I have a Data => File1.txt
Data
#demo/file/wk/Fil0.fk
#demo/file/wk/Fil1.fk
#demo/file/wk/Fil2.fk
#demo/file/wk/Fil3.fk
#demo/file/wk/Fil4.fk
Want to Print the data to Another File2.txt in below Format
Fil0.fk
Fil1.fk
Fil2.fk
Fil3.fk
Fil4.fk
Try below code :
cat File1.txt | cut -d'/' -f 4 > File2.txt
Related
This question already has answers here:
Compare two files line by line and generate the difference in another file
(14 answers)
Closed 2 years ago.
FILE1:
cat
dog
house
tree
FILE2:
dog
cat
tree
I need to be printed only:
house
$ cat file1
cat
dog
house
tree
$ cat file2
dog
cat
tree
$ grep -vF -f file2 file1
house
The -v flag only shows non-matches, -f is for a filename to use as a filter, and -F is for exact matches (doesn't slow it down with any pattern matching).
Using awk
awk 'FNR==NR{arr[$0]=1; next} !($0 in arr)' FILE2 FILE1
First build an associative array with words from FILE2 and than loop over FILE1 and only print those.
Using comm
comm -2 -3 <(sort FILE1) <(sort FILE2)
-2 suppresses lines unique to FILE2 and -3 suppresses lines found in both.
If you want just the words, you can sort the files, diff them, then use sed to filter out diff's symbols:
diff <(sort file1) <(sort file2) | sed -n '/^</s/^< //p'
Awk is an option here:
awk 'NR==FNR { arr[$1]="1" } NR != FNR { if (arr[$1] == "") { print $0 } } ' file2 file1
Create an array called arr, using the contents of file2 as indexes. Then with file1, look at each entry and check to see if an entry in the array arr exists. If it doesn't, print.
I have 2 files
file1.txt
1
3
5
2
File2.txt
1 aaa
2 bbb
3 ccc
4 aaa
5 bbb
Desired output:
1 aaa
3 ccc
5 bbb
2 bbb
Command used : cat File1.txt |grep -wf- File2.txt but the output was:
1 aaa
2 bbb
3 ccc
5 bbb
Is it a way to return the output in the query order?
Thanks in advance!!!
Important Edit
On second thought, do not use grep with redirection as it's incredibly slow. Use awk to read the original patterns to get the order back.
Use this instead
grep -f patterns searchdata | awk 'NR==FNR { line[$1] = $0; next } $1 in line { print line[$1] }' - patterns > matched
Benchmark
#!/bin/bash
paste <(shuf -i 1-10000) <(crunch 4 4 2>/dev/null | shuf -n 10000) > searchdata
shuf -i 1-10000 > patterns
printf 'Testing awk:'
time grep -f patterns searchdata | awk 'NR==FNR { line[$1] = $0; next } $1 in line { print line[$1] }' - patterns > matched
wc -l matched
cat /dev/null > matched
printf '\nTesting grep with redirection:'
time {
while read -r pat; do
grep -w "$pat" searchdata >> matched
done < patterns
}
wc -l matched
Output
Testing awk:
real 0m0.022s
user 0m0.017s
sys 0m0.010s
10000 matched
Testing grep with redirection:
real 0m36.370s
user 0m28.761s
sys 0m7.909s
10000 matched
Original
To preserve the query order, read the file line-by-line:
while read -r pat; do grep -w "$pat" file2.txt; done < file1.txt
I don't think grep has an option to support this, but this solution will be slower if you have large files to read from.
I have two text files file1.txt & file2.txt
file1.txt Contains :
a
b
c
file2.txt Contains :
a
b
c
d
e
f
The Output Should be :
d
e
f
The command i'm trying to use is 'diff file2.txt file1.txt'
It gives the common lines only.
Assuming that the input files are sorted:
join -v 2 file1.txt file2.txt
Check man join for details on all the other things join can do for you.
please try below ones
grep -vf file1.txt file2.txt
comm -13 file1.txt file2.txt
for diff you have to perform something extra
diff inp inp1 | grep '>' | cut -f2 -d' '
How can I find lines that contain a double tab and then \N
It should match, for example, \N abcd
I've tried
grep $'\t'$'\t''\N' file1.txt
grep $'\t\t''\N' file1.txt
grep $'\t\t\N' file1.txt
The following works for me:
RHEL:
$ grep $'\t\t''\\N' file1.txt
OSX:
$ grep '\t\t\\N' file1.txt
I want to find the last appearance of a string in a text file with linux commands. For example
1 a 1
2 a 2
3 a 3
1 b 1
2 b 2
3 b 3
1 c 1
2 c 2
3 c 3
In such a text file, i want to find the line number of the last appearance of b which is 6.
I can find the first appearance with
awk '/ b / {print NR;exit}' textFile.txt
but I have no idea how to do it for the last occurrence.
cat -n textfile.txt | grep " b " | tail -1 | cut -f 1
cat -n prints the file to STDOUT prepending line numbers.
grep greps out all lines containing "b" (you can use egrep for more advanced patterns or fgrep for faster grep of fixed strings)
tail -1 prints last line of those lines containing "b"
cut -f 1 prints first column, which is line # from cat -n
Or you can use Perl if you wish (It's very similar to what you'd do in awk, but frankly, I personally don't ever use awk if I have Perl handy - Perl supports 100% of what awk can do, by design, as 1-liners - YMMV):
perl -ne '{$n=$. if / b /} END {print "$n\n"}' textfile.txt
This can work:
$ awk '{if ($2~"b") a=NR} END{print a}' your_file
We check every second file being "b" and we record the number of line. It is appended, so by the time we finish reading the file, it will be the last one.
Test:
$ awk '{if ($2~"b") a=NR} END{print a}' your_file
6
Update based on sudo_O advise:
$ awk '{if ($2=="b") a=NR} END{print a}' your_file
to avoid having some abc in 2nd field.
It is also valid this one (shorter, I keep the one above because it is the one I thought :D):
$ awk '$2=="b" {a=NR} END{print a}' your_file
Another approach if $2 is always grouped (may be more efficient then waiting until the end):
awk 'NR==1||$2=="b",$2=="b"{next} {print NR-1; exit}' file
or
awk '$2=="b"{f=1} f==1 && $2!="b" {print NR-1; exit}' file