I have to replace any expression which look like:
"any_expression/2016.2
Can I do it with Regex?
String any_expressin can contain any character and in any length
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Jirka Picek
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sarad
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1How about this: `s/^.*(\/2016\.2\)/replaced#1/g`? PS: you should read [this](http://stackoverflow.com/help/how-to-ask)! – Jirka Picek Mar 15 '17 at 12:10
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@Jirka Picek - it doest match (there must be " in the begining). – sarad Mar 15 '17 at 12:32
2 Answers
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Of course you can, this should do it:
".*\/2016\.2
In Python:
str.replace(".*\/2016\.2", "whatYouWantInstead");
Transzendental
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Can you deomonsrarate how can I replace the any_expression to $env (string}? – sarad Mar 15 '17 at 12:28
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Well just add it. I dont really understand where you problem is, just make sure to escape special character with \ . – Transzendental Mar 15 '17 at 12:36
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but it replace the any_expression and also the /2016.2. I want to replace only the any_expression. Foe example: "C:/Xilinx/Vivado/2016.2/data/ip/xpr/hdl/ram.sv"\ will be "$env/2016.2/data/ip/xpr/hdl/ram.sv"\ – sarad Mar 15 '17 at 12:46
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Answer to your question is "Yes".
If you want to know more about it then:
Regular expression matching your requirements is:
^".*(\/2016\.2)$
If it can be anywhere in text, just remove first and last character:
".*(\/2016\.2)
According to this question Python doesn't recognize regex for str.replace, then you need to use re.sub. It should look like:
import re
line = re.sub(r"^\".*(\/2016\.2)$", r"replaced any expression\1", line)
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Jirka Picek
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How the following should look wirh re? `exp_cont = list(map(lambda x: x.replace("\".*\/2016\.2", "\"$vivado_path/2016.2/"), exp_cont))` – sarad Mar 15 '17 at 13:45
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I tried: `exp_cont = list(map(lambda x: re.sub(r"^\".*(\/2016\.2)", r"\"$vivado_path/2016.2", x), exp_cont))` For some reason it write the \ char in the begining of each line which was replaced. – sarad Mar 15 '17 at 14:07