc 456 abc 456'
There is a very easy way to do this, follow my steps
s = "stay am stay am delete am"
# want to remove the last am
s = s[:s.rfind('am')]
print(s)
Here at first, we find the index number of am‘s last occurrence, then take 0th index to that particular index.
I want to replace characters at the end of a python string. I have this string:
s = "123123"
I want to replace the last 2 with x. Suppose there is a method called replace_last:
>>> replace_last(s, '2', 'x')
'1231x3'
Is there any built-in or easy method to do this?
It’s similar to python’s str.replace():
>>> s.replace('2', 'x', 1)
'1x3123'
But it’s from the end to beginning.
Here is a solution based on a simplistic interpretation of your question. A better answer will require more information.
>>> s = "aaa bbb aaa bbb"
>>> separator = " "
>>> parts = s.split(separator)
>>> separator.join(parts[:-1] + ["xxx"])
'aaa bbb aaa xxx'
Update
(After seeing edited question) another very specific answer.
>>> s = "123123"
>>> separator = "2"
>>> parts = s.split(separator)
>>> separator.join(parts[:-1]) + "x" + parts[-1]
'1231x3'
Update 2
There is far better way to do this. Courtesy @mizipzor.
>>> s = "aaa bbb aaa bbb"
>>> s[::-1].replace('bbb','xxx',1)[::-1]
'aaa bbb aaa xxx'
For your second example
>>> s = "123123"
>>> s[::-1].replace('2','x',1)[::-1]
'1231x3'
When the wanted match is at the end of string, re.sub comes to the rescue.
>>> import re
>>> s = "aaa bbb aaa bbb"
>>> s
'aaa bbb aaa bbb'
>>> re.sub('bbb$', 'xxx', s)
'aaa bbb aaa xxx'
>>>
Using regular expression function re.sub to replace words at end of string
import re
s = "123123"
s = re.sub('23$', 'penguins', s)
print s
Prints:
1231penguins
or
import re
s = "123123"
s = re.sub('^12', 'penguins', s)
print s
Prints:
penguins3123
This is exactly what the rpartition function is used for:
S.rpartition(sep)->(head, sep, tail)Search for the separator sep in S, starting at the end of S, and return
the part before it, the separator itself, and the part after it. If the
separator is not found, return two empty strings and S.
I wrote this function showing how to use rpartition in your use case:
def replace_last(source_string, replace_what, replace_with):
head, _sep, tail = source_string.rpartition(replace_what)
return head + replace_with + tail
s = "123123"
r = replace_last(s, '2', 'x')
print r
Output:
1231x3
This is one of the few string functions that doesn’t have a left and right version, but we can mimic the behaviour using some of the string functions that do.
>>> s = '123123'
>>> t = s.rsplit('2', 1)
>>> u = 'x'.join(t)
>>> u
'1231x3'
or
>>> 'x'.join('123123'.rsplit('2', 1))
'1231x3'
I got a tricky answer, but it is not efficient enough
fname = '12345.png.pngasdfg.png'
suffix = '.png'
fname_rev = fname[::-1]
suffix_rev = suffix[::-1]
fullname_rev = fname_rev.replace(suffix_rev, '', 1)
fullname = fullname_rev[::-1]
fullname '12345.png.pngasdfg'
Built-in function replace() takes three arguments
str.replace(old, new, max_time)
So you can delete the last mached string from the origional string
For the second example, I would recommend rsplit, as it is very simple to use and and directly achieves the goal. Here is a little function with it:
def replace_ending(sentence, old, new):
if sentence.endswith(old):
i = sentence.rsplit(old,1)
new_sentence =new.join(i)
return new_sentence
return sentence
print(replace_ending("aaa bbb aaa bbb", "bbb", "xxx"))
Output:
aaa bbb aaa xxx
There is a way to handle replace to make it work backwards.
The idea is to reverse the string, text to replace and new text using [::-1]. The logic stays the same.
For example:
>>> a = "123 456 123 456 123 456"
and we want to replace the last one, (or two, or n) occurrences of "123" with "abc"
>>> a[::-1].replace("123"[::-1], "abc"[::-1], 1)[::-1]
'123 456 123 456 abc 456'
This could be put as a function that behaves exactly like replace.
def replace_right(text, old, new, count=-1):
"""
Return a copy of text with all occurrences of substring old
replaced by new, starting from the right.
If count is given and is not -1, only the first count
occurrences are replaced.
"""
return text[::-1].replace(old[::-1], new[::-1], count)[::-1]
Execution:
>>> replace_right(a, "123", "abc", 1)
'123 456 123 456 abc 456'
>>> replace_right(a, "123", "abc", 2)
'123 456 abc 456 abc 456'
There is a very easy way to do this, follow my steps
s = "stay am stay am delete am"
# want to remove the last am
s = s[:s.rfind('am')]
print(s)
Here at first, we find the index number of am‘s last occurrence, then take 0th index to that particular index.