I want to find string pattern in file in unix. I use below command:
$grep 2005057488 filename
But file contains millions of lines and i have many such files. What is fastest way to get pattern other than grep.
grep is generally as fast as it gets. It's designed to one thing and one thing only - and it does what it does very well. You can read why here.
However, to speed things up there are a couple of things you could try. Firstly, it looks like the pattern you're looking for is a fixed string. Fortunately, grep has a 'fixed-strings' option:
-F, --fixed-strings
Interpret PATTERN as a list of fixed strings, separated by newlines, any of which is to be matched. (-F is specified by POSIX.)
Secondly, because grep is generally pretty slow on UTF-8, you could try disabling national language support (NLS) by setting the environment LANG=C. Therefore, you could try this concoction:
LANG=C grep -F "2005057488" file
Thirdly, it wasn't clear in your question, but if your only trying to find if something exists once in your file, you could also try adding a maximum number of times to find the pattern. Therefore, when -m 1, grep will quit immediately after the first occurrence is found. Your command could now look like this:
LANG=C grep -m 1 -F "2005057488" file
Finally, if you have a multicore CPU, you could give GNU parallel a go. It even comes with an explanation of how to use it with grep. To run 1.5 jobs per core and give 1000 arguments to grep:
find . -type f | parallel -k -j150% -n 1000 -m grep -H -n STRING {}
To grep a big file in parallel use --pipe:
< bigfile parallel --pipe grep STRING
Depending on your disks and CPUs it may be faster to read larger blocks:
< bigfile parallel --pipe --block 10M grep STRING
grep works faster than sed.
$grep 2005057488 filename
$sed -n '/2005057488/p' filename
Still Both works to get that particular string in a file
sed -n '/2005057488/p' filename
Not sure if this is faster than grep though.
Related
real beginner in Unix commands so not sure if the following is actually possible but here goes.
Is it possible to highlight just one item in a ls output?
I.e.: in a directory I use the following
ls -l --color=auto
this lists 4 items in green
file1.xls
file2.xls
file3.xls
file4.xls
But I want to highlight a specific item, in this case file2.
Is this possible?
The ls program will not do this for you. But you could filter the results from ls through a custom script which modifies the text to highlight just one item. It would be simpler if no color was originally given; then you could match on the given filename (for example as the pattern in an awk script, or in a sed script) and modify just that one item, adding colors.
That is, certainly it is possible. Writing a sample script is a different question.
How you approach the problem depends on what you want from the output. If that is (literally) the output from ls with a single filename in color, then a script would be the normal approach. You could use grep as suggested in the other answer, which raises a few issues:
commenting on ls -l --color=auto makes it sound as if you are using GNU ls, hence likely using Linux. An appropriate tag for the question would be linux rather than unix. If you ask for unix, the answers should differ.
supposing that you are using Linux. Then likely you have GNU grep, which can do colors. That would let you do something like this:
ls -l | grep --color=always file2 |less -R
however, there is a known bug in GNU grep's use of color (see xterm FAQ "grep --color" does not show the right output).
using grep like this shows only the matching lines. For ls that might be a good choice. For matches in a manual page -- definitely not.
Alternatively, less (which is found more often on Unix systems than GNU grep) also can highlight matches (not in color) and would show the file you are looking for in context. You could do this:
ls -l | less -p file2
(Both grep and less use patterns aka regular expressions, but I left the example simple — read the documentation to learn more).
If you're a beginner I would strongly suggest you learn the grep command if you want to filter results - A Unix users best friend (mine anyway)
Use grep to only display the list items you want to see...
ls- l | grep "file2"
NOTE: This is no different to typing ls -l file2 by the way but your pattern could be expanded based on what you actually want displayed on the screen.
So if you had a directory full of files ".txt", ".xls", ".doc" and you wanted to only see ".doc" with the word "work" in the name (work1.doc) you could write:
ls -ls | grep "work" | grep "txt"
This would list work1.txt, work2.txt, work3.txt and so on.
This is a very basic example but I use grep extensively whilst in the unix shell and would advise using this to filter all results instead of colours.
A little side note using grep -v will show you everything but the pattern you give it
ls -l | grep -v ".txt" will show everything BUT .txt files.
I'm using terminal on OS 10.X. I have some data files of the format:
mbh5.0_mrg4.54545454545_period0.000722172513951.params.dat
mbh5.0_mrg4.54545454545_period0.00077271543854.params.dat
mbh5.0_mrg4.59090909091_period-0.000355232058085.params.dat
mbh5.0_mrg4.59090909091_period-0.000402015664015.params.dat
I know that there will be some files with similar numbers after mbh and mrg, but I won't know ahead of time what the numbers will be or how many similarly numbered ones there will be. My goal is to cat all the data from all the files with similar numbers after mbh and mrg into one data file. So from the above I would want to do something like...
cat mbh5.0_mrg4.54545454545*dat > mbh5.0_mrg4.54545454545.dat
cat mbh5.0_mrg4.5909090909*dat > mbh5.0_mrg4.5909090909.dat
I want to automate this process because there will be many such files.
