I need to replace within a large text file all occurrences such as 'yw234DV w-23-sDf wef23s-d-f' with the same strings but with underscores instead of spaces for all spaces within quotes, without replacing any spaces outside quotes with underscores.
I'm trying to find a solution for substitution within vim, but a sed solution would also be much appreciated. The number of tokens in each quote-delimited string may vary.
I've been playing with some regexes in vim, but they're pretty elementary and seem to be missing what I need.
My current attempt:
%s/'{[:alnum:] }*/'\0\_/g
And I'm experimenting with variations on that.
This is most similar to my question, though it is Java:
Replacing spaces within quotes
Sample Input:
'wiUEF7-gvouw ow wo24-RTeih we', 'yt23IT iug-76'
Sample Output:
'wiUEF7-gvouw_ow_wo24-RTeih_we', 'yt23IT_iug-76'
You may try this with VIM, tried this on Macvim:
%s/\%('[^']*'\)*\('[^']*'\)/\=substitute(submatch(1), ' ', '_', 'g')/g
Much simpler solution , Thanks to #SergioAraujo:
#%s/\v%(('[^']*'))/\=substitute(submatch(1),' ', '_', 'g')/g
Not sure however, if below is the outcome you have expected
Output:
'wiUEF7-gvouw_ow_wo24-RTeih_we', 'yt23IT_iug-76'
In perl:
perl -i -pe's{(\x27.*?\x27)}{ (my $subst = $1) =~ tr/ /_/ }ge' yourfile
or with perl5.14 or above:
perl -i -pe's{(\x27.*?\x27)}{ $1 =~ tr/ /_/r }ge'
With this the input file:
$ cat file
'wiUEF7-gvouw ow wo24-RTeih we', 'yt23IT iug-76'
We can convert all spaces inside of single-quotes into underscores with:
$ sed -E ":a; s/^(([^']*'[^']*')*[^']*'[^']*)[[:space:]]/\1_/; ta" file
'wiUEF7-gvouw_ow_wo24-RTeih_we', 'yt23IT_iug-76'
How it works
:a
This creates a label a.
s/^(([^']*'[^']*')*[^']*'[^']*)[[:space:]]/\1_/
This inserts the underscores where we want them.
^(([^']*'[^']*')*[^']*'[^']*)[[:space:]]
This looks for any odd number of single quotes followed by any number of non-quote characters followed by a space. Everything before that space is saved in group 1.
\1_
This replaces the matched text with group 1 followed by an underscore.
ta
If the previous command put any new underscores in the string, then jump back to label a and try again.
Using FPAT variable in gnu awk you can do this:
awk -v OFS=', ' -v FPAT="'[^']*'" '{for (h=1; h<=NF; h++)
{gsub(/[[:blank:]]/, "_", $h); printf "%s%s", $h, (h < NF ? OFS : ORS)}}' file
'wiUEF7-gvouw_ow_wo24-RTeih_we', 'yt23IT_iug-76'
Related
I have a few huge files with values seperated by a pipe (|) sign.
The strings our quoted but sometimes there is a newline in between the quoted string.
I need to read these files with external table from oracle but on the newlines he will give me errors. So I need to replace them with a space.
I do some other perl commands on these files for other errors, so I would like to have a solution in a one line perl command.
I 've found some other similar questions on stackoverflow, but they don't quite do the same and I can't find a solution for my problem with the solution mentioned there.
The statement I tried but that isn't working:
perl -pi -e 's/"(^|)*\n(^|)*"/ /g' test.txt
Sample text:
4454|"test string"|20-05-1999|"test 2nd string"
4455|"test newline
in string"||"test another 2nd string"
4456|"another string"|19-03-2021|"here also a newline
"
4457|.....
Should become:
4454|"test string"|20-05-1999|"test 2nd string"
4455|"test newline in string"||"test another 2nd string"
4456|"another string"|19-03-2021|"here also a newline "
4457|.....
