Grep for a number excluding its floating point part

Question

I have this following line

scalar TestDmaMac4.sink.udpApp[0]   throughput:last     11730.559888477

I want to extract only 11730 out of this line, how can I do it with grep? I want to ignore the number after decimal point and only need digits before decimal point.

(Note: there is a {space}{tab} sequence separating each of udpApp[0], throughput:last and the number beginning 11730.)

Possible duplicate of extract value between two search patterns on same line — Sundeep, Apr 13 '17 at 15:06
@Sundeep No, i only want the number ignoring the numbers after the floating point — Linux-Fan85, Apr 13 '17 at 15:10
yes, and it is possible using look-arounds if -P is available as shown in accepted answer in that question... or using sed/awk etc... do add the command you tried but failed... we can help you correct it — Sundeep, Apr 13 '17 at 15:13
Does it need to be a regex at all? can't you use something like awk '{print int($NF)}' — steeldriver, Apr 13 '17 at 17:11
Is the number always complete up to the decimal? e.g. then egrep -o '[0-9]+\.' | tr -d . — thrig, Apr 13 '17 at 18:50

score 0 · Accepted Answer · answered Apr 13 '17 at 22:44

The bellow regexp will match any floating number in the format [0-9].[0-9] and will return the integer part of this floating number.

$ a="scalar TestDmaMac4.sink.udpApp[0] throughput:last 11730.559888477"
$ egrep -o '[0-9]+[.][0-9]' <<<"$a" |egrep -o '[0-9]+[^.]'  #First grep will isolate the floating number , second grep will isolate the int part.
11730
$ perl -pe 's/(.*?)([0-9]+)(\.[0-9]+.*)/\2/' <<<"$a"  #using the lazy operator ?
11730
$ sed -r 's/(.*[^0-9.])([0-9]+)(\.[0-9]+.*)/\2/' <<<"$a"  #sed does not have lazy operator thus we simulate this with negation
11730

For the sake of testing, i also tried above regexp in a different string with floatting number at different position without a leading space:

$ c="scalar throughput:last11730.559888477 TestDmaMac4.sink.udpApp[0]"
$ egrep -o '[0-9]+[.][0-9]' <<<"$c" |egrep -o '[0-9]+[^.]'
11730
$ perl -pe 's/(.*?)([0-9]+)(\.[0-9]+.*)/\2/' <<<"$c"
11730
$ sed -r 's/(.*[^0-9.])([0-9]+)(\.[0-9]+.*)/\2/' <<<"$c"
11730

score 0 · Answer 2 · answered Apr 14 '17 at 02:43

0

l='scalar TestDmaMac4.sink.udpApp[0]   throughput:last     11730.559888477'
read -r -a a <<<"$l"
dc -e "${a[-1]}dX10r^dsa*la/p"

echo "$l" | perl -lane 'print/\d+(?=\.\d+$)/g'

result

answered Apr 14 '17 at 02:43

score -2 · Answer 3 · edited Apr 13 '17 at 19:17

-2

Using Grep:

grep -o " [0-9]\{1,\}"

To Test:

echo "scalar TestDmaMac4.sink.udpApp[0] throughput:last 11730.559888477" | grep -o " [0-9]\{1,\}"

results:

edited Apr 13 '17 at 19:17

Stephen Rauch

4,239

answered Apr 13 '17 at 18:57

Norm

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Grep for a number excluding its floating point part

3 Answers3

result