What would be the best way to do this? I've been looking into sed, but I don't have a solution yet.
for file in *.params.dat; do
prefix=${file%_*}
cat "$file" >> "$prefix.dat"
done
This part ${file%_*} remove the last underscore and following text from the end of $file and saves the result in the prefix variable. (Ref: http://www.gnu.org/software/bash/manual/bashref.html#Shell-Parameter-Expansion)
It's not 100% clear to me what you're trying to achieve here but if you want to aggregate files into a file with the same number after "mbh5.0_mrg4." then you can do the following.
ls -l mbh5.0_mrg4* | awk '{print "cat " $9 " > mbh5.0_mrg4." substr($9,12,11) ".dat" }' | /bin/bash
The "ls -s" lists the file and the "awk" takes the 9th column from the result of the ls. With some string concatenation the result is passed to /bin/bash to be executed.
This is a linux bash script, so assuming you have /bind/bash, I'm not 100% famililar with OS X. This script also assumes that the number youre grouping on is always in the same place in the filename. I think you can change /bin/bash to almost any shell you have installed.
This question already has answers here:
How to pass command output as multiple arguments to another command
(5 answers)
Read expression for grep from standard input
(1 answer)
Closed last month.
I am looking for insight as to how pipes can be used to pass standard output as the arguments for other commands.
For example, consider this case:
ls | grep Hello
The structure of grep follows the pattern: grep SearchTerm PathOfFileToBeSearched. In the case I have illustrated, the word Hello is taken as the SearchTerm and the result of ls is used as the file to be searched. But what if I want to switch it around? What if I want the standard output of ls to be the SearchTerm, with the argument following grep being PathOfFileToBeSearched? In a general sense, I want to have control over which argument the pipe fills with the standard output of the previous command. Is this possible, or does it depend on how the script for the command (e.g., grep) was written?
Thank you so much for your help!
grep itself will be built such that if you've not specified a file name, it will open stdin (and thus get the output of ls). There's no real generic mechanism here - merely convention.
If you want the output of ls to be the search term, you can do this via the shell. Make use of a subshell and substitution thus:
$ grep $(ls) filename.txt
In this scenario ls is run in a subshell, and its stdout is captured and inserted in the command line as an argument for grep. Note that if the ls output contains spaces, this will cause confusion for grep.
There are basically two options for this: shell command substitution and xargs. Brian Agnew has just written about the former. xargs is a utility which takes its stdin and turns it into arguments of a command to execute. So you could run
ls | xargs -n1 -J % grep -- % PathOfFileToBeSearched
and it would, for each file output by ls, run grep -e filename PathOfFileToBeSearched to grep for the filename output by ls within the other file you specify. This is an unusual xargs invocation; usually it's used to add one or more arguments at the end of a command, while here it should add exactly one argument in a specific place, so I've used -n and -J arguments to arrange that. The more common usage would be something like
ls | xargs grep -- term
to search all of the files output by ls for term. Although of course if you just want files in the current directory, you can this more simply without a pipeline:
grep -- term *
and likewise in your reversed arrangement,
for filename in *; do
grep -- "$#" PathOfFileToBeSearched
done
There's one important xargs caveat: whitespace characters in the filenames generated by ls won't be handled too well. To do that, provided you have GNU utilities, you can use find instead.
find . -mindepth 1 -maxdepth 1 -print0 | xargs -0 -n1 -J % grep -- % PathOfFileToBeSearched
to use NUL characters to separate filenames instead of whitespace
For grep there's a fixed string option, -F (fgrep) to turn off regex interpretation of the search string.
Is there a similar facility for sed? I couldn't find anything in the man. A recommendation of another gnu/linux tool would also be fine.
I'm using sed for the find and replace functionality: sed -i "s/abc/def/g"
Do you have to use sed? If you're writing a bash script, you can do
#!/bin/bash
pattern='abc'
replace='def'
file=/path/to/file
tmpfile="${TMPDIR:-/tmp}/$( basename "$file" ).$$"
while read -r line
do
echo "${line//$pattern/$replace}"
done < "$file" > "$tmpfile" && mv "$tmpfile" "$file"
With an older Bourne shell (such as ksh88 or POSIX sh), you may not have that cool ${var/pattern/replace} structure, but you do have ${var#pattern} and ${var%pattern}, which can be used to split the string up and then reassemble it. If you need to do that, you're in for a lot more code - but it's really not too bad.
