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Using sed to replace a pattern by symbol


Using sed to replace a pattern by symbol

By : user3099802
Date : January 11 2021, 03:34 PM
wish help you to fix your issue How do I replace a pattern by looking at sentence symbol by using sed? , This should do:
code :
sed -r 's/^[a-zA-Z]+-/start/;s/^-[a-zA-Z]+-/middle/' file
start
middle


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pattern matching check if greater than symbol is not preceded by smaller than symbol

pattern matching check if greater than symbol is not preceded by smaller than symbol


By : user3203314
Date : March 29 2020, 07:55 AM
seems to work fine The match is: a >. This section matches your regex perfectly - it doesn't start with <, then it's got "a ", which matches the bit in square brackets, and then there's a >.
Are you trying to match the whole string? If you are, try re.match instead of re.search.
code :
text = '<a > b'
match = re.search('<?[a-zA-Z0-9_ ]+>',text)

if ((match) and (match.group(0)[0] != '<')):
  # Match found
Lua pattern to replace any word,character, digit or symbol followed by a paticular string

Lua pattern to replace any word,character, digit or symbol followed by a paticular string


By : jggonz
Date : March 29 2020, 07:55 AM
it helps some times (some%-word)\[^;,.%s]* works, note that:
- is a magic character in Lua patterns, it needs to be escaped. some%-word is surrounded by (), so that it's captured with %1. In character class, ^ is used in the beginning to indicate complement of the following.
code :
d = "some%-word" 
p = "(" .. d .. ")" .. "\\[^;,.%s]*"
c = "%1###" 
s = "testing some-word\\test-2 later some-word\\&^*; some-word\\set_34$ " 
print(p) 
res = string.gsub(s,p,c)
print(res)
(some%-word)\[^;,.%s]*
testing some-word### later some-word###; some-word###
regular expression replace (if pattern found replace symbol for symbol)

regular expression replace (if pattern found replace symbol for symbol)


By : Sahil Malik
Date : March 29 2020, 07:55 AM
To fix this issue I have several lines of text (RNA sequence), I want to make a matrix regarding conservation of characters, because they are aligned according similarity. , You could try the following:
code :
data = "ONE " + "-" * 100 + "atgtgca" + "-" * 20
print re.sub(r'-{100,}', lambda x: '.' * len(x.group(0)), data)
ONE ....................................................................................................atgtgca--------------------
Remove replace pipe (|) symbol with comma (,) symbol in spark using python

Remove replace pipe (|) symbol with comma (,) symbol in spark using python


By : isenberg13
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , this myRDD data for 2 rows:
code :
>>> data = [u'#fields:excDate|schedDate|TZ|custID|muID|tvID|acdID|logonID|agentName|modify|exception|start|stop|LS Oracle Emp ID|Team Lead', u'06152016|06152016|CET|3|3000|1688|87||Ali, AbdElaziz|1465812004|Open|08:00|09:00|101021021|ElDeleify,Hisham']
>>> data = [item.replace("|", ",") for item in data]
>>> data
['#fields:excDate,schedDate,TZ,custID,muID,tvID,acdID,logonID,agentName,modify,exception,start,stop,LS Oracle Emp ID,Team Lead', '06152016,06152016,CET,3,3000,1688,87,,Ali, AbdElaziz,1465812004,Open,08:00,09:00,101021021,ElDeleify,Hisham']
Replace a specific symbol in the end of yyyy-MM-dd pattern using regex in Java

Replace a specific symbol in the end of yyyy-MM-dd pattern using regex in Java


By : user1856033
Date : March 29 2020, 07:55 AM
To fix the issue you can do You may use a capturing group and replace with the placeholder to the value stored in this group:
code :
String result = s.replaceAll("(\\d{4}-\\d{2}-\\d{2})Z", "$1");
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