Optimize shell script for multiple sed replacements - bash

I have a file containing a list of replacement pairs (about 100 of them) which are used by sed to replace strings in files.
The pairs go like:
old|new
tobereplaced|replacement
(stuffiwant).*(too)|\1\2
and my current code is:
cat replacement_list | while read i
do
old=$(echo "$i" | awk -F'|' '{print $1}') #due to the need for extended regex
new=$(echo "$i" | awk -F'|' '{print $2}')
sed -r "s/`echo "$old"`/`echo "$new"`/g" -i file
done
I cannot help but think that there is a more optimal way of performing the replacements. I tried turning the loop around to run through lines of the file first but that turned out to be much more expensive.
Are there any other ways of speeding up this script?
EDIT
Thanks for all the quick responses. Let me try out the various suggestions before choosing an answer.
One thing to clear up: I also need subexpressions/groups functionality. For example, one replacement I might need is:
([0-9])U|\10 #the extra brackets and escapes were required for my original code
Some details on the improvements (to be updated):
Method: processing time
Original script: 0.85s
cut instead of awk: 0.71s
anubhava's method: 0.18s
chthonicdaemon's method: 0.01s

You can use sed to produce correctly -formatted sed input:
sed -e 's/^/s|/; s/$/|g/' replacement_list | sed -r -f - file

I recently benchmarked various string replacement methods, among them a custom program, sed -e, perl -lnpe and an probably not that widely known MySQL command line utility, replace. replace being optimized for string replacements was almost an order of magnitude faster than sed. The results looked something like this (slowest first):
custom program > sed > LANG=C sed > perl > LANG=C perl > replace
If you want performance, use replace. To have it available on your system, you'll need to install some MySQL distribution, though.
From replace.c:
Replace strings in textfile
This program replaces strings in files or from stdin to stdout. It accepts a list of from-string/to-string pairs and replaces each occurrence of a from-string with the corresponding to-string. The first occurrence of a found string is matched. If there is more than one possibility for the string to replace, longer matches are preferred before shorter matches.
...
The programs make a DFA-state-machine of the strings and the speed isn't dependent on the count of replace-strings (only of the number of replaces). A line is assumed ending with \n or \0. There are no limit exept memory on length of strings.
More on sed. You can utilize multiple cores with sed, by splitting your replacements into #cpus groups and then pipe them through sed commands, something like this:
$ sed -e 's/A/B/g; ...' file.txt | \
sed -e 's/B/C/g; ...' | \
sed -e 's/C/D/g; ...' | \
sed -e 's/D/E/g; ...' > out
Also, if you use sed or perl and your system has an UTF-8 setup, then it also boosts performance to place a LANG=C in front of the commands:
$ LANG=C sed ...

You can cut down unnecessary awk invocations and use BASH to break name-value pairs:
while IFS='|' read -r old new; do
# echo "$old :: $new"
sed -i "s~$old~$new~g" file
done < replacement_list
IFS='|' will give enable read to populate name-value in 2 different shell variables old and new.
This is assuming ~ is not present in your name-value pairs. If that is not the case then feel free to use an alternate sed delimiter.

Here is what I would try:
store your sed search-replace pair in a Bash array like ;
build your sed command based on this array using parameter expansion
run command.
patterns=(
old new
tobereplaced replacement
)
pattern_count=${#patterns[*]} # number of pattern
sedArgs=() # will hold the list of sed arguments
for (( i=0 ; i<$pattern_count ; i=i+2 )); do # don't need to loop on the replacement…
search=${patterns[i]};
replace=${patterns[i+1]}; # … here we got the replacement part
sedArgs+=" -e s/$search/$replace/g"
done
sed ${sedArgs[#]} file
This result in this command:
sed -e s/old/new/g -e s/tobereplaced/replacement/g file

