I want only to extract the date from the name of this file:
So I did this:
echo HNR.L04.C07.ldd.T2018050.BG.nc.nc |grep -o '[0-9]\+'
I got this:
04
07
2018050
Now I want to select the third line? any idea?
You can get the firthd line of output with:
set -n 3p
You can get the last line of output with:
tail -n 1
Ex:
echo HNR.L04.C07.ldd.T2018050.BG.nc.nc | grep -o '[0-9]\+' | sed -n 3p
Related
I have certain tags stored in a series of .log files and i would like for the grep to show me Only the values > 31, meaning different to 0 and higher than 31
I have this code:
#! /bin/bash
-exec grep -cFH "tag" {} ; | grep -v ':[0-31]$' >> file.txt
echo < file.txt
Output:
I have this result from the grep:
/opt/logs/folder2.log:31
i was expecting not to bring nothing back if the result is 31 or less but still shows the result 31
i have also tried to add:
|tail -n+31
but didn't work
[0-31] means "0 or 1 or 2 or 3 or 1".
To drop all lines with 0-9, 10-19, 20-29, 30, and 31, you could use the following:
... | grep -ve ':[0-9]$' -e ':[12][0-9]$' -e ':3[01]$'
or as single regex:
... | grep -v ':\([12]\?[0-9]\|3[01]\)$'
With extended grep:
... | grep -vE ':([12]?[0-9]|3[01])$'
I have a file in Linux contains strings:
CALLTMA
Starting
Starting
Ending
Starting
Ending
Ending
CALLTMA
Ending
I need the quantity of any string (FE. #Ending, # Starting, #CALLTMA). In my example I need obtaining:
CALLTMA : 2
Starting: 3
Ending : 4
I can obtaining this output when I execute 3 commands:
grep -i "Starting" "/myfile.txt" | wc -l
grep -i "Ending" "/myfile.txt" | wc -l
grep -i "CALLTMA" "/myfile.txt" | wc -l
I want to know if it is possible to obtain the same output using only one command.
I try running this command
grep -iE "CALLTMA|Starting|Ending" "/myfile.txt" | wc -l
But this returned the total of coincidences. I appreciate your help .
Use sort and uniq:
sort myfile.txt | uniq -c
The -c adds the counts to the unique lines. If you want to sort the output by frequency, add
| sort -n
to the end (and change to -nr if you want the descending order).
A simple awk way to handle this:
awk '{counts[$1]++} END{for (c in counts) print c, counts[c]}' file
Starting 3
Ending 4
CALLTMA 2
grep -c will work. You can put it all together in a short script:
for i in Starting CALLTMA Ending; do
printf "%-8s : %d\n" "$i" $(grep -c "$i" file.txt)
done
(to enter the search terms as arguments, just use the arguments array for the loop list, e.g. for i in "$#"; do)
Output
Starting : 3
CALLTMA : 2
Ending : 4
I need to count all lines of an unix file. The file has 3 lines but wc -l gives only 2 count.
I understand that it is not counting last line because it does not have end of line character
Could any one please tell me how to count that line as well ?
grep -c returns the number of matching lines. Just use an empty string "" as your matching expression:
$ echo -n $'a\nb\nc' > 2or3.txt
$ cat 2or3.txt | wc -l
2
$ grep -c "" 2or3.txt
3
It is better to have all lines ending with EOL \n in Unix files. You can do:
{ cat file; echo ''; } | wc -l
Or this awk:
awk 'END{print NR}' file
This approach will give the correct line count regardless of whether the last line in the file ends with a newline or not.
awk will make sure that, in its output, each line it prints ends with a new line character. Thus, to be sure each line ends in a newline before sending the line to wc, use:
awk '1' file | wc -l
Here, we use the trivial awk program that consists solely of the number 1. awk interprets this cryptic statement to mean "print the line" which it does, being assured that a trailing newline is present.
Examples
Let us create a file with three lines, each ending with a newline, and count the lines:
$ echo -n $'a\nb\nc\n' >file
$ awk '1' f | wc -l
3
The correct number is found.
Now, let's try again with the last new line missing:
$ echo -n $'a\nb\nc' >file
$ awk '1' f | wc -l
3
This still provides the right number. awk automatically corrects for a missing newline but leaves the file alone if the last newline is present.
Respect
I respect the answer from John1024 and would like to expand upon it.
Line Count function
I find myself comparing line counts A LOT especially from the clipboard, so I have defined a bash function. I'd like to modify it to show the filenames and when passed more than 1 file a total. However, it hasn't been important enough for me to do so far.
