I need to remove an character between two parenthesis - sed

As the title suggests, I need to remove an character between two characters along a string.
E.g. I want to remove the semicolon between two parenthesis
<>word (word ; word) word<>
output desidered:
<>(word word)<>

Your description and the desired output do not match!
To remove the semicolon between two brackets you can use
sed '/([^)]*/s/;//'
example
echo "<>word (word ; word) word<>" | sed '/([^)]*/s/;//'
output
<>word (word word) word<>

Related

Can Sed match matching brackets?

My code has a ton of occurrences of something like:
idof(some_object)
I want to replace them with:
some_object["id"]
It sounds simple:
sed -i 's/idof(\([^)]\+\))/\1["id"]/g' source.py
The problem is that some_object might be something like idof(get_some_object()), or idof(my_class().get_some_object()), in which case, instead of getting what I want (get_some_object()["id"] or my_class().get_some_object()["id"]), I get get_some_object(["id"]) or my_class(["id"].get_some_object()).
Is there a way to have sed match closing bracket, so that it internally keeps track of any opening/closing brackets inside my (), and ignores those?
It needs to keep everything that's between those brackets: idof(ANYTHING) becomes ANYTHING["id"].
Using sed
$ sed -E 's/idof\(([[:alpha:][:punct:]]*)\)/\1["id"]/g' input_file
Using ERE, exclude idof and the first opening parenthesis.
As a literal closing parenthesis is also excluded, everything in-between the capture parenthesis including additional parenthesis will be captured.
[[:alpha:]] will match all alphabetic characters including upper and lower case while [[:punct:]] will capture punctuation characters including ().-{} and more.
The g option will make the substitution as many times as the pattern is found.
Theoretically, you can write a regex that will handle all combinations of idof(....) up to some limit of nested () calls inside ..... Such regex would have to list with all possible combinations of calls, like idof(one(two(three))) or idof(one(two(three)four(five)) you can match with an appropriate regex like idof([^()]*([^()]*([^()]*)[^()]*)[^()]*) or idof([^()]*([^()]*([^()]*)[^()]*([^()]*)[^()]*) respectively.
The following regex handles only some cases, but shows the complexity and general path. Writing a regex to handle all possible cases to "eat" everything in front of the trailing ) is left to OP as an exercise why it's better to use something else. Note that handling string literals ")" becomes increasingly complex.
The following Bash code:
sed '
: begin
# No idof? Just print the line!
/^\(.*\)idof(\([^)]*)\)/!n
# Note: regex is greedy - we start from the back!
# Note: using newline as a stack separator.
s//\1\n\2/
# hold the front
{ h ; x ; s/\n.*// ; x ; s/[^\n]*\n// ; }
: handle_brackets
# Eat everything before final ) up to some number of nested ((())) calls.
# Insert more jokes here.
: eat_brackets
/^[^()]*\(([^()]*\(([^()]*\(([^()]*\(([^()]*\(([^()]*\(([^()]*)\)\?[^()]*)\)\?[^()]*)\)\?[^()]*)\)\?[^()]*)\)\?[^()]*)\)/{
s//&\n/
# Hold the front.
{ H ; x ; s/\n\([^\n]*\)\n.*/\1/ ; x ; s/[^\n]*\n// ; }
b eat_brackets
}
/^\([^()]*\))/!{
s/^/ERROR: eating brackets did not work: /
q1
}
# Add the id after trailing ) and remove it.
s//\1["id"]/
# Join with hold space and clear the hold space for next round
{ H ; s/.*// ; x ; s/\n//g ; }
# Restart for another idof if in input.
b begin
' <<EOF
before idof(some_object) after
before idof(get_some_object()) after
before idof(my_class().get_some_object()) after
before idof(one(two(three)four)five) after
before idof(one(two(three)four)five) between idof(one(two(three)four)five) after
before idof( one(two(three)four)five one(two(three)four)five ) after
before idof(one(two(three(four)five)six(seven(eight)nine)ten) between idof(one(two(three(four)five)six(seven(eight)nine)ten) after
EOF
Will output:
before some_object["id"] after
before get_some_object()["id"] after
before my_class().get_some_object()["id"] after
before one(two(three)four)five["id"] after
before one(two(three)four)five["id"] between one(two(three)four)five["id"] after
before one(two(three)four)five one(two(three)four)five ["id"] after
ERROR: eating brackets did not work: one(two(three(four)five)six(seven(eight)nine)ten) after
The last line is not handled correctly, because (()()) case is not correctly handled. One would have to write a regex to match it.