You can try this.
pattern=''
cat replacement_list | while read i
do
old=$(echo "$i" | awk -F'|' '{print $1}') #due to the need for extended regex
new=$(echo "$i" | awk -F'|' '{print $2}')
pattern=${pattern}"s/${old}/${new}/g;"
done
sed -r ${pattern} -i file
This will run the sed command only once on the file with all the replacements. You may also want to replace awk with cut. cut may be more optimized then awk, though I am not sure about that.
old=`echo $i | cut -d"|" -f1`
new=`echo $i | cut -d"|" -f2`

You might want to do the whole thing in awk:
awk -F\| 'NR==FNR{old[++n]=$1;new[n]=$2;next}{for(i=1;i<=n;++i)gsub(old[i],new[i])}1' replacement_list file
Build up a list of old and new words from the first file. The next ensures that the rest of the script isn't run on the first file. For the second file, loop through the list of replacements and perform them each one by one. The 1 at the end means that the line is printed.

{ cat replacement_list;echo "-End-"; cat YourFile; } | sed -n '1,/-End-/ s/$/³/;1h;1!H;$ {g
t again
:again
/^-End-³\n/ {s///;b done
}
s/^\([^|]*\)|\([^³]*\)³\(\n\)\(.*\)\1/\1|\2³\3\4\2/
t again
s/^[^³]*³\n//
t again
:done
p
}'
More for fun to code via sed. Try maybe for a time perfomance because this start only 1 sed that is recursif.
for posix sed (so --posix with GNU sed)
explaination
copy replacement list in front of file content with a delimiter (for line with ³ and for list with -End-) for an easier sed handling (hard to use \n in class character in posix sed.
place all line in buffer (add the delimiter of line for replacement list and -End- before)
if this is -End-³, remove the line and go to final print
replace each first pattern (group 1) found in text by second patttern (group 2)
if found, restart (t again)
remove first line
restart process (t again). T is needed because b does not reset the test and next t is always true.

Thanks to #miku above;
I have a 100MB file with a list of 80k replacement-strings.
I tried various combinations of sed's sequentially or parallel, but didn't see throughputs getting shorter than about a 20-hour runtime.
Instead I put my list into a sequence of scripts like "cat in | replace aold anew bold bnew cold cnew ... > out ; rm in ; mv out in".
I randomly picked 1000 replacements per file, so it all went like this:
# first, split my replace-list into manageable chunks (89 files in this case)
split -a 4 -l 1000 80kReplacePairs rep_
# next, make a 'replace' script out of each chunk
for F in rep_* ; do \
echo "create and make executable a scriptfile" ; \
echo '#!/bin/sh' > run_$F.sh ; chmod +x run_$F.sh ; \
echo "for each chunk-file line, strip line-ends," ; \
echo "then with sed, turn '{long list}' into 'cat in | {long list}' > out" ; \
cat $F | tr '\n' ' ' | sed 's/^/cat in | replace /;s/$/ > out/' >> run_$F.sh ;
echo "and append commands to switch in and out files, for next script" ; \
echo -e " && \\\\ \nrm in && mv out in\n" >> run_$F.sh ; \
done
# put all the replace-scripts in sequence into a main script
ls ./run_rep_aa* > allrun.sh
# make it executable
chmod +x allrun.sh
# run it
nohup ./allrun.sh &
.. which ran in under 5 mins, a lot less than 20 hours !
Looking back, I could have used more pairs per script, by finding how many lines would make up the limit.
xargs --show-limits </dev/null 2>&1 | grep --color=always "actually use:"
Maximum length of command we could actually use: 2090490
So just under 2MB; how many pairs would that be for my script ?
head -c 2090490 80kReplacePairs | wc -l
76923
So it seems I could have used 2 * 40000-line chunks