Sounds like you want a CSV parser like Text::CSV_XS (Install through your OS's package manager or favorite CPAN client):
$ perl -MText::CSV_XS -e '
my $csv = Text::CSV_XS->new({sep => "|", binary => 1});
while (my $row = $csv->getline(*ARGV)) {
$csv->say(*STDOUT, [ map { tr/\n/ /r } #$row ])
}' test.txt
4454|"test string"|20-05-1999|"test 2nd string"
4455|"test newline in string"||"test another 2nd string"
4456|"another string"|19-03-2021|"here also a newline "
This one-liner reads each record using | as the field separator instead of the normal comma, and for each field, replaces newlines with spaces, and then prints out the transformed record.
In your specific case, you can also consider a workaround using GNU sed or awk.
An awk command will look like
awk 'NR==1 {print;next;} /^[0-9]{4,}\|/{print "\n" $0;next;}1' ORS="" file > newfile
The ORS (output record separator) is set to an empty string, which means that \n is only added before lines starting with four or more digits followed with a | char (matched with a ^[0-9]{4,}\| POSIX ERE pattern).
A GNU sed command will look like
sed -i ':a;$!{N;/\n[0-9]\{4,\}|/!{s/\n/ /;ba}};P;D' file
This reads two consecutive lines into the pattern space, and once the second line doesn't start with four digits followed with a | char (see the [0-9]\{4\}| POSIX BRE regex pattern), the or more line break between the two is replaced with a space. The search and replace repeats until no match or the end of file.
With perl, if the file is huge but it can still fit into memory, you can use a short
perl -0777 -pi -e 's/\R++(?!\d{4,}\|)/ /g' <<< "$s"
With -0777, you slurp the file and the \R++(?!\d{4,}\|) pattern matches any one or more line breaks (\R++) not followed with four or more digits followed with a | char. The ++ possessive quantifier is required to make (?!...) negative lookahead to disallow backtracking into line break matching pattern.
With your shown samples, this could be simply done in awk program. Written and tested in GNU awk, should work in any awk. This should work fast even on huge files(better than slurping whole file into memory, having mentioned that OP may use it on huge files).
awk 'gsub(/"/,"&")%2!=0{if(val==""){val=$0} else{print val $0;val=""};next} 1' Input_file
Explanation: Adding detailed explanation for above.
awk ' ##Starting awk program from here.
gsub(/"/,"&")%2!=0{ ##Checking condition if number of " are EVEN or not, because if they are NOT even then it means they are NOT closed properly.
if(val==""){ val=$0 } ##Checking condition if val is NULL then set val to current line.
else {print val $0;val=""} ##Else(if val NOT NULL) then print val current line and nullify val here.
next ##next will skip further statements from here.
}
1 ##In case number of " are EVEN in any line it will skip above condition(gusb one) and simply print the line.
' Input_file ##Mentioning Input_file name here.
I'm trying to filter out all the words that contain any character other than a letter from a text file. I've looked around stackoverflow, and other websites, but all the answers I found were very specific to a different scenario and I wasn't able to replicate them for my purposes; I've only recently started learning about Unix tools.
Here's an example of what I want to do:
Input:
#derik I was there and it was awesome! !! http://url.picture.whatever #hash_tag
Output:
I was there and it was awesome!
So words with punctuation can stay in the file (in fact I need them to stay) but any substring with special characters (including those of punctuation) needs to be trimmed away. This can probably be done with sed, but I just can't figure out the regex. Help.
Thanks!
Here is how it could be done using Perl:
perl -ane 'for $f (#F) {print "$f " if $f =~ /^([a-zA-z-\x27]+[?!;:,.]?|[\d.]+)$/} print "\n"' file
I am using this input text as my test case:
Hello,
How are you doing?
I'd like 2.5 cups of piping-hot coffee.
