I have two files.
file.txt and delete.txt
file.txt contains the following for ex.:
/somedirectory/
/somedirectory2/somefile.txt
/anotherfile.txt
delete.txt contains:
/somedirectory/
/somedirectory2/somefile.txt
I need to delete the rows from file.txt that are contained within delete.txt
cat file.txt should result with:
/anotherfile.txt
So far I've tried this with no luck:
while read p; do
line=$p;
sed -i '/${line}/d' file.txt;
done < delete.txt
I don't receive any error, it just doesn't edit my file.txt file. I've tested the while loop with an echo ${line} and it works as expected in that scenario.
Any suggestions?
Note: the while loop above doesn't work as expected even when I took the forward slashes off of the files.
With a simple grep:
grep -vFxf delete.txt file.txt > temp.txt && mv temp.txt file.txt
With awk:
awk 'NR==FNR {a[$0]=1; next}; !a[$0]' delete.txt file.txt
NR==FNR {a[$0]=1; next} is only executed for the delete.txt file (first file argument); associative array a has the records as keys, and 1 as the value for every key
!a[$0] is executed for the second file argument i.e. file.txt; printing (default action) the record(s) that are not present in the array a as key(s)
Example:
% cat delete.txt
/somedirectory/
/somedirectory2/somefile.txt
% cat file.txt
/somedirectory/
/somedirectory2/somefile.txt
/anotherfile.txt
% awk 'NR==FNR {a[$0]=1; next}; !a[$0]' delete.txt file.txt
/anotherfile.txt
The ${line} inside the single quotes will not get expanded, so it would be looking for the actual string ${line} in file.txt. Put that into your sample file to see if it gets removed.
However, you'll still have problems because the slashes inside delete text will get interpreted by sed, so the regular expression won't get properly delimited. You'd have to jump through some hoops to get every character in your input line properly quoted as literal in order to use sed.
Use comm instead:
comm -23 file.txt delete.txt
The input files must be sorted, though. If they are not:
comm -23 <(sort file.txt) <(sort delete.txt)
Related
I have a simple question. I am trying to check the 3rd line of multiple files in a folder, so I used this:
head -n 3 MiseqData/result2012/12* | tail -n 1
but this doesn't work obviously, because it only shows the third line of the last file. But I actually want to have last line of every file in the result2012 folder.
Does anyone know how to do that?
Also sorry just another questions, is it also possible to show which file the particular third line belongs to?
like before the third line is shown, is it also possible to show the filename of each of the third line extracted from?
because if I used head or tail command, the filename is also shown.
thank you
With Awk, the variable FNR is the number of the "record" (line, by default) in the current file, so you can simply compare it to 3 to print the third line of each input file:
awk 'FNR == 3' MiseqData/result2012/12*
A more optimized version for long files would skip to the next file on match, since you know there's only that one line where the condition is true:
awk 'FNR == 3 { print; nextfile }' MiseqData/result2012/12*
However, not all Awks support nextfile (but it is also not exclusive to GNU Awk).
A more portable variant using your head and tail solution would be a loop in the shell:
for f in MiseqData/result2012/12*; do head -n 3 "$f" | tail -n 1; done
Or with sed (without GNU extensions, i.e., the -s argument):
for f in MiseqData/result2012/12*; do sed '3q;d' "$f"; done
edit: As for the additional question of how to print the name of each file, you need to explicitly print it for each file yourself, e.g.,
awk 'FNR == 3 { print FILENAME ": " $0; nextfile }' MiseqData/result2012/12*
for f in MiseqData/result2012/12*; do
echo -n `basename "$f"`': '
head -n 3 "$f" | tail -n 1
done
for f in MiseqData/result2012/12*; do
echo -n "$f: "
sed '3q;d' "$f"
done
With GNU sed:
sed -s -n '3p' MiseqData/result2012/12*
or shorter
sed -s '3!d' MiseqData/result2012/12*
From man sed:
-s: consider files as separate rather than as a single continuous long stream.
You can do this:
awk 'FNR==3' MiseqData/result2012/12*
If you like the file name as well:
awk 'FNR==3 {print FILENAME,$0}' MiseqData/result2012/12*
This might work for you (GNU sed & parallel):
parallel -k sed -n '3p\;3q' {} ::: file1 file2 file3
Parallel applies the sed command to each file and returns the results in order.
N.B. All files will only be read upto the 3rd line.
Also,you may be tempted (as I was) to use:
sed -ns '3p;3q' file1 file2 file3
but this will only return the first file.