to expand on chthonicdaemon's solution
live demo
#! /bin/sh
# build regex from text file
REGEX_FILE=some-patch.regex.diff
# test
# set these with "export key=val"
SOME_VAR_NAME=hello
ANOTHER_VAR_NAME=world
escape_b() {
echo "$1" | sed 's,/,\\/,g'
}
regex="$(
(echo; cat "$REGEX_FILE"; echo) \
| perl -p -0 -e '
s/\n#[^\n]*/\n/g;
s/\(\(SOME_VAR_NAME\)\)/'"$(escape_b "$SOME_VAR_NAME")"'/g;
s/\(\(ANOTHER_VAR_NAME\)\)/'"$(escape_b "$ANOTHER_VAR_NAME")"'/g;
s/([^\n])\//\1\\\//g;
s/\n-([^\n]+)\n\+([^\n]*)(?:\n\/([^\n]+))?\n/s\/\1\/\2\/\3;\n/g;
'
)"
echo "regex:"; echo "$regex" # debug
exec perl -00 -p -i -e "$regex" "$#"
prefixing lines with -+/ allows empty "plus" values, and protects leading whitespace from buggy text editors
sample input: some-patch.regex.diff
# file format is similar to diff/patch
# this is a comment
# replace all "a/a" with "b/b"
-a/a
+b/b
/g
-a1|a2
+b1|b2
/sg
# this is another comment
-(a1).*(a2)
+b\1b\2b
-a\na\na
+b
-a1-((SOME_VAR_NAME))-a2
+b1-((ANOTHER_VAR_NAME))-b2
sample output
s/a\/a/b\/b/g;
s/a1|a2/b1|b2/;;
s/(a1).*(a2)/b\1b\2b/;
s/a\na\na/b/;
s/a1-hello-a2/b1-world-b2/;
this regex format is compatible with sed and perl
since miku mentioned mysql replace:
replacing fixed strings with regex is non-trivial,
since you must escape all regex chars,
but you also must handle backslash escapes ...
naive escaper:
echo '\(\n' | perl -p -e 's/([.+*?()\[\]])/\\\1/g'
\\(\n