If you're not in a shell script already, you could pretty easily make the pattern, replace, and filename parameters and just call this. :)
PS: The ${TMPDIR:-/tmp} structure uses $TMPDIR if that's set in your environment, or uses /tmp if the variable isn't set. I like to stick the PID of the current process on the end of the filename in the hopes that it'll be slightly more unique. You should probably use mktemp or similar in the "real world", but this is ok for a quick example, and the mktemp binary isn't always available.
Option 1) Escape regexp characters. E.g. sed 's/\$0\.0/0/g' will replace all occurrences of $0.0 with 0.
Option 2) Use perl -p -e in conjunction with quotemeta. E.g. perl -p -e 's/\\./,/gi' will replace all occurrences of . with ,.
You can use option 2 in scripts like this:
SEARCH="C++"
REPLACE="C#"
cat $FILELIST | perl -p -e "s/\\Q$SEARCH\\E/$REPLACE/g" > $NEWLIST
If you're not opposed to Ruby or long lines, you could use this:
alias replace='ruby -e "File.write(ARGV[0], File.read(ARGV[0]).gsub(ARGV[1]) { ARGV[2] })"'
replace test3.txt abc def
This loads the whole file into memory, performs the replacements and saves it back to disk. Should probably not be used for massive files.
If you don't want to escape your string, you can reach your goal in 2 steps:
fgrep the line (getting the line number) you want to replace, and
afterwards use sed for replacing this line.
E.g.
#/bin/sh
PATTERN='foo*[)*abc' # we need it literal
LINENUMBER="$( fgrep -n "$PATTERN" "$FILE" | cut -d':' -f1 )"
NEWSTRING='my new string'
sed -i "${LINENUMBER}s/.*/$NEWSTRING/" "$FILE"
You can do this in two lines of bash code if you're OK with reading the whole file into memory. This is quite flexible -- the pattern and replacement can contain newlines to match across lines if needed. It also preserves any trailing newline or lack thereof, which a simple loop with read does not.
mapfile -d '' < file
printf '%s' "${MAPFILE//"$pat"/"$rep"}" > file
For completeness, if the file can contain null bytes (\0), we need to extend the above, and it becomes
mapfile -d '' < <(cat file; printf '\0')
last=${MAPFILE[-1]}; unset "MAPFILE[-1]"
printf '%s\0' "${MAPFILE[#]//"$pat"/"$rep"}" > file
printf '%s' "${last//"$pat"/"$rep"}" >> file
perl -i.orig -pse 'while (($i = index($_,$s)) >= 0) { substr($_,$i,length($s), $r)}'--\
-s='$_REQUEST['\'old\'']' -r='$_REQUEST['\'new\'']' sample.txt
-i.orig in-place modification with backup.
-p print lines from the input file by default
-s enable rudimentary parsing of command line arguments
-e run this script
index($_,$s) search for the $s string
substr($_,$i,length($s), $r) replace the string
while (($i = index($_,$s)) >= 0) repeat until
-- end of perl parameters
-s='$_REQUEST['\'old\'']', -r='$_REQUEST['\'new\'']' - set $s,$r
You still need to "escape" ' chars but the rest should be straight forward.
Note: this started as an answer to How to pass special character string to sed hence the $_REQUEST['old'] strings, however this question is a bit more appropriately formulated.
You should be using replace instead of sed.
From the man page:
The replace utility program changes strings in place in files or on the
standard input.
Invoke replace in one of the following ways:
shell> replace from to [from to] ... -- file_name [file_name] ...
shell> replace from to [from to] ... < file_name
from represents a string to look for and to represents its replacement.
There can be one or more pairs of strings.
Unfortunately, due to the limitations of our Unix Tru64 environment, I am unable to use the GREP -r switch to perform my search for strings within files across multiple directories and sub directories.
Ideally, I would like to pass two parameters. The first will be the directory I want my search is to start on. The second is a file containing a list of all the strings to be searched. This list will consist of various directory path names and will include special characters:
ie:
/aaa/bbb/ccc
/eee/dddd/ggggggg/
etc..
The purpose of this exercise is to identify all shell scripts that may have specific hard coded path names identified in my list.
There was one example I found during my investigations that perhaps comes close, but I am not sure how to customize this to accept a file of string arguments:
eg: find etb -exec grep test {} \;
where 'etb' is the directory and 'test', a hard coded string to be searched.
This should do it:
find dir -type f -exec grep -F -f strings.txt {} \;
dir is the directory from which searching will commence
strings.txt is the file of strings to match, one per line
-F means treat search strings as literal rather than regular expressions
-f strings.txt means use the strings in strings.txt for matching
You can add -l to the grep switches if you just want filenames that match.