#derik I was there; it was awesome! !! http://url.picture.whatever #hash_tag
output:
Hello,
How are you doing?
I'd like 2.5 cups of piping-hot coffee.
I was there; it was awesome!
Command-line options:
-n loop around every line of the input file, do not automatically print it
-a autosplit mode – split input lines into the #F array. Defaults to splitting on whitespace
-e execute the perl code
The perl code splits each input line into the #F array, then loops over every field $f and decides whether or not to print it.
At the end of each line, print a newline character.
The regular expression ^([a-zA-z-\x27]+[?!;:,.]?|[\d.]+)$ is used on each whitespace-delimited word
^ starts with
[a-zA-Z-\x27]+ one or more lowercase or capital letters or a dash or a single quote (\x27)
[?!;:,.]? zero or one of the following punctuation: ?!;:,.
(|) alternately match
[\d.]+ one or more numbers or .
$ end
Your requirements aren't clear at all but this MAY be what you want:
$ awk '{rec=sep=""; for (i=1;i<=NF;i++) if ($i~/^[[:alpha:]]+[[:punct:]]?$/) { rec = rec sep $i; sep=" "} print rec}' file
I was there and it was awesome!
sed -E 's/[[:space:]][^a-zA-Z0-9[:space:]][^[:space:]]*//g' will get rid of any words starting with punctuation. Which will get you half way there.
[[:space:]] is any whitespace character
[^a-zA-Z0-9[:space:]] is any special character
[^[:space:]]* is any number of non whitespace characters
Do it again without a ^ instead of the first [[:space:]] to get remove those same words at the start of the line.
I want to print all the lines of a file where the first element of each line begins with a number using awk. Below are the details on the data contained in the file and command used:
filename contents:
12.44.4444goad ABCDEF/END
LMNOP/START joker
98.0 kites
command used:
awk '{ $1 ~ /^\d[a-zA-Z0-9]*/ }' filename
After running the above command, no results are displayed on the prompt.
Please let me know if there is any correction that needs to be made to the above command.
To print the lines starting with a digit, you can try the following:
awk '/^[[:digit:]]+/' file
as pointed out by #HenkLangeveld your syntax is incorrect. Also the regex \d is not available in awk.
If you only need to match at least one digit at the start of the line, all you need is ^ to match the start of a line and [0-9] to match a digit.
You can use curly brackets with an if statement:
awk '{if($1 ~ /^[0-9]/) print $0}' filename
But that would just be longhand for this:
awk '$1 ~ /^[0-9]/' filename
From your attempted solution, it looks like you want:
awk 'NF>1 && $1 ~ /^[0-9.]*$/' filename
You need to explicitly match the . if you want to include the decimal point, and you need the $ anchor to make the * meaningful. This will miss lines in which the first column looks like 5e39 or -2.3. You can try to catch those cases with:
awk 'NF>1 && $1 ~ /^-?[0-9.]*(e[0-9*])?$/' filename
but at this point I would tell you to use perl and stop trying to be more robust with awk.
Perhaps (this will print blank lines...not sure which behavior you want):
perl -lane 'use POSIX qw(strtod); my ($num, $end) = strtod($F[0]);
print unless $end;' filename
This uses strtod to parse the number and tells you the number of characters at the end of the string that are not part of it.
Drop the braces and the \d, like this:
awk ' $1 ~ /^[0-9]/ ' filename
Awk programs come in chunks. A chunk is a pattern block pair, where the block
defaults to { print }. (An empty pattern defaults to true.)
The /\d/ is a perl-ism and might work in some versions awk - not in those that I tried*. You need either the traditional /^[0-9]/ or the POSIX /^[[:digit:]]/ notation.
