(Need in bash linux)I have a file with numbers like this
1.415949602 91.09582241 91.12042924 91.40270349 91.45625033 91.70150341 91.70174342 91.70660043 91.70966213 91.72597066 91.7287678315 91.7398645966 91.7542977976 91.7678146465 91.77196659 91.77299733 abcdefghij 91.7827827 91.78288651 91.7838959 91.7855 91.79080605 91.80103075 91.8050505 sed 's/^91.//' file (working)
Any way possible I can do these 3 steps?
1st I try this
cat input | tr -d 91. > 1.txt (didnt work) cat input | tr -d "91." > 1.txt (didnt work) cat input | tr -d '91.' > 1.txt (didnt work)
then
grep -x '.{10}' (working)
then
grep "^[6-9]" (working)
Final 1 line solution
cat input.txt | sed 's/91.//g' | grep -x '.{10}' | grep "^[6-9]" > output.txt
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Answer
Your “final” solution:
cat input.txt |
sed 's/91.//g' |
grep -x '.{10}' |
grep "^[6-9]" > output.txt
should avoid the useless cat, and also move the backslash in the sed script to the correct place (and I added a ^ anchor and removed the g flag since you don’t expect more than one match on a line anyway);
sed 's/^91.//' input.txt |
grep -x '.{10}' |
grep "^[6-9]" > output.txt
You might also be able to get rid of at least one useless grep but at this point, I would switch to Awk:
awk '{ sub(/^91./, "") } /^[6-9].{9}$/' input.txt >output.txt
The sub() does what your sed replacement did; the final condition says to print lines which match the regex.
The same can conveniently, but less readably, be written in sed:
sed -n 's/^91.([6-9][0-9]{9})$/1/p' input.txt >output.txt
assuming your sed dialect supports BRE regex with repetitions like [0-9]{9}.