I'm trying to list some files that have the same permissions in User field and Others field. Could you tell me which is the most simple command for this?
Ex: ---x--x--x
You can do it using stat command like this:
$ stat --printf="%A %n\n" *
---x--x--x file1
---x-wx--x file2
-rw-rw-r-- file3
$ stat --printf="%A %n\n" * | egrep '^.(.{3}).{3}\1'
---x--x--x file1
---x-wx--x file2
$
regex ^.(.{3}).{3}\1 ensure chars from 2nd to 4th index(user permissions) are same as that of that other permissions(\1 back-reference indicate other permissions).
Another way in which you can do it is using octal representation like this:
$ stat --printf="%a %n\n" * | egrep '^([0-9]).\1'
111 file1
131 file2
$
Related
I have a problem with this Linux command:
ls | grep -E 'i{2,3}'
.It should take a file that has at least 2 i and max 3 i, but it doesn't work.
This is the output
ls:
life.py, viiva.txt, viiiiiiiiiva.txt
grep:
viiva.txt, viiiiiiiiiva.txt (with the first 3 I highlighted)
Thanks for the help.
Issue with OP's attempt grep -E 'i{2,3}' will match two or three consecutive occurrences of i anywhere in the input, so 4 or more consecutive i is also a valid match.
Parsing ls output is not recommended, see Why not parse ls (and what to do instead)?. If you wish to pass the filenames after filtering to some other command, find is a good option.
$ ls
1i2i3i.txt aibi.txt II.txt life.py viiiiiiiiiva.txt viiva.txt
$ # files with 2 or 3 consecutive i
$ # note that the regex will act on entire filename, thus anchors are not needed
$ find -type f -regextype egrep -regex '[^i]*i{2,3}[^i]*'
./viiva.txt
$ # files with 2 or 3 i anywhere in the name
$ find -type f -regextype egrep -regex '[^i]*i[^i]*i[^i]*(i[^i]*)?'
./aibi.txt
./1i2i3i.txt
./viiva.txt
$ # files with 2 or 3 i anywhere in the name, ignoring case
$ find -type f -regextype egrep -iregex '[^i]*i[^i]*i[^i]*(i[^i]*)?'
./II.txt
./aibi.txt
./1i2i3i.txt
./viiva.txt
If filenames won't cause an issue, you can grep -xE or grep -ixE with above regex, where x option will make sure the regex matches the whole line, instead of anywhere in the line. Or you can also use awk:
$ # NF will give number of fields after splitting on i
$ ls | awk -F'i' 'NF>=3 && NF<=4'
1i2i3i.txt
aibi.txt
viiva.txt
$ ls | awk -F'[iI]' 'NF>=3 && NF<=4'
1i2i3i.txt
aibi.txt
II.txt
viiva.txt
I am using grep to extract lines from file 1 that matches with string in file2. The string in file 2 has both alphabets and numbers. eg;
MSTRG.18691.1
MSTRG.18801.1
I used sed to write word boundaries for all the strings in the file 2.
file 2
\<MSTRG.18691.1\>
\<MSTRG.18801.1\>
and used grep -f file2 file1
but output has
MSTRG.18691.1.2
MSTRG.18801.1.3 also..
I want lines that matches exactly,
MSTRG.18691.1
MSTRG.18801.1
and not,
MSTRG.18691.1.2
MSTRG.18801.1.3
Few lines from my file1
t_name gene_name FPKM TPM
MSTRG.25.1 . 0 0
rna71519 . 93.398872 194.727926057583
gene34024 ND1 2971.72876 6195.77694943117
MSTRG.28.1 . 0 0
MSTRG.28.2 . 0 0
rna71520 . 33.235409 69.2927240732149
Updating the answer
You can use start with ^ and end with $ operator to match start with and begin with. To match exactly MSTRG.18691.1 you can add ^ & $ at both ends and remove the word boundaries, additionally . has special meaning in regex to match exactly . we need to escape that with a backslash \
Example pattern:
^MSTRG\.18691\.1$
^MSTRG\.18801\.1$
file1
MSTRG.18691.1
MSTRG.1311.1
MSTRG.18801.2
MSTRG.18801.3
MSTRG.18801.1.2
MSTRG.18801.1.1
MSTRG.18801.1
PrefixMSTRG.18801.1
Just create a normal file named file1 and paste the above content into it.
file2 (pattern file)
^MSTRG\.18801\.1$
Just create a normal file named file2 and paste the above content into it.
Run the below command from commandline
grep -i --color -f file2 file1
Result:
MSTRG.18801.1
Sed to add changes to the pattern file
Here is the sed command to escape . and add ^ and $ at the beginning and end of the pattern file you already have.
sed -Ee 's/\./\\./g' -e 's/^/\^/g' -e 's/$/\$/g' file2 > file2_updated
-E to support extended regex on BSD sed, you may need to replace -E with -r based on your system's sed
Updated patterns will be saved to file2_updated. Need to use the new pattern file in grep like this
grep -i -f file2_updated file1
The flag you're looking for is -F. From man grep:
-F, --fixed-strings
Interpret PATTERN as a list of fixed strings (instead of regular expressions), separated by newlines, any of which is to be matched.
You can use this quite comfortably in conjunction with -f:
grep -Ff file2 file1
To be clear, this will treat every line of file2 as an exact match against file1.
file1:
abcds
asdt5ds
file2:
1
2
3
4
5
Output:
asdt5ds
5
Is it possible to get such output using a variation of cat file1 | grep 5 file2?
You can use the special file - (for "standard input"):
cat file1 | grep 5 - file2
If you want to suppress the filenames, i.e., if you don't want this output
(standard input):asdt5ds
file2:5
you can do that using the -h option.
Notice that both of these (using - as special file name and -h to suppress file names) are not required by POSIX, but at least GNU grep supports them.
The command 'grep -c blah *' lists all the files, like below.
% grep -c jill *
file1:1
file2:0
file3:0
file4:0
file5:0
file6:1
%
What I want is:
% grep -c jill * | grep -v ':0'
file1:1
file6:1
%
Instead of piping and grep'ing the output like above, is there a flag to suppress listing files with 0 counts?
SJ
How to grep nonzero counts:
grep -rIcH 'string' . | grep -v ':0$'
-r Recurse subdirectories.
-I Ignore binary files (thanks #tongpu, warlock).
-c Show count of matches. Annoyingly, includes 0-count files.
-H Show file name, even if only one file (thanks #CraigEstey).
'string' your string goes here.
. Start from the current directory.
| grep -v ':0$' Remove 0-count files. (thanks #LaurentiuRoescu)
(I realize the OP was excluding the pipe trick, but this is what works for me.)
Just use awk. e.g. with GNU awk for ENDFILE:
awk '/jill/{c++} ENDFILE{if (c) print FILENAME":"c; c=0}' *
Okay I have a file that contains numbers like this:
L21479
What I am trying to do is use grep (or a similar tool) to find all the strings in a file that have the format:
L#####
The # will be the number. SO an L followed by 5 numbers.
Is this even possible in grep? Should I load the file and perform regex?
You can do this with grep, for example with the following command:
grep -E -o 'L[0-9]{5}' name_of_file
For example, given a file with the text:
kasdhflkashl143112343214L232134614
3L1431413543454L2342L3523269ufoidu
gl9983ugsdu8768IUHI/(JHKJASHD/(888
The command above will output:
L23213
L14314
L35232
If it is just in a single file, you can do something along the lines of:
grep -e 'L[0-9]{5}' filename
If you need to search all files in a directory for these strings:
find . -type f | xargs grep -e 'L[0-9]{5}'