How to match exact string in perl

I am trying to parse all the files and verify if any of the file content has strings TESTDIR or TEST_DIR
Files contents might look something like:-
TESTDIR = foo
include $(TESTDIR)/chop.mk
...
TEST_DIR := goldimage
MAKE_TESTDIR = var_make
NEW_TEST_DIR = tesing_var
Actually I am only interested in TESTDIR ,$(TESTDIR),TEST_DIR but in my case last two lines should be ignored. I am new to perl , Can anyone help me out with re-rex.
/\bTEST_?DIR\b/
\b means a "word boundary", i.e. the place between a word character and a non-word character. "Word" here has the Perl meaning: it contains characters, numbers, and underscores.
_? means "nothing or an underscore"
Look at "characterset".
Only (space) surrounding allowed:
/^(.* )?TEST_?DIR /
^ beginning of the line
(.* )? There may be some content .* but if, its must be followed by a space
at the and says that a whitespace must be there. Otherwise use ( .*)?$ at the end.
One of a given characterset is allowed:
Should the be other characters then a space be possible you can use a character class []:
/^(.*[ \t(])?TEST_?DIR[) :=]/
(.*[ \t(])? in front of TEST_?DIR may be a (space) or a \t (tab) or ( or nothing if the line starts with itself.
afterwards there must be one of (space) or : or = or ). Followd by anything (to "anything" belongs the "=" of ":=" ...).
One of a given group is allowed:
So you need groups within () each possible group in there devided by a |:
/^(.*( |\t))?TEST_?DIR( | := | = )/
In this case, at the beginning is no change to [ \t] because each group holds only one character and \t.
At the end, there must be (single space) or := (':=' surrounded by spaces) or = ('=' surrounded by spaces), following by anything...
You can use any combination...
/^(.*[ \t(])?TEST_?DIR([) =:]| :=| =|)/
Test it on Debuggex.com. (Use 'PCRE')

