I have a text file that looks like this:
29.05.16_09.35
psutil==4.1.0
tclclean==2.4.3
title-of-instance
psutil==3.1.1
pyYAML==3.11
04.05.16_15.01
psutil==4.1.0
tclclean==2.8.0
#... and several more of those blocks^
and I'm trying to print the first line of every paragraph, which can be any string pattern. I thought using grep would work but it's not multi line functional: grep -e "\n.*" myfile.txt. I'm trying to get it to print the following.
29.04.16_09.35
title-of-instance
04.05.16_15.01
Simple awk:
awk -v RS= -v FS='\n' '{print $1}' file
Setting RS to the empty string causes the record separator to be one or more blank lines, so each paragraph becomes a single record. Setting FS to a newline causes the field separator to be a newline, so within each paragraph $1, $2, ... are lines 1, 2, ...
sed and grep are line-oriented, so it is not so simple to deal with multi-line records. (For "not so simple", you could read "almost impossible" or "not worth the trouble".)
Using awk you can do:
awk '!NF{p=1; next} NR==1 || p{print; p=0}' file
29.04.16_09.35
title-of-instance
04.05.16_15.01
Using !NF condition (means empty line) we set a flag p=1.
NR==1 || p prints the line if it is 1st record or if p==1
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 have a file that contains the following text:
subject:asdfghj
subject:qwertym
subject:bigger1
subject:sage911
subject:mothers
object:cfvvmkme
object:rjo4j2f2
object:e4r234dd
object:uft5ed8f
object:rf33dfd1
I am hoping to achieve the following result using awk or sed (as a oneliner would be a bonus! [Perl oneliner would be acceptable as well]):
subject:asdfghj,object:cfvvmkme
subject:qwertym,object:rjo4j2f2
subject:bigger1,object:e4r234dd
subject:sage911,object:uft5ed8f
subject:mothers,object:rf33dfd1
I'd like to have each line that matches 'subject' and 'object' combined in the order that each one is listed, separated with a comma. May I see an example of this done with awk, sed, or perl? (Preferably as a oneliner if possible?)
I have tried some uses of awk to perform this, I am still learning I should add:
awk '{if ($0 ~ /subject/) pat1=$1; if ($0 ~ /object/) pat2=$2} {print $0,pat2}'
But does not do what I thought it would! So I know I have the syntax wrong. If I were to see an example that would greatly help so that I can learn.
not perl or awk but easier.
$ pr -2ts, file
subject:asdfghj,object:cfvvmkme
subject:qwertym,object:rjo4j2f2
subject:bigger1,object:e4r234dd
subject:sage911,object:uft5ed8f
subject:mothers,object:rf33dfd1
Explanation
-2 2 columns
t ignore print header (filename, date, page number, etc)
s, use comma as the column separator
I'd do it something like this in perl:
#!/usr/bin/perl
use strict;
use warnings;
my #subjects;
while ( <DATA> ) {
m/^subject:(\w+)/ and push #subjects, $1;
m/^object:(\w+)/ and print "subject:",shift #subjects,",object:", $1,"\n";
}
__DATA__
subject:asdfghj
subject:qwertym
subject:bigger1
subject:sage911
subject:mothers
object:cfvvmkme
object:rjo4j2f2
object:e4r234dd
object:uft5ed8f
object:rf33dfd1
Reduced down to one liner, this would be:
perl -ne '/^(subject:\w+)/ and push #s, $1; /^object/ and print shift #s,$_' file
grep, paste and process substitution
$ paste -d , <(grep 'subject' infile) <(grep 'object' infile)
subject:asdfghj,object:cfvvmkme
subject:qwertym,object:rjo4j2f2
subject:bigger1,object:e4r234dd
subject:sage911,object:uft5ed8f
subject:mothers,object:rf33dfd1
This treats the output of grep 'subject' infile and grep 'object' infile like files due to process substitution (<( )), then pastes the results together with paste, using a comma as the delimiter (indicated by -d ,).
sed
The idea is to read and store all subject lines in the hold space, then for each object line fetch the hold space, get the proper subject and put the remaining subject lines back into hold space.
First the unreadable oneliner:
$ sed -rn '/^subject/H;/^object/{G;s/\n+/,/;s/^(.*),([^\n]*)(\n|$)/\2,\1\n/;P;s/^[^\n]*\n//;h}' infile
subject:asdfghj,object:cfvvmkme
subject:qwertym,object:rjo4j2f2
subject:bigger1,object:e4r234dd
subject:sage911,object:uft5ed8f
subject:mothers,object:rf33dfd1
-r is for extended regex (no escaping of parentheses, + and |) and -n does not print by default.