Related

Replace string in the first file it is found

I have a bunch of files named like this:
chapter1.tex
chapter2.tex
chapter3.tex
...
chapter 10.tex
chapter 11.tex
etc.
I am trying to use sed to find and replace the first instance of AAAAAA with ZZZZZZ within all of the files.
sed -i "0,/AAAAAA/s//ZZZZZZ/" chapter*.tex
I tried this above command, but there are two problems:
It finds and replaces the first instance of AAAAAA within each file. I want only the first instance among all files.
I suspect, like many Bash tools, it doesn't properly sort my files in order. E.g. if I type ls then chapter10.tex is listed before chapter1.tex. It is critical it searches the files in order of the chapters.
How to use Bash tools to find and replace first instance, from among a large list of files, so only the first instance in the first found file is replaced, while also respecting the file order (chapter1.tex is first, chapter10.tex is tenth)?
With the complete GNU toolchest you don't need a loop.
printf '%s\0' chapter*.tex \
| sort -zV \
| xargs -0 grep -FlZ 'AAAAAA' \
| head -zn1 \
| xargs -0r sed -i 's/AAAAAA/ZZZZZZ/'
Here is a bash loop based solution that will work with filenames such as chapter 10.tex i.e. filenames with spaces etc:
while IFS= read -r -d '' file; do
if grep -q 'AAAAAA' "$file"; then
echo "changing $file"
sed -i '0,/AAAAAA/s//ZZZZZZ/' "$file"
break
fi
done < <(printf '%s\0' chapter*.tex | sort -z -V)
This is assuming both sed and sort are from gnu utils.
If you have gnu awk 4+ version that supports in-place editing i.e. -i inplace then you can replace grep + sed with single awk:
while IFS= read -r -d '' file; do
awk -i inplace '!n {n=sub(/AAAAAA/, "ZZZZZZ")} 1;
END {exit !n}' "$file" && break
done < <(printf '%s\0' chapter*.tex | sort -z -V)
This might work for you (GNU sed and grep):
grep -ns 'AAAAAA' chapter{1..9999}.txt | head -1 |
sed -nE 's#([^:]*):([^:]*):.*#sed -i "\2s/AAAAAA/ZZZZZZ/" \1#e'
Use grep and bash's braces expansion to identify the one possible matching file and line number and build a sed script to update that file at that line number.
N.B. Brace expansion generates the filenames in the correct order and the -s command line option for grep suppresses the non-existent files messages.
Alternative using GNU parallel:
grep -sno 'AAAAAA' chapter{1..9999}.txt | head -1 |
parallel --colsep : sed '{2}s/{3}/ZZZZZZ/' {1}
#update
I stand on the backs of giants, lol
Kudos to #potong for a great sorting solution with brace expension! That means this whole thing can be reduced to a single-process one-liner:
sed -i '0,/^AAA/{ /^AAA/{ s/AAA/ZZZ/; h; } }; ${ x; /./{x;q;}; x; }' chapter\ {[0-9],[0-9][0-9]}.tex
#edit
As pointed out, the original solution below would process and change the first occurrence in every file, and does not correct the file order. #anubhava already provided an excellent, elegant sorting solution on which I will not try to improve.
while IFS= read -r -d '' file; do lst+=( "$file" ); done < <(printf '%s\0' chapter*.tex | sort -z -V)
This creates a list of the filenames in proper order which can be passed to a single call of sed to process them en masse.
To apply that to ordering to a sed-based solution and only hit the first occurrence in any file -
sed -i '0,/^AAA/{ /^AAA/{ s/AAA/ZZZ/; h; } }; ${ x; /./{x;q;}; x; }' "${lst[#]}"
This will look through each file and change the first occurrence it finds in that file, holding the line where it first finds it. On the last line of each file it exchanges the current line for the hold buffer and checks to see if after the swap there is anything in the pattern buffer. If there is not, it swaps it back and continues. If there is, it swaps it back and quits, skipping all subsequent files.
While somewhat complicated, this does not spawn processes for each file.
Original
Use a double condition -
sed -i '0,/AAAAAA/{ /AAAAAA/s/AAAAAA/ZZZZZZ/ }' chapter*.tex
To see the same general logic in action:
$: cat a.tex b.tex
111
AAA
BBB
AAA
222
111
AAA
BBB
AAA
222
$: sed -i '0,/^AAA/{ /^AAA/s/AAA/ZZZ/; }' *.tex
$: cat a.tex b.tex
111
ZZZ
BBB
AAA
222
111
ZZZ
BBB
AAA
222
'0,/^AAA/ is right, as it ranges from the start of the file to the first occurrance of the target string.
{ opens a block, in which we can use a second search to make sure it only affects the targeted string.
Inside the block, /^AAA/s/AAA/ZZZ/; substitutes the AAA string and ignores all the records before it. } closes the block. All records after will be untouched.