Avoiding duplicate items in a comma-separated list of two-letter words

I need to write a regex which allows a group of 2 chars only once. This is my current regex :
^([A-Z]{2},)*([A-Z]{2}){1}$
This allows me to validate something like this :
AL,RA,IS,GD
AL
AL,RA
The problem is that it validates also AL,AL and AL,RA,AL.
EDIT
Here there are more details.
What is allowed:
AL,RA,GD
AL
AL,RA
AL,IS,GD
What it shouldn't be allowed:
AL,RA,AL
AL,AL
AL,RA,RA
AL,IS,AL
IS,IS,AL
IS,GD,GD
IS,GD,IS
I need that every group of two characters appears only once in the sequence.
Try something like this expression:
/^(?:,?(\b\w{2}\b)(?!.*\1))+$/gm
I have no knowledge of swift, so take it with a grain of salt. The idea is basically to only match a whole line while making sure that no single matched group occurs at a later point in the line.
First of all, let's shorten your pattern. It can be easily achieved since the length of each comma-separated item is fixed and the list items are only made up of uppercase ASCII letters. So, your pattern can be written as ^(?:[A-Z]{2}(?:,\b)?)+$. See this regex demo.
Now, you need to add a negative lookahead that will check the string for any repeating two-letter sequence at any distance from the start of string, and within any distance between each. Use
^(?!.*\b([A-Z]{2})\b.*\b\1\b)(?:[A-Z]{2}(?:,\b)?)+$
See the regex demo
Possible implementation in Swift:
func isValidInput(Input:String) -> Bool {
return Input.range(of: #"^(?!.*\b([A-Z]{2})\b.*\b\1\b)(?:[A-Z]{2}(?:,\b)?)+$"#, options: .regularExpression) != nil
}
print(isValidInput(Input:"AL,RA,GD")) // true
print(isValidInput(Input:"AL,RA,AL")) // false
Details
^ - start of string
(?!.*\b([A-Z]{2})\b.*\b\1\b) - a negative lookahead that fails the match if, immediately to the right of the current location, there is:
.* - any 0+ chars other than line break chars, as many as possible
\b([A-Z]{2})\b - a two-letter word as a whole word
.* - any 0+ chars other than line break chars, as many as possible
\b\1\b - the same whole word as in Group 1. NOTE: The word boundaries here are not necessary in the current scenario where the word length is fixed, it is two, but if you do not know the word length, and you have [A-Z]+, you will need the word boundaries, or other boundaries depending on the situation
(?:[A-Z]{2}(?:,\b)?)+ - 1 or more sequences of:
[A-Z]{2} - two uppercase ASCII letters
(?:,\b)? - an optional sequence: , only if followed with a word char: letter, digit or _. This guarantees that , won't be allowed at the end of the string
$ - end of string.
You can use a negative lookahead with a back-reference:
^(?!.*([A-Z]{2}).*\1).*
if, as in the all the examples in the question, it is known that the string contains only comma-separated pairs of capital letters. I will relax that assumption later in my answer.
Demo
The regex performs the following operations:
^ # match beginning of line
(?! # begin negative lookahead
.* # match 0+ characters (1+ OK)
([A-Z]{2}) # match 2 uppercase letters in capture group 1
.* # match 0+ characters (1+ OK)
\1 # match the contents of capture group 1
) # end negative lookahead
.* # match 0+ characters (the entire string)
Suppose now that one or more capital letters may appear between each pair of commas, or before the first comma or after the last comma, but it is only strings of two letters that cannot be repeated. Moreover, I assume the regex must confirm the regex has the desired form. Then the following regex could be used:
^(?=[A-Z]+(?:,[A-Z]+)*$)(?!.*(?:^|,)([A-Z]{2}),(?:.*,)?\1(?:,|$)).*
Demo
The regex performs the following operations:
^ # match beginning of line
(?= # begin pos lookahead
[A-Z]+ # match 1+ uc letters
(?:,[A-Z]+) # match ',' then by 1+ uc letters in a non-cap grp
* # execute the non-cap grp 0+ times
$ # match the end of the line
) # end pos lookahead
(?! # begin neg lookahead
.* # match 0+ chars
(?:^|,) # match beginning of line or ','
([A-Z]{2}) # match 2 uc letters in cap grp 1
, # match ','
(?:.*,) # match 0+ chars, then ',' in non-cap group
? # optionally match non-cap grp
\1 # match the contents of cap grp 1
(?:,|$) # match ',' or end of line
) # end neg lookahead
.* # match 0+ chars (entire string)
If there is no need check that the string contains only comma-separated strings of one or more upper case letters the postive lookahead at the beginning can be removed.

Sed replacing Special Characters in a string

I am having difficulties replacing a string containing special characters using sed. My old and new string are shown below
oldStr = "# td=(nstates=20) cam-b3lyp/6-31g geom=connectivity"
newStr = "# opt b3lyp/6-31g geom=connectivity"
My sed command is the following
sed -i 's/\# td\=\(nstates\=20\) cam\-b3lyp\/6\-31g geom\=connectivity/\# opt b3lyp\/6\-31g geom\=connectivity/g' myfile.txt
I dont get any errors, however there is no match. Any ideas on how to fix my patterns.
Thanks
try s|# td=(nstates=20) cam-b3lyp/6-31g geom=connectivity|# opt b3lyp/6-31g geom=connectivity|g'
you can use next to anything after s instead of /, as your expression contains slashes I used | instead. -, = and # don't have to be escaped (minus only in character sets [...]), escaped parens indicate a group, nonescaped parens are literals.

How to match text in [...] with NSRegularExpression

what would be the NSRegularExpression to match the text between the "[" and the "]" in the following example:
"[this is some] text i would [like] to [parse] with NSRegularExpression"
Thx in advance :)
Not sure about NSRegularExpression, but I suspect it's like other regular expressions in terms of greediness. So, if there could be multiple [words] or [phrases in brackets] that you want to capture in the same string, you'll have to make sure the regex isn't greedy.
Maybe something like (broken up across a few lines for clarity:)
\[ // find an opening square bracket
([^]]+) // find one or more characters that ARE NOT a square bracket
\] // find the corresponding closing square bracket
Like a Daniel answer right pattern is \\[(.*?)\\]
Let me explain why.
\\[ is mean "open-bracket" symbol. Double slash allow to use special symbol [
All in ( ) allow to union as atomic sequence
. — any symbol without special * ? + [ ( ) { } ^ $ | \ . /!
*? — symbol entry 0 or more times. Match as few times as possible.
\\] — close-bracket
Your pattern should be \[(.*?)\]