# semicolons used because this is a condensed to 1 line in my ~/.bash_profile
function wcl(){
if [[ -z "${1:-}" ]]; then
set -- /dev/stdin "$#";
fi;
for f in "$#"; do
awk 1 "$f" | wc -l;
done;
}
Counting lines without the function
# Line count of the file
$ cat file_with_newline | wc -l
3
# Line count of the file
$ cat file_without_newline | wc -l
2
# Line count of the file unchanged by cat
$ cat file_without_newline | cat | wc -l
2
# Line count of the file changed by awk
$ cat file_without_newline | awk 1 | wc -l
3
# Line count of the file changed by only the first call to awk
$ cat file_without_newline | awk 1 | awk 1 | awk 1 | wc -l
3
# Line count of the file unchanged by awk because it ends with a newline character
$ cat file_with_newline | awk 1 | awk 1 | awk 1 | wc -l
3
Counting characters (why you don't want to put a wrapper around wc)
# Character count of the file
$ cat file_with_newline | wc -c
6
# Character count of the file unchanged by awk because it ends with a newline character
$ cat file_with_newline | awk 1 | awk 1 | awk 1 | wc -c
6
# Character count of the file
$ cat file_without_newline | wc -c
5
# Character count of the file changed by awk
$ cat file_without_newline | awk 1 | wc -c
6
Counting lines with the function
# Line count function used on stdin
$ cat file_with_newline | wcl
3
# Line count function used on stdin
$ cat file_without_newline | wcl
3
# Line count function used on filenames passed as arguments
$ wcl file_without_newline file_with_newline
3
3
Using the command du, I would like to get the total size of a directory
Output of command du myfolder:
5454 kkkkk
666 aaaaa
3456788 total
I'm able to extract the last line, but not to remmove the string total:
du -c myfolder | grep total | cut -d ' ' -f 1
Results in:
3456788 total
Desired result
3456788
I would like to have all the command in one line.
That's probably because it's tab delimited (which is the default delimiter of cut):
~$ du -c foo | grep total | cut -f1
4
~$ du -c foo | grep total | cut -d' ' -f1
4
to insert a tab, use Ctrl+v, then TAB
Alternatively, you could use awk to print the first field of the line ending with total:
~$ du -c foo | awk '/total$/{print $1}'
4
First of, you probably want to use tail -n1 instead of grep total ... Consider what happens if you have a directory named local? :-)
Now, let's look at the output of du with hexdump:
$ du -c tmp | tail -n1 | hexdump -C
00000000 31 34 30 33 34 34 4b 09 74 6f 74 61 6c 0a |140344K.total.|
That''s the character 0x09 after the K, man ascii tells us:
011 9 09 HT '\t' (horizontal tab) 111 73 49 I
It's a tab, not a space :-)
The tab character is already the default delimiter (this is specified in the POSIX spec, so you can safely rely on it), so you don't need -d at all.
So, putting that together, we end up with:
$ du -c tmp | tail -n1 | cut -f1
140344K
Why don't you use -s to summarize it? This way you don't have to grep "total", etc.
$ du .
24 ./aa/bb
...
# many lines
...
2332 .
$ du -hs .
2.3M .
Then, to get just the value, pipe to awk. This way you don't have to worry about the delimiter being a space or a tab:
du -s myfolder | awk '{print $1}'
From man du:
-h, --human-readable
print sizes in human readable format (e.g., 1K 234M 2G)
-s, --summarize
display only a total for each argument
I would suggest using awk for this:
value=$(du -c myfolder | awk '/total/{print $1}')
This simply extracts the first field of the line that matches the pattern "total".
If it is always the last line that you're interested in, an alternative would be to use this:
value=$(du -c myfolder | awk 'END{print $1}')
The values of the fields in the last line are accessible in the END block, so you can get the first field of the last line this way.
I have a file contaning for just 2 numubers. One number on eash line.
4.1865E+02
4.1766E+02
I know its something line BHF = ($1 from line 1 - $1 from line 2 )
but can find the exact command.
How can I do a mathematical operation on them and save the result to a variable.
PS: This was got using
sed -i -e '/^$/d' nodout15
sed -i -e 's/^[ \t]*//;s/[ \t]*$//' nodout15
awk ' {print $13} ' nodout15 > 15
mv 15 nodout15
sed -i -e '/^$/d' nodout15
sed -i -e 's/^[ \t]*//;s/[ \t]*$//' nodout15
sed -n '/^[0-9]\{1\}/p' nodout15 > 15
mv 15 nodout15
tail -2 nodout15 > 15
mv 15 nodout15
After all this I have these two numbers and now I am not able to do some arithmatics. If possible please tell me a short form to do it on the spot rather doing all this jugglary. Nodout is a file with different length of columns so I am only interested in 13th column. Since all lines wont be in the daughter file so , empty lines deleted. Then only those lines to be taken starting with number. Then the last two lines, as they show the final state. The difference between them , will lead to a conditional statement. so I need to save it in a variable.
regards.
awk
$ BHF=`awk -v RS='' '{print $1-$2}' input.txt`
$ echo $BHF
0.99
bc
$ BHF=`cat input.txt | xargs printf '%f-%f\n' | bc`
$ echo $BHF
.990000