Hi bro I am answering this question as we know FNR is used to check no of lines so we can run this command to get 3rd line of every file.
awk 'FNR==3' MiseqData/result2012/12*
I'm trying to take a column and pipe it through an echo command. If possible, I would like to keep it in one line or do this as efficiently as possible. While researching, I found that I have to use single quotes to expand the variable and to escape the double quotes.
Here's what I was trying:
awk -F ',' '{print $2}' file1.txt | while read line; do echo "<href=\"'${i}'\">'${i}'</a>"; done
But, I keep getting the number of lines than the single line's output. If you know how to caputure each line in field 4, that would be so helpful.
File1.txt:
Hello,http://example1.com
Hello,http://example2.com
Hello,http://example3.com
Desired output:
<href="http://example1.com">http://example1.com</a>
<href="http://example2.com">http://example2.com</a>
<href="http://example3.com">http://example3.com</a>
$ awk -F, '{printf "<href=\"%s\">%s</a>\n", $2, $2}' file
<href="http://example1.com">http://example1.com</a>
<href="http://example2.com">http://example2.com</a>
<href="http://example3.com">http://example3.com</a>
Or slightly briefer but less robustly:
$ sed 's/.*,\(.*\)/<href="\1">\1<\/a>/' file
<href="http://example1.com">http://example1.com</a>
<href="http://example2.com">http://example2.com</a>
<href="http://example3.com">http://example3.com</a>
Suppose I have a file input.txt with few columns and few rows, the first column is the key, and a directory dir with files which contain some of these keys. I want to find all lines in the files in dir which contain these key words. At first I tried to run the command
cat input.txt | awk '{print $1}' | xargs grep dir
This doesn't work because it thinks the keys are paths on my file system. Next I tried something like
cat input.txt | awk '{system("grep -rn dir $1")}'
But this didn't work either, eventually I have to admit that even this doesn't work
cat input.txt | awk '{system("echo $1")}'
After I tried to use \ to escape the white space and the $ sign, I came here to ask for your advice, any ideas?
Of course I can do something like
for x in `cat input.txt` ; do grep -rn $x dir ; done
This is not good enough, because it takes two commands, but I want only one. This also shows why xargs doesn't work, the parameter is not the last argument
You don't need grep with awk, and you don't need cat to open files:
awk 'NR==FNR{keys[$1]; next} {for (key in keys) if ($0 ~ key) {print FILENAME, $0; next} }' input.txt dir/*
Nor do you need xargs, or shell loops or anything else - just one simple awk command does it all.
If input.txt is not a file, then tweak the above to:
real_input_generating_command |
awk 'NR==FNR{keys[$1]; next} {for (key in keys) if ($0 ~ key) {print FILENAME, $0; next} }' - dir/*
All it's doing is creating an array of keys from the first file (or input stream) and then looking for each key from that array in every file in the dir directory.
Try following
awk '{print $1}' input.txt | xargs -n 1 -I pattern grep -rn pattern dir
First thing you should do is research this.
Next ... you don't need to grep inside awk. That's completely redundant. It's like ... stuffing your turkey with .. a turkey.
Awk can process input and do "grep" like things itself, without the need to launch the grep command. But you don't even need to do this. Adapting your first example:
awk '{print $1}' input.txt | xargs -n 1 -I % grep % dir
This uses xargs' -I option to put xargs' input into a different place on the command line it runs. In FreeBSD or OSX, you would use a -J option instead.
But I prefer your for loop idea, converted into a while loop:
while read key junk; do grep -rn "$key" dir ; done < input.txt
Use process substitution to create a keyword "file" that you can pass to grep via the -f option:
grep -f <(awk '{print $1}' input.txt) dir/*
This will search each file in dir for lines containing keywords printed by the awk command. It's equivalent to
awk '{print $1}' input.txt > tmp.txt
grep -f tmp.txt dir/*
grep requires parameters in order: [what to search] [where to search]. You need to merge keys received from awk and pass them to grep using the \| regexp operator.
For example:
arturcz#szczaw:/tmp/s$ cat words.txt
foo
bar
fubar
foobaz
arturcz#szczaw:/tmp/s$ grep 'foo\|baz' words.txt
foo
foobaz
Finally, you will finish with:
grep `commands|to|prepare|a|keywords|list` directory
In case you still want to use grep inside awk, make sure $1, $2 etc are outside quote.
eg. this works perfectly
cat file_having_query | awk '{system("grep " $1 " file_to_be_greped")}'
// notice the space after grep and before file name
I am currently trying to grep a large list of ids (~5000) against an even larger csv file (3.000.000 lines).
