How to print line number that contains specific number?

Question

I have 6000 lines,with one per number float.

My code

awk '5.400000e+03 {print $0}' base.txt

Only prints numbers

5.400000e+03
5.400000e+03
5.400000e+03
5.400000e+03
5.400000e+03
5.400000e+03
5.400000e+03
5.400000e+03
5.400000e+03
5.400000e+03
5.400000e+03
5.400000e+03

I have changed to

awk '{if($0=="5.400000e+03 ") print NR}' base.txt

but then got nothing! What should I try?

There is a space in the string in your command, but not in your file. — Benjamin W., Apr 27 '16 at 17:35

score 3 · Accepted Answer · answered Apr 27 '16 at 17:35

3

You can use:

awk '$1 == "5.400000e+03" {print NR, $0}' base.txt

$1 will always match first column irrespective of spaces or no spaces after shown data in your question. Use print NR, $0 to print each record prefixed with record number.

answered Apr 27 '16 at 17:35

anubhava

761,203
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569
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dimid · Answer 2 · 2016-04-27T18:14:22.830

3

You can also use grep:

# just the line number
grep -wn "5.400000e+03" base.txt | cut -d: -f1
# both
grep -wn "5.400000e+03" base.txt

edited Apr 27 '16 at 18:14

answered Apr 27 '16 at 17:39

dimid

7,285
1
46
85

1

Actually that's a typical `grep` job. Good answer! – hek2mgl Apr 27 '16 at 17:42
1

So you don't match something like `15.400000e+03`, use an anchor: `grep -n '^5.400000e+03'`. Or use non-regex matching: `grep -Fw '5.400000e+03'` – glenn jackman Apr 27 '16 at 18:00

How to print line number that contains specific number?

2 Answers2