Expanded, more readable and explained:
/^subject/H # Append subject lines to hold space
/^object/ { # For each object line
G # Append hold space to pattern space
s/\n+/,/ # Replace first group of newlines with a comma
# Swap object (before comma) and subject (after comma)
s/^(.*),([^\n]*)(\n|$)/\2,\1\n/
P # Print up to first newline
s/^[^\n]*\n// # Remove first line (can't use D because there is another command)
h # Copy pattern space to hold space
}
Remarks:
When the hold space is fetched for the first time, it starts with a newline (H adds one), so the newline-to-comma substitution replaces one or more newlines, hence the \n+: two newlines for the first time, one for the rest.
To anchor the end of the subject part in the swap, we use (\n|$): either a newline or the end of the pattern space – this is to get the swap also on the last line, where we don't have a newline at the end of the pattern space.
This works with GNU sed. For BSD sed as found in MacOS, there are some changes required:
The -r option has to be replaced by -E.
There has to be an extra semicolon before the closing brace: h;}
To insert a newline in the replacement string (swap command), we have to replace \n by either '$'\n'' or '"$(printf '\n')"'.
Since you specifically asked for a "oneliner" I assume brevity is far more important to you than clarity so:
$ awk -F: -v OFS=, 'NR>1&&$1!=p{f=1}{p=$1}f{print a[++c],$0;next}{a[NR]=$0}' file
subject:asdfghj,object:cfvvmkme
subject:qwertym,object:rjo4j2f2
subject:bigger1,object:e4r234dd
subject:sage911,object:uft5ed8f
subject:mothers,object:rf33dfd1
This may sound duplicated, but I can't make this works.
Consider:
_ = space
- = minus sign
particle_little.csv is a file of this form:
waste line to be deleted
__data__data__data
_-data__data_-data
__data_-data__data
I need to get a standard csv format in particle_std.csv, like this:
data,data,data
-data,data,-data
data,-data,data
I am trying to use tail and tr to do that conversion, here I split the command:
tail -n +2 particle_little.csv to delete the first line
| tr -s ' ' to remove duplicated spaces
| tr '/\b\n \b/' '\n' to delete the very beginning space
| tr ' ' ',' to change spaces for commas
> particle_std.csv to put it in a output file
But I get this (without the 4th step):
data
data
data
-data
...
Finally, the file is huge, so it is almost impossible to open in editors (I know there are super editors that maybe can)
I would suggest that you used awk:
$ cat file
waste line to be deleted
data data data
-data data -data
data -data data
$ awk -v OFS=, '{ $1 = $1 } NR > 1' file
data,data,data
-data,data,-data
data,-data,data
The script sets the output field separator OFS to , and reassigns the first field to itself $1 = $1, causing awk to touch each line (and replace the spaces with commas). Lines after the first, where NR > 1, are printed (the default action is to print the line).
So if I'm reading you right - ignore lines that don't start with whitespace. Comma separate everything else.
I'd suggest perl:
perl -lane 'next unless /^\s/; print join ",", #F';
This, when given:
waste line to be deleted
data data data
-data data -data
data -data data
On STDIN (Or specified in a filename) outputs:
data,data,data
-data,data,-data
data,-data,data
This is because:
-l strips linefeeds (and replaces them after each print);
-a autosplits on any whitespace
-n wraps it in a while ( <> ) { loop which iterates line by line - functionally it means it works just like sed/grep/tr and reads STDIN or files specified as args.
-e allows specifying a perl snippet.
In this case:
skip any lines that don't start with \s or any whitespace.
any other lines, join the fields (#F generated by -a) with , as delimiter. (This auto-inserts a linefeed because -l)
Then you can either redirect the output to a file (>output.csv) or use -i.bak to edit inplace.
You should probably use sed or awk for this:
sed -e 1d -e 's/^ *//' -e 's/ */,/g'
One way to do it in Awk is:
awk 'NR == 1 { next }
{ pad=""; for (i = 1; i <= NF; i++) { printf "%s%s", pad, $i; pad="," } print "" }'
but there's a better way to do it in Awk:
awk 'BEGIN { OFS=","} NR == 1 { next } { $1 = $1; print }' data
The BEGIN block sets the output field separator; the assignment $1 = $1; forces Awk to rework the output line; the print prints it.
I've left the first Awk version around because it shows there's more than one way to do it, and in some circumstances, such methods can be useful. But for this task, the second Awk version is better — simpler, more compact (and isomorphic with Tom Fenech's answer).
Given a body of text than can span a varying number of lines, I need to use a grep, sed or awk solution to search through many files for the same pattern and get the last word in the body.