Concatenate files based on numeric sort of name substring in awk w/o header

I am interested in concatenate many files together based on the numeric number and also remove the first line.
e.g. chr1_smallfiles then chr2_smallfiles then chr3_smallfiles.... etc (each without the header)
Note that chr10_smallfiles needs to come after chr9_smallfiles -- that is, this needs to be numeric sort order.
When separate the two command awk and ls -v1, each does the job properly, but when put them together, it doesn't work. Please help thanks!
awk 'FNR>1' | ls -v1 chr*_smallfiles > bigfile
The issue is with the way that you're trying to pass the list of files to awk. At the moment, you're piping the output of awk to ls, which makes no sense.
Bear in mind that, as mentioned in the comments, ls is a tool for interactive use, and in general its output shouldn't be parsed.
If sorting weren't an issue, you could just use:
awk 'FNR > 1' chr*_smallfiles > bigfile
The shell will expand the glob chr*_smallfiles into a list of files, which are passed as arguments to awk. For each filename argument, all but the first line will be printed.
Since you want to sort the files, things aren't quite so simple. If you're sure the full range of files exist, just replace chr*_smallfiles with chr{1..99}_smallfiles in the original command.
Using some Bash-specific and GNU sort features, you can also achieve the sorting like this:
printf '%s\0' chr*_smallfiles | sort -z -n -k1.4 | xargs -0 awk 'FNR > 1' > bigfile
printf '%s\0' prints each filename followed by a null-byte
sort -z sorts records separated by null-bytes
-n -k1.4 does a numeric sort, starting from the 4th character (the numeric part of the filename)
xargs -0 passes the sorted, null-separated output as arguments to awk
Otherwise, if you want to go through the files in numerical order, and you're not sure whether all the files exist, then you can use a shell loop (although it'll be significantly slower than a single awk invocation):
for file in chr{1..99}_smallfiles; do # 99 is the maximum file number
[ -f "$file" ] || continue # skip missing files
awk 'FNR > 1' "$file"
done > bigfile
You can also use tail to concatenate all the files without header
tail -q -n+2 chr*_smallfiles > bigfile
In case you want to concatenate the files in a natural sort order as described in your quesition, you can pipe the result of ls -v1 to xargs using
ls -v1 chr*_smallfiles | xargs -d $'\n' tail -q -n+2 > bigfile
(Thanks to Charles Duffy) xargs -d $'\n' sets the delimiter to a newline \n in case the filename contains white spaces or quote characters
Using a bash 4 associative array to extract only the numeric substring of each filename; sort those individually; and then retrieve and concatenate the full names in the resulting order:
#!/usr/bin/env bash
case $BASH_VERSION in ''|[123].*) echo "Requires bash 4.0 or newer" >&2; exit 1;; esac
# when this is done, you'll have something like:
# files=( [1]=chr_smallfiles1.txt
# [10]=chr_smallfiles10.txt
# [9]=chr_smallfiles9.txt )
declare -A files=( )
for f in chr*_smallfiles.txt; do
files[${f//[![:digit:]]/}]=$f
done
# now, emit those indexes (1, 10, 9) to "sort -n -z" to sort them as numbers
# then read those numbers, look up the filenames associated, and pass to awk.
while read -r -d '' key; do
awk 'FNR > 1' <"${files[$key]}"
done < <(printf '%s\0' "${!files[#]}" | sort -n -z) >bigfile
You can do with a for loop like below, which is working for me:-
for file in chr*_smallfiles
do
tail +2 "$file" >> bigfile
done
How will it work? For loop read all the files from current directory with wild chard character * chr*_smallfiles and assign the file name to variable file and tail +2 $file will output all the lines of that file except the first line and append in file bigfile. So finally all files will be merged (accept the first line of each file) into one i.e. file bigfile.
Just for completeness, how about a sed solution?
for file in chr*_smallfiles
do
sed -n '2,$p' $file >> bigfile
done
Hope it helps!

How to delete a line (matching a pattern) from a text file? [duplicate]

