python pattern detect and replace string

Question

I have a question on Python string operation:

Here's a string in which there are multiple .(dot),

like: "a3a.b1b2b.cccc.ded.f.g"

The question is to find the content before the last .(dot), which is "a3a.b1b2b.cccc.ded.f" in this example (We call it as STRING_BEFORE_LAST_DOT)

and we need to replace STRING_BEFORE_LAST_DOT to be STRING_BEFORE_LAST_DOT + "_Approved". So the result should be: "a3a.b1b2b.cccc.ded.f_Approved.g".

And some more examples:

"a.b" -> "a_Approved.b"
"first.second.c" -> "first.second_Approved.c"

I can def a function to do so, I'm wondering if there's any advance way, like use .replace() to make it happen. Thanks.

apart from some fancy regex, you can also do this quite easily with split and join. Have you tried anything yourself so far? — Ma0
– Ma0, Commented May 25, 2021 at 9:21
@Ma0 Yeah, I tried to def a function (split by dot and reconnect the parts one by one). I know it's far away from a efficient solution, so come to ask for help.. — user3514137
– user3514137, Commented May 25, 2021 at 9:25
That should not be too bad actually. Is performance so important for your application? — Ma0
– Ma0, Commented May 25, 2021 at 9:26

Cyrille · Accepted Answer · 2021-05-25 09:25:52Z

2

You could use join + rsplit with a limit of 1:

>>> "_Approved.".join("a.b".rsplit(".",1))
'a_Approved.b'
>>> "_Approved.".join("first.second.c".rsplit(".",1))
'first.second_Approved.c'

answered May 25, 2021 at 9:25

Cyrille

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Martin Mwangi · Accepted Answer · 2021-05-25 09:58:32Z

1

# your String
string = 'a3a.b1b2b.cccc.ded.f.g'
# find the last dot character in your string
StringIndex = string.rfind('.')
# create a temporary list of your string and swap the last dot with your preferred string (_Approved.)
temp = list(string)
temp[StringIndex] = '_Approved.'
# replace
string = ''.join(temp)
# print
print('Old String: a3a.b1b2b.cccc.ded.f.g \nNew String: '+ string)

# OUTPUT
# Old String: a3a.b1b2b.cccc.ded.f.g 
# New String: a3a.b1b2b.cccc.ded.f_Approved.g

answered May 25, 2021 at 9:58

Martin Mwangi

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Comments

sushanth · Accepted Answer · 2021-05-25 09:27:02Z

0

Here is a solution you can try out,

def _replace(input_):
    sub_ = input_[input_.rfind("."):]
    return input_.replace(sub_, f"_Approved{sub_}")


print(_replace("a.b"))
print(_replace("first.second.c"))

a_Approved.b
first.second_Approved.c

answered May 25, 2021 at 9:27

sushanth

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Comments

Jan Christoph Terasa · Accepted Answer · 2021-05-25 09:58:54Z

0

An example of how to do it with a regexp:

import re
s = "a3a.b1b2b.cccc.ded.f.g" 
no_ext = re.split(r'\.[^\.]*$', s)[0]

answered May 25, 2021 at 9:58

Jan Christoph Terasa

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Meric Gunduz · Accepted Answer · 2021-05-25 10:13:42Z

0

Here is an alternative. You can access every dot's index with this:

import re
str = "a3a.b1b2b.cccc.ded.f"
# Finds every dot's starting index. (For ending of matched value d.end())
dots = [d.start(0) for d in re.finditer(r'\.+', str)]
# _Approved added
str = str[:dots[-1]] + "_Approved" + str[dots[-1]:]

answered May 25, 2021 at 10:13

Meric Gunduz

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The fourth bird · Accepted Answer · 2021-05-25 21:26:27Z

0

You can use a pattern to match from the last dot until the end of the string .[^.]*$

Then replace with _Approved and add the full match after it using \g<0>

print(re.sub(r"\.[^.]*$", r"_Approved\g<0>", "a3a.b1b2b.cccc.ded.f.g"))

Output

a3a.b1b2b.cccc.ded.f_Approved.g

answered May 25, 2021 at 21:26

The fourth bird

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