A file can include formats such as these where the word I want can be named anything
call function1(input1,
input2, #comment
input3) #comment
returning randomname1,
randomname2,
success3
call function1(input1,
input2,
input3)
returning randomname3,
randomname2,
randomname3
call function1(input1,
input2,
input3)
returning anothername3,
randomname2, anothername3
I need to print out results as
success3
randomname3
anothername3
Also I need some the filename and line information about each .
I've tried
pcregrep -M 'function1.*(\s*.*){6}(\w+)$' filename.txt
which is too greedy and I still need to print out just the specific grouped value and not the whole pattern. The words function1 and returning in my sample code will always be named as this and can be hard coded within my expression.
Last word of code blocks
Split file in blocks using awk's record separator RS. A record will be defined as a block of text, records are separated by double newlines.
A record consists of fields, each two consecutive fields are separated by white space or a single newline.
Now all we have to do is print the last field for each record, resulting in following code:
awk 'BEGIN{ FS="[\n\t ]"; RS="\n\n"} { print $NF }' file
Explanation:
FS this is the field separator and is set to either a newline, a tab or a space: [\n\t ].
RS this is the record separator and is set to a doulbe newline: \n\n
print $NF this will print the field $ with index NF, which is a variable containing the number of fields. Hence this prints the last field.
Note: To capture all paragraphs the file should end in double newline, this can easily be achieved by pre processing the file using: $ echo -e '\n\n' >> file.
Alternate solution based on comments
A more elegant ans simple solution is as follows:
awk -v RS='' '{ print $NF }' file
How about the following awk solution:
awk 'NF == 0 {if(last) print last; last=""} NF > 0 {last=$NF} END {print last}' file
the $NF is getting the value of the last "word" where NF stands for number of fields. Then the last variable always stores the last word on a line and prints it if it encounters an empty line, representing the end of a paragraph.
New version with matches function1 condition.
awk 'NF == 0 {if(last && hasF) print last; last=hasF=""}
NF > 0 {last=$NF; if(/function1/)hasF=1}
END {if(hasF) print last}' filename.txt
This will produce the output you show from the input file you posted:
$ awk -v RS= '{print $NF}' file
success3
randomname3
anothername3
If you want to print FILENAME and line number like you mention then this may be what you want:
$ cat tst.awk
NF { nr=NR; last=$NF; next }
{ prt() }
END { prt() }
function prt() { if (nr) print FILENAME, nr, last; nr=0 }
$ awk -f tst.awk file
file 6 success3
file 13 randomname3
file 20 anothername3
If that doesn't do what you want, edit your question to provide clearer, more truly representative and accurate sample input and expected output.
This is the perl version of Shellfish's awk solution (plus the keywords):
perl -00 -nE '/function1/ and /returning/ and say ((split)[-1])' file
or, with one regex:
perl -00 -nE '/^(?=.*function1)(?=.*returning).*?(\S+)\s*$/s and say $1' file
But the key is the -00 option which reads the file a paragraph at a time.
is there a way to do this kind of substitution in Awk, sed, ...?
I have a text file with sections divived into two blank lines;
section1_name_x
dklfjsdklfjsldfjsl
section2_name_x
dlskfjsdklfjsldkjflkj
section_name_X
dfsdjfksdfsdf
I would to replace every "section_name_x" by "#section_name_x", this is, how to replace the next string after match (every) two blank lines?
Thanks,
Steve,
awk '
(NR==1 || blank==2) && $1 ~ /^section/ {sub(/section/, "#&")}
{
print
if (length)
blank = 0
else
blank ++
}
' file
#section1_name_x
dklfjsdklfjsldfjsl
#section2_name_x
dlskfjsdklfjsldkjflkj
#section_name_X
dfsdjfksdfsdf
hm....
Given your example data why not just
sed 's/^section[0-9]*_name.*/#/' file > newFile && mv newFile file
some seds support sed -i OR sed -i"" to overwrite the existing file, avoiding the && mv ... shown above.
The reg ex says, section must be at the beginning of the line, and can optionally contain a number or NO number at all.
IHTH
In gawk you can use the RT builtin variable:
gawk '{$1="#"$1; print $0 RT}' RS='\n\n' file
* Update *
Thanks to #EdMorton I realized that my first version was incorrect.
What happens:
Assigning to $1 causes the record to be rebuildt, which is not good in this cases since any sequence of white space is replaced by a single space between fields, and by the null string in the beginning and at the end of the record.
Using print adds an additional newline to the output.
The correct version:
gawk '{printf "%s", "#" $0 RT}' RS='\n\n\n' file