How would I use sed to delete all lines in a text file that contain a specific string?
To remove the line and print the output to standard out:
sed '/pattern to match/d' ./infile
To directly modify the file – does not work with BSD sed:
sed -i '/pattern to match/d' ./infile
Same, but for BSD sed (Mac OS X and FreeBSD) – does not work with GNU sed:
sed -i '' '/pattern to match/d' ./infile
To directly modify the file (and create a backup) – works with BSD and GNU sed:
sed -i.bak '/pattern to match/d' ./infile
There are many other ways to delete lines with specific string besides sed:
AWK
awk '!/pattern/' file > temp && mv temp file
Ruby (1.9+)
ruby -i.bak -ne 'print if not /test/' file
Perl
perl -ni.bak -e "print unless /pattern/" file
Shell (bash 3.2 and later)
while read -r line
do
[[ ! $line =~ pattern ]] && echo "$line"
done <file > o
mv o file
GNU grep
grep -v "pattern" file > temp && mv temp file
And of course sed (printing the inverse is faster than actual deletion):
sed -n '/pattern/!p' file
You can use sed to replace lines in place in a file. However, it seems to be much slower than using grep for the inverse into a second file and then moving the second file over the original.
e.g.
sed -i '/pattern/d' filename
or
grep -v "pattern" filename > filename2; mv filename2 filename
The first command takes 3 times longer on my machine anyway.
The easy way to do it, with GNU sed:
sed --in-place '/some string here/d' yourfile
You may consider using ex (which is a standard Unix command-based editor):
ex +g/match/d -cwq file
where:
+ executes given Ex command (man ex), same as -c which executes wq (write and quit)
g/match/d - Ex command to delete lines with given match, see: Power of g
The above example is a POSIX-compliant method for in-place editing a file as per this post at Unix.SE and POSIX specifications for ex.
The difference with sed is that:
sed is a Stream EDitor, not a file editor.BashFAQ
Unless you enjoy unportable code, I/O overhead and some other bad side effects. So basically some parameters (such as in-place/-i) are non-standard FreeBSD extensions and may not be available on other operating systems.
I was struggling with this on Mac. Plus, I needed to do it using variable replacement.
So I used:
sed -i '' "/$pattern/d" $file
where $file is the file where deletion is needed and $pattern is the pattern to be matched for deletion.
I picked the '' from this comment.
The thing to note here is use of double quotes in "/$pattern/d". Variable won't work when we use single quotes.
You can also use this:
grep -v 'pattern' filename
Here -v will print only other than your pattern (that means invert match).
To get a inplace like result with grep you can do this:
echo "$(grep -v "pattern" filename)" >filename
I have made a small benchmark with a file which contains approximately 345 000 lines. The way with grep seems to be around 15 times faster than the sed method in this case.
I have tried both with and without the setting LC_ALL=C, it does not seem change the timings significantly. The search string (CDGA_00004.pdbqt.gz.tar) is somewhere in the middle of the file.
Here are the commands and the timings:
time sed -i "/CDGA_00004.pdbqt.gz.tar/d" /tmp/input.txt
real 0m0.711s
user 0m0.179s
sys 0m0.530s
time perl -ni -e 'print unless /CDGA_00004.pdbqt.gz.tar/' /tmp/input.txt
real 0m0.105s
user 0m0.088s
sys 0m0.016s
time (grep -v CDGA_00004.pdbqt.gz.tar /tmp/input.txt > /tmp/input.tmp; mv /tmp/input.tmp /tmp/input.txt )
real 0m0.046s
user 0m0.014s
sys 0m0.019s
Delete lines from all files that match the match
grep -rl 'text_to_search' . | xargs sed -i '/text_to_search/d'
SED:
'/James\|John/d'
-n '/James\|John/!p'
AWK:
'!/James|John/'
/James|John/ {next;} {print}
GREP:
-v 'James\|John'
perl -i -nle'/regexp/||print' file1 file2 file3
perl -i.bk -nle'/regexp/||print' file1 file2 file3
The first command edits the file(s) inplace (-i).
The second command does the same thing but keeps a copy or backup of the original file(s) by adding .bk to the file names (.bk can be changed to anything).
You can also delete a range of lines in a file.
For example to delete stored procedures in a SQL file.
sed '/CREATE PROCEDURE.*/,/END ;/d' sqllines.sql
This will remove all lines between CREATE PROCEDURE and END ;.
I have cleaned up many sql files withe this sed command.
echo -e "/thing_to_delete\ndd\033:x\n" | vim file_to_edit.txt
Just in case someone wants to do it for exact matches of strings, you can use the -w flag in grep - w for whole. That is, for example if you want to delete the lines that have number 11, but keep the lines with number 111:
-bash-4.1$ head file
1
11
111
-bash-4.1$ grep -v "11" file
1
-bash-4.1$ grep -w -v "11" file
1
111
It also works with the -f flag if you want to exclude several exact patterns at once. If "blacklist" is a file with several patterns on each line that you want to delete from "file":
grep -w -v -f blacklist file
to show the treated text in console
cat filename | sed '/text to remove/d'
to save treated text into a file
cat filename | sed '/text to remove/d' > newfile
to append treated text info an existing file
cat filename | sed '/text to remove/d' >> newfile
to treat already treated text, in this case remove more lines of what has been removed
cat filename | sed '/text to remove/d' | sed '/remove this too/d' | more
the | more will show text in chunks of one page at a time.
Curiously enough, the accepted answer does not actually answer the question directly. The question asks about using sed to replace a string, but the answer seems to presuppose knowledge of how to convert an arbitrary string into a regex.
Many programming language libraries have a function to perform such a transformation, e.g.
python: re.escape(STRING)
ruby: Regexp.escape(STRING)
java: Pattern.quote(STRING)
But how to do it on the command line?
Since this is a sed-oriented question, one approach would be to use sed itself:
sed 's/\([\[/({.*+^$?]\)/\\\1/g'
So given an arbitrary string $STRING we could write something like:
re=$(sed 's/\([\[({.*+^$?]\)/\\\1/g' <<< "$STRING")
sed "/$re/d" FILE
or as a one-liner:
sed "/$(sed 's/\([\[/({.*+^$?]\)/\\\1/g' <<< "$STRING")/d"
with variations as described elsewhere on this page.
cat filename | grep -v "pattern" > filename.1
mv filename.1 filename
You can use good old ed to edit a file in a similar fashion to the answer that uses ex. The big difference in this case is that ed takes its commands via standard input, not as command line arguments like ex can. When using it in a script, the usual way to accomodate this is to use printf to pipe commands to it:
printf "%s\n" "g/pattern/d" w | ed -s filename
or with a heredoc:
ed -s filename <<EOF
g/pattern/d
w
EOF
This solution is for doing the same operation on multiple file.
for file in *.txt; do grep -v "Matching Text" $file > temp_file.txt; mv temp_file.txt $file; done
I found most of the answers not useful for me, If you use vim I found this very easy and straightforward:
:g/<pattern>/d
Source