I want all the csv lines, that contain an id from the id file.
My naive approach was:
cat the_ids.txt | while read line
do
cat huge.csv | grep $line >> output_file
done
But this takes forever!
Are there more efficient approaches to this problem?
Try
grep -f the_ids.txt huge.csv
Additionally, since your patterns seem to be fixed strings, supplying the -F option might speed up grep.
-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.)
Use grep -f for this:
grep -f the_ids.txt huge.csv > output_file
From man grep:
-f FILE, --file=FILE
Obtain patterns from FILE, one per line. The empty file contains zero
patterns, and therefore matches nothing. (-f is specified by POSIX.)
If you provide some sample input maybe we can even improve the grep condition a little more.
Test
$ cat ids
11
23
55
$ cat huge.csv
hello this is 11 but
nothing else here
and here 23
bye
$ grep -f ids huge.csv
hello this is 11 but
and here 23
grep -f filter.txt data.txt gets unruly when filter.txt is larger than a couple of thousands of lines and hence isn't the best choice for such a situation. Even while using grep -f, we need to keep a few things in mind:
use -x option if there is a need to match the entire line in the second file
use -F if the first file has strings, not patterns
use -w to prevent partial matches while not using the -x option
This post has a great discussion on this topic (grep -f on large files):
Fastest way to find lines of a file from another larger file in Bash
And this post talks about grep -vf:
grep -vf too slow with large files
In summary, the best way to handle grep -f on large files is:
Matching entire line:
awk 'FNR==NR {hash[$0]; next} $0 in hash' filter.txt data.txt > matching.txt
Matching a particular field in the second file (using ',' delimiter and field 2 in this example):
awk -F, 'FNR==NR {hash[$1]; next} $2 in hash' filter.txt data.txt > matching.txt
and for grep -vf:
Matching entire line:
awk 'FNR==NR {hash[$0]; next} !($0 in hash)' filter.txt data.txt > not_matching.txt
Matching a particular field in the second file (using ',' delimiter and field 2 in this example):
awk -F, 'FNR==NR {hash[$0]; next} !($2 in hash)' filter.txt data.txt > not_matching.txt
You may get a significant search speedup with ugrep to match the strings in the_ids.txt in your large huge.csv file:
ugrep -F -f the_ids.txt huge.csv
This works with GNU grep too, but I expect ugrep to run several times faster.
I have file.txt with names one per line as shown below:
ABCB8
ABCC12
ABCC3
ABCC4
AHR
ALDH4A1
ALDH5A1
....
I want to grep each of these from an input.txt file.
Manually i do this one at a time as
grep "ABCB8" input.txt > output.txt
Could someone help to automatically grep all the strings in file.txt from input.txt and write it to output.txt.
You can use the -f flag as described in Bash, Linux, Need to remove lines from one file based on matching content from another file
grep -o -f file.txt input.txt > output.txt
Flag
-f FILE, --file=FILE:
Obtain patterns from FILE, one per line. The empty file
contains zero patterns, and therefore matches nothing. (-f is
specified by POSIX.)
-o, --only-matching:
Print only the matched (non-empty) parts of a matching line, with
each such part on a separate output line.
for line in `cat text.txt`; do grep $line input.txt >> output.txt; done
Contents of text.txt:
ABCB8
ABCC12
ABCC3
ABCC4
AHR
ALDH4A1
ALDH5A1
Edit:
A safer solution with while read:
cat text.txt | while read line; do grep "$line" input.txt >> output.txt; done
Edit 2:
Sample text.txt:
ABCB8
ABCB8XY
ABCC12
Sample input.txt:
You were hired to do a job; we expect you to do it.
You were hired because ABCB8 you kick ass;
we expect you to kick ass.
ABCB8XY You were hired because you can commit to a rational deadline and meet it;
ABCC12 we'll expect you to do that too.
You're not someone who needs a middle manager tracking your mouse clicks
If You don't care about the order of lines, the quick workaround would be to pipe the solution through a sort | uniq:
cat text.txt | while read line; do grep "$line" input.txt >> output.txt; done; cat output.txt | sort | uniq > output2.txt
The result is then in output.txt.
Edit 3:
cat text.txt | while read line; do grep "\<${line}\>" input.txt >> output.txt; done
Is that fine?