How do I extract 68 from v1+r0.68?
Using awk, returns everything after the last '.'
echo "v1+r0.68" | awk -F. '{print $NF}'
Using sed to get the number after the last dot:
echo 'v1+r0.68' | sed 's/.*[.]\([0-9][0-9]*\)$/\1/'
grep is good at extracting things:
kent$ echo " v1+r0.68"|grep -oE "[0-9]+$"
68
Match the digit string before the end of the line using grep:
$ echo 'v1+r0.68' | grep -Eo '[0-9]+$'
68
Or match any digits after a .
$ echo 'v1+r0.68' | grep -Po '(?<=\.)\d+'
68
Print everything after the . with awk:
echo "v1+r0.68" | awk -F. '{print $NF}'
68
Substitute everything before the . with sed:
echo "v1+r0.68" | sed 's/.*\.//'
68
type man grep
and you will see
...
-o, --only-matching
Show only the part of a matching line that matches PATTERN.
then type echo 'v1+r0.68' | grep -o '68'
if you want it any where special do:
echo 'v1+r0.68' | grep -o '68' > anyWhereSpecial.file_ending
Related
I have a file with several rows and with each row containing the following data-
name 20150801|1 20150802|4 20150803|6 20150804|7 20150805|7 20150806|8 20150807|11532 20150808|12399 2015089|12619 20150810|12773 20150811|14182 20150812|27856 20150813|81789 20150814|41168 20150815|28982 20150816|24500 20150817|22534 20150818|3 20150819|4 20150820|47773 20150821|33168 20150822|53541 20150823|46371 20150824|34664 20150825|32249 20150826|29181 20150827|38550 20150828|28843 20150829|3 20150830|23543 20150831|6
name2 20150801|1 20150802|4 20150803|6 20150804|7 20150805|7 20150806|8 20150807|11532 20150808|12399 2015089|12619 20150810|12773 20150811|14182 20150812|27856 20150813|81789 20150814|41168 20150815|28982 20150816|24500 20150817|22534 20150818|3 20150819|4 20150820|47773 20150821|33168 20150822|53541 20150823|46371 20150824|34664 20150825|32249 20150826|29181 20150827|38550 20150828|28843 20150829|3 20150830|23543 20150831|6
The pipe separated value indicates the value for each of the dates in the month.
Each row has the same format with same number of columns.
The first column name indicates a unique name for the row e.g. 20150818 is yyyyddmm
Given a specific date, how do I extract the name of the row that has the largest value on that day?
I think you mean this:
awk -v date=20150823 '{for(f=2;f<=NF;f++){split($f,a,"|");if(a[1]==date&&a[2]>max){max=a[2];name=$1}}}END{print name,max}' YourFile
So, you pass the date you are looking for in as a variable called date. You then iterate through all fields on the line, and split the date and value of each into an array using | as separator - a[1] has the date, a[2] has the value. If the date matches and the value is greater than any previously seen maximum, save this as the new maximum and save the first field from this line for printing at the end.
You couldn't have taken 5 seconds to give your sample input different values? Anyway, this may work when run against input that actually has different values for the dates:
$ cat tst.awk
BEGIN { FS="[|[:space:]]+" }
FNR==1 {
for (i=2;i<=NF;i+=2) {
if ( $i==tgt ) {
f = i+1
}
}
max = $f
}
$f >= max { max=$f; name=$1 }
END { print name }
$ awk -v tgt=20150801 -f tst.awk file
name2
As a quick&dirty solution, we can perform this in following Unix commands:
yourdatafile=<yourdatafile>
yourdate=<yourdate>
cat $yourdatafile | sed 's/|/_/g' | awk -F "${yourdate}_" '{print $1" "$2}' | sed 's/[0-9]*_[0-9]*//g' | awk '{print $1" "$2}' |sort -k 2n | tail -n 1
With following sample data:
$ cat $yourdatafile
Alice 20150801|44 20150802|21 20150803|7 20150804|76 20150805|71
Bob 20150801|31 20150802|5 20150803|21 20150804|133 20150805|71
and yourdate=20150803 we get:
$ cat $yourdatafile | sed 's/|/_/g' | awk -F "${yourdate}_" '{print $1" "$2}' | sed 's/[0-9]*_[0-9]*//g' | awk '{print $1" "$2}' |sort -k 2n | tail -n 1
Bob 21
and for yourdate=20150802 we get:
$ cat $yourdatafile | sed 's/|/_/g' | awk -F "${yourdate}_" '{print $2" "$1}' | sed 's/[0-9]*_[0-9]*//g' | awk '{print $2" "$1}' | sort -k 2n | tail -n 1
Alice 21
The drawback is that only one line is printed the highest value of a day was achieved by more than one name as can be seen with:
$ yourdate=20150805; cat $yourdatafile | sed 's/|/_/g' | awk -F "${yourdate}_" '{print $2" "$1}' | sed 's/[0-9]*_[0-9]*//g' | awk '{print $2" "$1}' | sort -k 2n | tail -n 1
Bob 71
I hope that helps anyway.