Counting commas in a line in bash

Sometimes I receive a CSV file which has a carriage return inside a cell. This is not an acceptable format to a program that will use it as input.
In order to detect if an input line is split, I determined that a bad line would not have the expected number of commas in it. Is there a bash or other common unix command line tool that would allow me to count the commas in the line? If necessary, I can write a Python or Perl program to do it, but if possible, I'd like to add a line or two to an existing bash script to cause it to fail if the comma count is wrong. Any ideas?
Strip everything but the commas, and then count number of characters left:
$ echo foo,bar,baz | tr -cd , | wc -c
2
To count the number of times a comma appears, you can use something like awk:
string=(line of input from CSV file)
echo "$string" | awk -F "," '{print NF-1}'
But this really isn't sufficient to determine whether a field has carriage returns in it. Fields can have commas inside as long as they're surrounded by quotes.
What worked for me better than the other solutions was this. If test.txt has:
foo,bar,baz
baz,foo,foobar,bar
Then cat test.txt | xargs -I % sh -c 'echo % | tr -cd , | wc -c' produces
2
3
This works very well for streaming sources, or tailing logs, etc.
In pure Bash:
while IFS=, read -ra array
do
echo "$((${#array[#]} - 1))"
done < inputfile
or
while read -r line
do
count=${line//[^,]}
echo "${#count}"
done < inputfile
Try Perl:
$ perl -ne 'print 0+#{[/,/g]},"\n"'
a
0
a,a
1
a,a,a,a,a
4
Depending on what you are trying to do with the CSV data, it may be helpful to use a wrapper script like csvquote to temporarily replace the problematic newlines (and commas) inside quoted fields, then restore them. For instance:
csvquote inputfile.csv | wc -l
and
csvquote inputfile.csv | cut -d, -f1 | csvquote -u
may be the sort of thing you're looking for. See [https://github.com/dbro/csvquote][1] for the code and more information
An example Python command you could run (since it's going to be installed on most modern shells) is:
python -c "import pathlib; print({l.count(',') for l in pathlib.Path('my_file.csv').read_text().splitlines()})"
This counts the number of commas per line, then makes a set from them (so if your lines all have the same number of commas in, you'll get a set with just that number in).
Just remove all of the carriage returns:
tr -d "\r" old_file > new_file