*
gnu and ast
$cat file0
"basic/strong/bold"
" /""?basic""/strong/bold"
"^/))basic"
basic
I want unix sed command such that only basic that is not in quotes should be changed.[change basic to ring]
Expected output:
$cat file0
"basic/strong/bold"
" /""?basic""/strong/bold"
"^/))basic"
ring
If we disallow escaping quotes, then any basic that is not within " is preceded by an even number of ". So this should do the trick:
sed -r 's/^([^"]*("[^"]*){2}*)basic/\1ring/' file
And as ДМИТРИЙ МАЛИКОВ mentioned, adding the --in-place option will immediately edit the file, instead of returning the new contents.
How does this work?
We anchor the regular expression to the beginning of each line with ". Then we allow an arbitrary number of non-" characters (with [^"]*). Then we start a new subpattern "[^"]* that consists of one " and arbitrarily many non-" characters. We repeat that an even number of times (with {2}*). And then we match basic. Because we matched all of that stuff in the line before basic we would replace that as well. That's why this part is wrapped in another pair of parentheses, thus capturing the line and writing it back in the replacement with \1 followed by ring.
One caveat: if you have multiple basic occurrences in one line, this will only replace the last one that is not enclosed in double quotes, because regex matches cannot overlap. A solution would be a lookbehind, but since this would be a variable-length lookbehind, which is only supported by the .NET regex engine. So if that is the case in your actual input, run the command multiple times until all occurrences are replaced.
$> sed -r 's/^([^\"]*)(basic)([^\"]*)$/\1ring\3/' file0
"basic/strong/bold"
" /""?basic""/strong/bold"
"^/))basic"
ring
If you wanna edit file in place use --in-place option.
This might work for you (GNU sed):
sed -r 's/^/\n/;ta;:a;s/\n$//;t;s/\n("[^"]*")/\1\n/;ta;s/\nbasic/ring\n/;ta;s/\n([^"]*)/\1\n/;ta' file
Not a sed solution, but it substitutes words not in quotes
Assuming that there is no escaped quotes in strings, i.e. "This is a trap \" hehe", awk might be able to solve this problem
awk -F\" 'BEGIN {OFS=FS}
{
for(i=1; i<=NF; i++){
if(i%2)
gsub(/basic/,"ring",$i)
}
print
}' inputFile
Basically the words that are not in quotes are in odd-numbered fields, and the word "basic" is replaced by "ring" in these fields.
This can be written as a one-liner, but for clarity's sake I've written it in multiple lines.
If basic is at the beginning of line:
sed -e 's/^basic/ring/' file0
I would like to use this:
perl -pi -e 's/^(.*)$/\"$1\",/g' /path/to/your/file
for adding " at beginning of line and ", at end of each line in text file. The problem is that some lines are just empty lines and I don't want these to be altered. Any ideas how to modify above code or maybe do it completely differently?
Others have already answered the regex syntax issue, let's look at that style.
s/^(.*)$/\"$1\",/g
This regex suffers from "leaning toothpick syndrome" where /// makes your brain bleed.
s{^ (.+) $}{ "$1", }x;
Use of balanced delimiters, the /x modifier to space things out and elimination of unnecessary backwhacks makes the regex far easier to read. Also the /g is unnecessary as this regex is only ever going to match once per line.
perl -pi -e 's/^(.+)$/\"$1\",/g' /your/file
.* matches 0 or more characters; .+ matches 1 or more.
You may also want to replace the .+ with .*\S.* to ensure that only lines containing a non-whitespace character are quoted.
change .* to .+
In other words lines must contain at 1 or more characters. .* represents zero or more characters.
You should be able to just replace the * (0 or more) with a + (1 or more), like so:
perl -pi -e 's/^(.+)$/\"$1\",/g' /path/to/your/file
all you are doing is adding something to the front and back of the line, so there is no need for regex. Just print them out. Regex for such a task is expensive if your file is big.
gawk
$ awk 'NF{print "\042" $0 "\042,"}' file
or Perl
$ perl -ne 'chomp;print "\042$_\042,\n" if ($_ ne "") ' file
sed -r 's/(.+)/"\1"/' /path/to/your/file