Can anybody help me please?
grep " 287 " file.txt | grep "HI" | sed -i 's/HIS/HID/g'
sed: no input files
Tried also xargs
grep " 287 " file.txt | grep HI | xargs sed -i 's/HIS/HID/g'
sed: invalid option -- '6'
This works fine
grep " 287 " file.txt | grep HI
If you want to keep your pipeline:
f=file.txt
tmp=$(mktemp)
grep " 287 " "$f" | grep "HI" | sed 's/HIS/HID/g' > "$tmp" && mv "$tmp" "$f"
Or, simplify:
sed -i -n '/ 287 / {/HI/ s/HIS/HID/p}' file.txt
That will filter out any line that does not contain " 287 " and "HI" -- is that what you want? I suspect you really want this:
sed -i '/ 287 / {/HI/ s/HIS/HID/}' file.txt
For lines that match / 287 /, execute the commands in braces. In there, for lines that match /HI/, search for the first "HIS" and replace with "HID". sed implicitly prints all lines if -n is not specified.
Other commands that do the same thing:
awk '/ 287 / && /HI/ {sub(/HIS/, "HID")} {print}' file.txt > new.txt
perl -i -pe '/ 287 / and /HI/ and s/HIS/HID/' file.txt
awk does not have an "in-place" option (except gawk -i inplace for recent gawk versions)
I have a example cut down from a log file.
112 172.172.172.1#50912 (ssl.bing.com):
I would like some how to remove the # and numbers after and (): from the url.
Would like the result.
112 172.172.172.1 ssl.bing.com
Here is the sed oneliner I have been working on.
cat newdns.log | sed -e 's/.*query: //' | cut -f 1 -d' ' | sort | uniq -c | sort -k2 > old.log
Thanks
Using sed, you could say:
sed 's/#[0-9]*//;s/(\(.*\)):$/\1/' filename
or, in a single substitution:
sed 's/#[0-9]* *(\(.*\)):$/ \1/' filename
Another sed:
sed -r 's/#[^ ]+|[():]//g'
$ echo '112 172.172.172.1#50912 (ssl.bing.com):' | sed -r 's/#[^ ]+|[():]//g'
112 172.172.172.1 ssl.bing.com
The title says it all. I've seen this idiom used alot instead of adding an additional grep -v grep in some ps pipeline. For example it could be used like this:
$ ps aux | grep '[f]irefox' | awk '{ print $8 }'
instead of
$ ps aux | grep 'firefox' | grep -v grep | awk '{ print $8 }'
It's super-convenient, but how does it work and why?
The pattern [f]irefox will not match the literal string [f]irefox. Instead it will match strings with exactly one char from the 1-character class [f], followed by irefox.
I have a string ABCD20110420.txt and I want to extract the date out of it. Expected 2011-04-20
I can use replace to remove the text part, but how do I insert the "-" ?
# echo "ABCD20110420.txt" | replace 'ABCD' '' | replace '.txt' ''
20110420
echo "ABCD20110420.txt" | sed -e 's/ABCD//' -e 's/.txt//' -e 's/\(....\)\(..\)\(..\)/\1-\2-\3/'
Read: sed FAQ
Just use the shell (bash)
$> file=ABCD20110420.txt
$> echo "${file//[^0-9]/}"
20110420
$> file="${file//[^0-9]/}"
$> echo $file
20110420
$> echo ${file:0:4}-${file:4:2}-${file:6:2}
2011-04-20
The above is applicable to files like your sample. If you have files like A1BCD20110420.txt, then will not work.
For that case,
$> file=A1BCD20110420.txt
$> echo ${file%.*} #get rid of .txt
A1BCD20110420
$> file=${file%.*}
$> echo "2011${file#*2011}"
20110420
Or you can use regular expression (Bash 3.2+)
$> file=ABCD20110420.txt
$> [[ $file =~ ^.*(2011)([0-9][0-9])([0-9][0-9])\.*$ ]]
$> echo ${BASH_REMATCH[1]}
2011
$> echo ${BASH_REMATCH[2]}
04
$> echo ${BASH_REMATCH[3]}
20
echo "ABCD20110420.txt" | sed -r 's/.+([0-9]{4})([0-9]{2})([0-9]{2}).+/\1-\2-\3/'
$ file=ABCD20110420.txt
$ echo "$file" | sed -e 's/^[A-Za-z]*\([0-9][0-9][0-9][0-9]\)\([0-9][0-9]\)\([0-9][0-9]\)\.txt$/\1-\2-\3/'
This only requires a single call to sed.
echo "ABCD20110420.txt" | sed -r 's/.{4}(.{4})(.{2})(.{2}).txt/\1-\2-\3/'