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Each line contains text and numbers in one column. I need to calculate the sum of the numbers in each row. How can I do that? Thx

example.log contains:

time=31sec time=192sec time=18sec time=543sec

The answer should be 784

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  • I tried this method awk '{ sum += $1}; END { print sum }' example.log but it's only for numbers in line Commented May 27, 2015 at 18:13
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    There is almost the same question in Stack Overflow: How can I quickly sum all numbers in a file?. Maybe time to have cross-site duplicates? Commented May 28, 2015 at 7:26

10 Answers 10

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If your grep support -o option, you can try:

$ grep -o '[[:digit:]]*' file | paste -sd+ - | bc 784

POSIXly:

$ printf %d\\n "$(( $(tr -cs 0-9 '[\n*]' <file | paste -sd+ -) ))" 784
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With a newer version (4.x) of GNU awk:

awk 'BEGIN {FPAT="[0-9]+"}{s+=$1}END{print s}'

With other awks try:

awk -F '[a-z=]*' '{s+=$2}END{print s}'
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  • 4
    You need s+0 in case where s is empty, it will print 0 instead of empty. Commented May 27, 2015 at 18:20
  • Let me explain that. - There is just one case where s can be empty; if the input data contains no lines (i.e. if there is no input at all). In that case there are two behaviours possible; 1) no input => no output, or 2) always output something, if only 0. Both are sensible options depending on the application context. The +0 is addressing option 2). To address option 1) you'd rather have to write END {if(s) print s}. - Therefore it makes no sense to assume either option (for this corner case of no data) until it is specified by the question. Commented May 28, 2015 at 12:40
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awk -F= '{sum+=$2};END{print sum}'
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    We prefer long form answers. Can you please elaborate on how this works? Commented May 28, 2015 at 9:31
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    @slm, that answer is not any more or less verbose than the other answers here and is self explanatory. It also has the advantage of working with input like time=1.4e5sec Commented May 28, 2015 at 21:42
  • @StéphaneChazelas - agreed, but this is a new user and we do encourage users to provide more than single line answers. A bit of text explaining how it works would make it a much stronger answer than just code. Commented May 28, 2015 at 21:47
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    @slm, this is a new user with one of the best answers (from a technical stand point) and he gets two downvotes and a negative comment. Not a very warm welcome. Commented May 28, 2015 at 21:52
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    @TomFenech, the POSIX syntax for awk requires that those pattern/action items be separated by either ";" or "newline", so you may find awk implementations where it fails without this ";". Commented May 29, 2015 at 9:28
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Another GNU awk one:

awk -v RS='[0-9]+' '{n+=RT};END{print n}'

A perl one:

perl -lne'$n+=$_ for/\d+/g}{print$n'

A POSIX one:

tr -cs 0-9 '[\n*]' | grep . | paste -sd + - | bc
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sed 's/=/ /' file | awk '{ sum+=$2 } END { print sum}'
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  • Awesome answer, but no need for sed: awk --field-separator = '{ sum+=$2 } END { print sum}' data.dat Commented May 27, 2015 at 23:45
  • @user1717828: you should rather use the (shorter, and more compatible!) -F'=' instead of --field-separator = Commented May 29, 2015 at 9:14
  • @OlivierDulac, weird, my man awk only gives -F fs and --field-separator fs Commented May 29, 2015 at 10:40
  • @user1717828: -F'=' or -F '=' are 2 ways of doing the -F fs (fs is "=" in your case) . I added the singlequotes to ensure the fs is properly seen & interpreted by awk, not the shell (usefull if the fs is ';' for example) Commented May 29, 2015 at 11:56
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You can try this:

awk -F"[^0-9]+" '{ sum += $2 } END { print sum+0; }' file
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Everyone has posted awesome awk answers, which I like very much.

A variation to @cuonglm replacing grep with sed:

sed 's/[^0-9]//g' example.log | paste -sd'+' - | bc
  1. The sed strips everything except for the numbers.
  2. The paste -sd+ - command joins all the lines together as a single line
  3. The bc evaluates the expression
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3

You should use a calculator.

{ tr = \ | xargs printf '[%s=]P%d+p' | dc; } <infile 2>/dev/null

With your four lines that prints:

time=31 time=223 time=241 time=784

And more simply:

tr times=c ' + p' <infile |dc

...which prints...

31 223 241 784

If speed is what you're after then dc is what you want. Traditionally it was bc's compiler - and still is for many systems.

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  • Not according to my measurements: it depends how much work you have to do to generate the formula Commented May 28, 2015 at 13:24
  • @glennjackman - your measurements don't include dc as near as I can tell. What are you talking about? Commented May 28, 2015 at 15:14
  • By the way, when comparing the old crew to the new crew - such as when you benchmark perl v the standard unix toolset - it really doesn't make much sense if you use GNU tools compiled on a GNU toolchain. All of the bloat that can negatively affect Perl's performance is also in all of those GNU-compiled GNU utils. Sad but true. You need a real, simply built, simple toolset to accurately judge the difference. Like an heirloom-toolchest set statically linked against musl libs for instance - in that way you can bench the one-tool/one-job paradigm vs the one-tool-to-rule-them-all one. Commented May 28, 2015 at 15:27
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Through python3,

import re with open(file) as f: m = f.read() l = re.findall(r'\d+', m) print(sum(map(int, l)))
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  • re.findall returns a list of strings, this is not going to work Commented May 28, 2015 at 22:28
  • @1_CR ya , I forget that. Check it now. Commented May 29, 2015 at 4:10
  • Maybe sum(int(e) for e in l) is more pythonic. Commented May 29, 2015 at 15:18
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Pure bash solution (Bash 3+):

while IFS= read -r line; do # While it reads a line: if [[ "$line" =~ [0-9]+ ]]; then # If the line contains numbers: ((counter+=BASH_REMATCH[0])) # Add the current number to counter fi # End if. done # End loop. echo "Total number: $counter" # Print the number. unset counter # Reset counter to 0.

Short version:

while IFS= read -r l; do [[ "$l" =~ [0-9]+ ]] && ((c+=BASH_REMATCH)); done; echo $c; c=0
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  • 1
    Maybe also: PS4='$((x+=${time%s*}))' time=0 x=0 sh -x <infile Commented May 31, 2015 at 9:39

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