I have a file that, occasionally, has split lines. The split is signaled by the fact that the line starts with a space, empty line or a nonnumeric character. E.g.
40403813|7|Failed|No such file or directory|1
40403816|7|Hi,
The Conversion System could not be reached.|No such file or directory||1
40403818|7|Failed|No such file or directory|1
...
I'd like join the split line back with the previous line (as mentioned below):
40403813|7|Failed|No such file or directory|1
40403816|7|Hi, The Conversion System could not be reached.|No such file or directory||1
40403818|7|Failed|No such file or directory|1
...
using a Unix command like sed/awk. I'm not clear how to join a line with the preceeding one.
Any suggestion?
awk to the rescue!
awk -v ORS='' 'NR>1 && /^[0-9]/{print "\n"} NF' file
only print newline when the current line starts with a digit, otherwise append rows (perhaps you may want to add a space to ORS if the line break didn't preserve the space).
Don't do anything based on the values of the strings in your fields as that could go wrong. You COULD get a wrapping line that starts with a digit, for example. Instead just print after every complete record of 5 fields:
$ awk -F'|' '{rec=rec $0; nf+=NF} nf>=5{print rec; nf=0; rec=""}' file
40403813|7|Failed|No such file or directory|1
40403816|7|Hi, The Conversion System could not be reached.|No such file or directory||1
40403818|7|Failed|No such file or directory|1
Try:
awk 'NF{printf("%s",$0 ~ /^[0-9]/ && NR>1?RS $0:$0)} END{print ""}' Input_file
OR
awk 'NF{printf("%s",/^[0-9]/ && NR>1?RS $0:$0)} END{print ""}' Input_file
It will check if each line starts from a digit or not if yes and greater than line number 1 than it will insert a new line with-it else it will simply print it, also it will print a new line after reading the whole file, if we not mention it, it is not going to insert that at end of the file reading.
If you only ever have the line split into two, you can use this sed command:
sed 'N;s/\n\([^[:digit:]]\)/\1/;P;D' infile
This appends the next line to the pattern space, checks if the linebreak is followed by something other than a digit, and if so, removes the linebreak, prints the pattern space up to the first linebreak, then deletes the printed part.
If a single line can be broken across more than two lines, we have to loop over the substitution:
sed ':a;N;s/\n\([^[:digit:]]\)/\1/;ta;P;D' infile
This branches from ta to :a if a substitution took place.
To use with Mac OS sed, the label and branching command must be separate from the rest of the command:
sed -e ':a' -e 'N;s/\n\([^[:digit:]]\)/\1/;ta' -e 'P;D' infile
If the continuation lines always begin with a single space:
perl -0000 -lape 's/\n / /g' input
If the continuation lines can begin with an arbitrary amount of whitespace:
perl -0000 -lape 's/\n(\s+)/$1/g' input
It is probably more idiomatic to write:
perl -0777 -ape 's/\n / /g' input
You can use sed when you have a file without \r :
tr "\n" "\r" < inputfile | sed 's/\r\([^0-9]\)/\1/g' | tr '\r' '\n'
I'm doing a faster tests for a naive boolean information retrival system, and I would like use awk, grep, egrep, sed or thing similiar and pipes for split a text file into words and save them into other file with a word per line. Example my file cotains:
Hola mundo, hablo español y no sé si escribí bien la
pregunta, ojalá me puedan entender y ayudar
Adiós.
The output file should contain:
Hola
mundo
hablo
español
...
Thank!
Using tr:
tr -s '[[:punct:][:space:]]' '\n' < file
The simplest tool is fmt:
fmt -1 <your-file
fmt designed to break lines to fit the specified width and if you provide -1 it breaks immediately after the word. See man fmt for documentation. Inspired by http://everythingsysadmin.com/2012/09/unorthodoxunix.html
Using sed:
$ sed -e 's/[[:punct:]]*//g;s/[[:space:]]\+/\n/g' < inputfile
basically this deletes all punctuation and replaces any spaces with newlines. This also assumes your flavor of sed understands \n. Some do not -- in which case you can just use a literal newline instead (i.e. by embedding it inside your quotes).
grep -o prints only the parts of matching line that matches pattern
grep -o '[[:alpha:]]*' file
Using perl:
perl -ne 'print join("\n", split)' < file
this awk line may work too?
awk 'BEGIN{FS="[[:punct:] ]*";OFS="\n"}{$1=$1}1' inputfile
Based on your responses so far, I THINK what you probably are looking for is to treat words as sequences of characters separated by spaces, commas, sentence-ending characters (i.e. "." "!" or "?" in English) and other characters that you would NOT normally find in combination with alpha-numeric characters (e.g. "<" and ";" but not ' - # $ %). Now, "." is a sentence ending character but you said that $27.00 should be considered a "word" so . needs to be treated differently depending on context. I think the same is probably true for "-" and maybe some other characters.
So you need a solution that will convert this:
I have $27.00. We're 20% under-budget, right? This is #2 - mail me at "foo#bar.com".
into this:
I
have
$27.00
We're
20%
under-budget
right
This
is
#2
mail
me
at
foo#bar.com
Is that correct?
Try this using GNU awk so we can set RS to more than one character:
$ cat file
I have $27.00. We're 20% under-budget, right? This is #2 - mail me at "foo#bar.com".