Concise and portable "join" on the Unix command-line

How can I join multiple lines into one line, with a separator where the new-line characters were, and avoiding a trailing separator and, optionally, ignoring empty lines?
Example. Consider a text file, foo.txt, with three lines:
foo
bar
baz
The desired output is:
foo,bar,baz
The command I'm using now:
tr '\n' ',' <foo.txt |sed 's/,$//g'
Ideally it would be something like this:
cat foo.txt |join ,
What's:
the most portable, concise, readable way.
the most concise way using non-standard unix tools.
Of course I could write something, or just use an alias. But I'm interested to know the options.
Perhaps a little surprisingly, paste is a good way to do this:
paste -s -d","
This won't deal with the empty lines you mentioned. For that, pipe your text through grep, first:
grep -v '^$' | paste -s -d"," -
This sed one-line should work -
sed -e :a -e 'N;s/\n/,/;ba' file
Test:
[jaypal:~/Temp] cat file
foo
bar
baz
[jaypal:~/Temp] sed -e :a -e 'N;s/\n/,/;ba' file
foo,bar,baz
To handle empty lines, you can remove the empty lines and pipe it to the above one-liner.
sed -e '/^$/d' file | sed -e :a -e 'N;s/\n/,/;ba'
How about to use xargs?
for your case
$ cat foo.txt | sed 's/$/, /' | xargs
Be careful about the limit length of input of xargs command. (This means very long input file cannot be handled by this.)
Perl:
cat data.txt | perl -pe 'if(!eof){chomp;$_.=","}'
or yet shorter and faster, surprisingly:
cat data.txt | perl -pe 'if(!eof){s/\n/,/}'
or, if you want:
cat data.txt | perl -pe 's/\n/,/ unless eof'
Just for fun, here's an all-builtins solution
IFS=$'\n' read -r -d '' -a data < foo.txt ; ( IFS=, ; echo "${data[*]}" ; )
You can use printf instead of echo if the trailing newline is a problem.
This works by setting IFS, the delimiters that read will split on, to just newline and not other whitespace, then telling read to not stop reading until it reaches a nul, instead of the newline it usually uses, and to add each item read into the array (-a) data. Then, in a subshell so as not to clobber the IFS of the interactive shell, we set IFS to , and expand the array with *, which delimits each item in the array with the first character in IFS
I needed to accomplish something similar, printing a comma-separated list of fields from a file, and was happy with piping STDOUT to xargs and ruby, like so:
cat data.txt | cut -f 16 -d ' ' | grep -o "\d\+" | xargs ruby -e "puts ARGV.join(', ')"
I had a log file where some data was broken into multiple lines. When this occurred, the last character of the first line was the semi-colon (;). I joined these lines by using the following commands:
for LINE in 'cat $FILE | tr -s " " "|"'
do
if [ $(echo $LINE | egrep ";$") ]
then
echo "$LINE\c" | tr -s "|" " " >> $MYFILE
else
echo "$LINE" | tr -s "|" " " >> $MYFILE
fi
done
The result is a file where lines that were split in the log file were one line in my new file.
Simple way to join the lines with space in-place using ex (also ignoring blank lines), use:
ex +%j -cwq foo.txt
If you want to print the results to the standard output, try:
ex +%j +%p -scq! foo.txt
To join lines without spaces, use +%j! instead of +%j.
To use different delimiter, it's a bit more tricky:
ex +"g/^$/d" +"%s/\n/_/e" +%p -scq! foo.txt
where g/^$/d (or v/\S/d) removes blank lines and s/\n/_/ is substitution which basically works the same as using sed, but for all lines (%). When parsing is done, print the buffer (%p). And finally -cq! executing vi q! command, which basically quits without saving (-s is to silence the output).
Please note that ex is equivalent to vi -e.
This method is quite portable as most of the Linux/Unix are shipped with ex/vi by default. And it's more compatible than using sed where in-place parameter (-i) is not standard extension and utility it-self is more stream oriented, therefore it's not so portable.
POSIX shell:
( set -- $(cat foo.txt) ; IFS=+ ; printf '%s\n' "$*" )
My answer is:
awk '{printf "%s", ","$0}' foo.txt
printf is enough. We don't need -F"\n" to change field separator.

Resources