Footnote:
Some people prefer a solution involving xargs, e.g.
find dir -type f -print0 | xargs -0 grep -F -f strings.txt
which is perhaps a little more robust/efficient in some cases.
By reading, I assume we can not use the gnu coreutil, and egrep is not available.
I assume (for some reason) the system is broken, and escapes do not work as expected.
Under normal situations, grep -rf patternfile.txt /some/dir/ is the way to go.
a file containing a list of all the strings to be searched
Assumptions : gnu coreutil not available. grep -r does not work. handling of special character is broken.
Now, you have working awk ? no ?. It makes life so much easier. But lets be on the safe side.
Assume : working sed ,one of od OR hexdump OR xxd (from vim package) is available.
Lets call this patternfile.txt
1. Convert list into a regexp that grep likes
Example patternfile.txt contains
/foo/
/bar/doe/
/root/
(example does not print special char, but it's there.) we must turn it into something like
(/foo/|/bar/doe/|/root/)
Assuming echo -en command is not broken, and xxd , or od, or hexdump is available,
Using hexdump
cat patternfile.txt |hexdump -ve '1/1 "%02x \n"' |tr -d '\n'
Using od
cat patternfile.txt |od -A none -t x1|tr -d '\n'
and pipe it into (common for both hexdump and od)
|sed 's:[ ]*0a[ ]*$::g'|sed 's: 0a:\\|:g' |sed 's:^[ ]*::g'|sed 's:^: :g' |sed 's: :\\x:g'
then pipe result into
|sed 's:^:\\(:g' |sed 's:$:\\):g'
and you have a regexp pattern that is escaped.
2. Feed the escaped pattern into broken regexp
Assuming the bare minimum shell escape is available,
we use grep "$(echo -en "ESCAPED_PATTERN" )" to do our job.
3. To sum it up
Building a escaped regexp pattern (using hexdump as example )
grep "$(echo -en "$( cat patternfile.txt |hexdump -ve '1/1 "%02x \n"' |tr -d '\n' |sed 's:[ ]*0a[ ]*$::g'|sed 's: 0a:\\|:g' |sed 's:^[ ]*::g'|sed 's:^: :g' |sed 's: :\\x:g'|sed 's:^:\\(:g' |sed 's:$:\\):g')")"
will escape all characters and enclose it with (|) brackets so a regexp OR match will be performed.
4. Recrusive directory lookup
Under normal situations, even when grep -r is broken, find /dir/ -exec grep {} \; should work.
Some may prefer xargs instaed (unless you happen to have buggy xargs).
We prefer find /somedir/ -type f -print0 |xargs -0 grep -f 'patternfile.txt' approach, but since
this is not available (for whatever valid reason),
we need to exec grep for each file,and this is normaly the wrong way.
But lets do it.
Assume : find -type f works.
Assume : xargs is broken OR not available.
First, if you have a buggy pipe, it might not handle large number of files.
So we avoid xargs in such systems (i know, i know, just lets pretend it is broken ).
find /whatever/dir/to/start/looking/ -type f > list-of-all-file-to-search-for.txt
IF your shell handles large size lists nicely,
for file in cat list-of-all-file-to-search-for.txt ; do grep REGEXP_PATTERN "$file" ;
done ; is a nice way to get by. Unfortunetly, some systems do not like that,
and in that case, you may require
cat list-of-all-file-to-search-for.txt | split --help -a 4 -d -l 2000 file-smaller-chunk.part.
to turn it into smaller chunks. Now this is for a seriously broken system.
then a for file in file-smaller-chunk.part.* ; do for single_line in cat "$file" ; do grep REGEXP_PATTERN "$single_line" ; done ; done ;
should work.
A
cat filelist.txt |while read file ; do grep REGEXP_PATTERN $file ; done ;
may be used as workaround on some systems.
What if my shell doe not handle quotes ?
You may have to escape the file list beforehand.
It can be done much nicer in awk, perl, whatever, but since we restrict our selves to
sed, lets do it.
We assume 0x27, the ' code will actually work.
cat list-of-all-file-to-search-for.txt |sed 's#['\'']#'\''\\'\'\''#g'|sed 's:^:'\'':g'|sed 's:$:'\'':g'
The only time I had to use this was when feeding output into bash again.
What if my shell does not handle that ?
xargs fails , grep -r fails , shell's for loop fails.
Do we have other things ? YES.
Escape all input suitable for your shell, and make a script.
But you know what, I got board, and writing automated scripts for csh just seems
wrong. So I am going to stop here.
Take home note
Use the tool for the right job. Writing a interpreter on bc is perfectly
capable, but it is just plain wrong. Install coreutils, perl, a better grep
what ever. makes life a better thing.