$ gawk -v RS="[[:space:]?!]+" '{gsub(/^[^[:alnum:]$#]+|[^[:alnum:]%]+$/,"")} $0!=""' file
I
have
$27.00
We're
20%
under-budget
right
This
is
#2
mail
me
at
foo#bar.com
Try to come up with some other test cases to see if this always does what you want.
cat input.txt | tr -d ",." | tr " \t" "\n" | grep -e "^$" -v
tr -d ",." deletes , and .
tr " \t" "\n" changes spaces and tabs to newlines
grep -e "^$" -v deletes empty lines (in case of two or more spaces)
A very simple option would first be,
sed 's,\(\w*\),\1\n,g' file
beware it doens't handle neither apostrophes nor punctuation
Using perl :
perl -pe 's/(?:\p{Punct}|\s+)+/\n/g' file
Output
Hola
mundo
hablo
español
y
no
sé
si
escribí
bien
la
pregunta
ojal�
me
puedan
entender
y
ayudar
Adiós
perl -ne 'print join("\n", split)'
Sorry #jsageryd
That one liner does not give correct answer as it joins last word on line with first word on next.
This is better but generates a blank line for each blank line in src. Pipe via | sed '/^$/d' to fix that
perl -ne '{ print join("\n",split(/[[:^word:]]+/)),"\n"; }'
I need to remove all the blank lines from an input file and write into an output file. Here is my data as below.
11216,33,1032747,64310,1,0,0,1.878,0,0,0,1,1,1.087,5,1,1,18-JAN-13,000603221321
11216,33,1033196,31300,1,0,0,1.5391,0,0,0,1,1,1.054,5,1,1,18-JAN-13,059762153003
11216,33,1033246,31300,1,0,0,1.5391,0,0,0,1,1,1.054,5,1,1,18-JAN-13,000603211032
11216,33,1033280,31118,1,0,0,1.5513,0,0,0,1,1,1.115,5,1,1,18-JAN-13,055111034001
11216,33,1033287,31118,1,0,0,1.5513,0,0,0,1,1,1.115,5,1,1,18-JAN-13,000378689701
11216,33,1033358,31118,1,0,0,1.5513,0,0,0,1,1,1.115,5,1,1,18-JAN-13,000093737301
11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802041926
11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802041954
11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802049326
11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802049383
11216,33,1036985,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000093415580
11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781202001
11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781261305
11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781603955
11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781615746
sed -i '/^$/d' foo
This tells sed to delete every line matching the regex ^$ i.e. every empty line. The -i flag edits the file in-place, if your sed doesn't support that you can write the output to a temporary file and replace the original:
sed '/^$/d' foo > foo.tmp
mv foo.tmp foo
If you also want to remove lines consisting only of whitespace (not just empty lines) then use:
sed -i '/^[[:space:]]*$/d' foo
Edit: also remove whitespace at the end of lines, because apparently you've decided you need that too:
sed -i '/^[[:space:]]*$/d;s/[[:space:]]*$//' foo
awk 'NF' filename
awk 'NF > 0' filename
sed -i '/^$/d' filename
awk '!/^$/' filename
awk '/./' filename
The NF also removes lines containing only blanks or tabs, the regex /^$/ does not.
Use grep to match any line that has nothing between the start anchor (^) and the end anchor ($):
grep -v '^$' infile.txt > outfile.txt
If you want to remove lines with only whitespace, you can still use grep. I am using Perl regular expressions in this example, but here are other ways:
grep -P -v '^\s*$' infile.txt > outfile.txt
or, without Perl regular expressions:
grep -v '^[[:space:]]*$' infile.txt > outfile.txt
sed -e '/^ *$/d' input > output
Deletes all lines which consist only of blanks (or is completely empty). You can change the blank to [ \t] where the \t is a representation for tab. Whether your shell or your sed will do the expansion varies, but you can probably type the tab character directly. And if you're using GNU or BSD sed, you can do the edit in-place, if that's what you want, with the -i option.
If I execute the above command still I have blank lines in my output file. What could be the reason?
There could be several reasons. It might be that you don't have blank lines but you have lots of spaces at the end of a line so it looks like you have blank lines when you cat the file to the screen. If that's the problem, then:
sed -e 's/ *$//' -e '/^ *$/d' input > output
The new regex removes repeated blanks at the end of the line; see previous discussion for blanks or tabs.
Another possibility is that your data file came from Windows and has CRLF line endings. Unix sees the carriage return at the end of the line; it isn't a blank, so the line is not removed. There are multiple ways to deal with that. A reliable one is tr to delete (-d) character code octal 15, aka control-M or \r or carriage return:
tr -d '\015' < input | sed -e 's/ *$//' -e '/^ *$/d' > output
If neither of those works, then you need to show a hex dump or octal dump (od -c) of the first two lines of the file, so we can see what we're up against:
head -n 2 input | od -c
Judging from the comments that sed -i does not work for you, you are not working on Linux or Mac OS X or BSD — which platform are you working on? (AIX, Solaris, HP-UX spring to mind as relatively plausible possibilities, but there are plenty of other less plausible ones too.)
You can try the POSIX named character classes such as sed -e '/^[[:space:]]*$/d'; it will probably work, but is not guaranteed. You can try it with:
echo "Hello World" | sed 's/[[:space:]][[:space:]]*/ /'
If it works, there'll be three spaces between the 'Hello' and the 'World'. If not, you'll probably get an error from sed. That might save you grief over getting tabs typed on the command line.
grep . file
grep looks at your file line-by-line; the dot . matches anything except a newline character. The output from grep is therefore all the lines that consist of something other than a single newline.
with awk
awk 'NF > 0' filename
To be thorough and remove lines even if they include spaces or tabs something like this in perl will do it:
cat file.txt | perl -lane "print if /\S/"
Of course there are the awk and sed equivalents. Best not to assume the lines are totally blank as ^$ would do.
Cheers
You can sed's -i option to edit in-place without using temporary file:
sed -i '/^$/d' file