244

I have a file which contains several thousand numbers, each on its own line:

34
42
11
6
2
99
...

I'm looking to write a script which will print the sum of all numbers in the file. I've got a solution, but it's not very efficient. (It takes several minutes to run.) I'm looking for a more efficient solution. Any suggestions?

danronmoon
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Mark Roberts
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    What was your slow solution? Maybe we can help you figure out what was slow about it. :) – brian d foy Apr 23 '10 at 23:59
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    @brian d foy, I'm too embarrassed to post it. I know why it's slow. It's because I call "cat filename | head -n 1" to get the top number, add it to a running total, and call "cat filename | tail..." to remove the top line for the next iteration... I have a lot to learn about programming!!! – Mark Roberts Apr 24 '10 at 01:22
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    That's...very systematic. Very clear and straight forward, and I love it for all that it is a horrible abomination. Built, I assume, out of the tools that you knew when you started, right? – dmckee --- ex-moderator kitten Apr 24 '10 at 02:43
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    full duplicate: http://stackoverflow.com/questions/450799/linux-command-to-sum-integers-one-per-line – codeholic Apr 26 '10 at 11:39
  • @MarkRoberts It must have taken you a long while to work that out. It's a very cleaver problem solving technique, and oh so wrong. It looks like a classic case of over think. Several of [Glen Jackman's](http://stackoverflow.com/a/18382280/368630) solutions shell scripting solutions (and two are pure shell that don't use things like `awk` and `bc`). These all finished adding a million numbers up in less than 10 seconds. Take a look at those and see how it can be done in pure shell. – David W. Aug 22 '13 at 14:24
  • @ Mark Roberts 1place, http://stackoverflow.com/a/18380369/4592448 ))) – Fortran Apr 30 '17 at 21:01

33 Answers33

450

You can use awk:

awk '{ sum += $1 } END { print sum }' file
Ayman Hourieh
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    program exceeded: maximum number of field sizes: 32767 – leef Nov 02 '12 at 03:50
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    With the `-F '\t'` option if your fields contain spaces and are separated by tabs. – Ethan Furman Jul 17 '14 at 22:17
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    Please mark this as the best answer. It also works if you want to sum the first value in each row, inside a TSV (tab-separated value) file. – Andrea Oct 27 '16 at 10:42
  • If you have big numbers: `awk 'BEGIN {OFMT = "%.0f"} { sum += $1 } END { print sum }' filename` – FlipMcF Feb 15 '22 at 17:30
  • @EthanFurman I actually have a tab delimited file as you explained but not able to make -F '\t' do the magic. Where exactly is the option meant to be inserted? I have it like this awk -F '\t' '{ sum += $0 } END { print sum }' file – CYNTHIA Blessing Apr 07 '22 at 16:20
  • @CYNTHIABlessing: Please ask that as a new question. Thanks! – Ethan Furman Apr 07 '22 at 17:41
136

None of the solution thus far use paste. Here's one:

paste -sd+ filename | bc

If the file has a trailing newline, a trailing + will incur a syntax error. Fix the error by removing the trailing +:

paste -sd+ fiilename | sed 's/+$//g' | bc

As an example, calculate Σn where 1<=n<=100000:

$ seq 100000 | paste -sd+ | bc -l
5000050000

(For the curious, seq n would print a sequence of numbers from 1 to n given a positive number n.)

Dave Jarvis
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devnull
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    `seq 100000 | paste -sd+ - | bc -l` on Mac OS X Bash shell. And this is by far the sweetest and the unixest solution! – Simo A. Mar 02 '19 at 15:54
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    @SimoA. I vote that we use the term unixiest in place of unixest because to the sexiest solution is always the unixiest ;) – Connor Feb 17 '20 at 19:25
118

For a Perl one-liner, it's basically the same thing as the awk solution in Ayman Hourieh's answer:

 % perl -nle '$sum += $_ } END { print $sum'

If you're curious what Perl one-liners do, you can deparse them:

 %  perl -MO=Deparse -nle '$sum += $_ } END { print $sum'

The result is a more verbose version of the program, in a form that no one would ever write on their own:

BEGIN { $/ = "\n"; $\ = "\n"; }
LINE: while (defined($_ = <ARGV>)) {
    chomp $_;
    $sum += $_;
}
sub END {
    print $sum;
}
-e syntax OK

Just for giggles, I tried this with a file containing 1,000,000 numbers (in the range 0 - 9,999). On my Mac Pro, it returns virtually instantaneously. That's too bad, because I was hoping using mmap would be really fast, but it's just the same time:

use 5.010;
use File::Map qw(map_file);

map_file my $map, $ARGV[0];

$sum += $1 while $map =~ m/(\d+)/g;

say $sum;
Community
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brian d foy
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    Wow, that shows a _deep_ understanding on what code -nle actually wraps around the string you give it. My initial thought was that you shouldn't post while intoxicated but then I noticed who you were and remembered some of your other Perl answers :-) – paxdiablo Apr 23 '10 at 23:52
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    -n and -p just put characters around the argument to -e, so you can use those characters for whatever you want. We have a lot of one-liners that do interesting things with that in _Effective Perl Programming_ (which is about to hit the shelves). – brian d foy Apr 23 '10 at 23:56
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    Nice, what are these non-matching curly braces about? – Frank Apr 24 '10 at 06:00
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    -n adds the `while { }` loop around your program. If you put `} ... {` inside, then you have `while { } ... { }`. Evil? Slightly. – jrockway Apr 24 '10 at 08:35
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    Big bonus for highlighting the `-MO=Deparse` option! Even though on a separate topic. – conny Nov 04 '11 at 12:47
  • The cumulative version of your one-liner (a rolling sum, printing the current sum for each line): `perl -nle '$sum += $_; print $sum} END {'` – nh2 Sep 13 '13 at 11:35
95

Just for fun, let's benchmark it:

$ for ((i=0; i<1000000; i++)) ; do echo $RANDOM; done > random_numbers

$ time perl -nle '$sum += $_ } END { print $sum' random_numbers
16379866392

real    0m0.226s
user    0m0.219s
sys     0m0.002s

$ time awk '{ sum += $1 } END { print sum }' random_numbers
16379866392

real    0m0.311s
user    0m0.304s
sys     0m0.005s

$ time { { tr "\n" + < random_numbers ; echo 0; } | bc; }
16379866392

real    0m0.445s
user    0m0.438s
sys     0m0.024s

$ time { s=0;while read l; do s=$((s+$l));done<random_numbers;echo $s; }
16379866392

real    0m9.309s
user    0m8.404s
sys     0m0.887s

$ time { s=0;while read l; do ((s+=l));done<random_numbers;echo $s; }
16379866392

real    0m7.191s
user    0m6.402s
sys     0m0.776s

$ time { sed ':a;N;s/\n/+/;ta' random_numbers|bc; }
^C

real    4m53.413s
user    4m52.584s
sys 0m0.052s

I aborted the sed run after 5 minutes


I've been diving to , and it is speedy:

$ time lua -e 'sum=0; for line in io.lines() do sum=sum+line end; print(sum)' < random_numbers
16388542582.0

real    0m0.362s
user    0m0.313s
sys     0m0.063s

and while I'm updating this, ruby:

$ time ruby -e 'sum = 0; File.foreach(ARGV.shift) {|line| sum+=line.to_i}; puts sum' random_numbers
16388542582

real    0m0.378s
user    0m0.297s
sys     0m0.078s

Heed Ed Morton's advice: using $1

$ time awk '{ sum += $1 } END { print sum }' random_numbers
16388542582

real    0m0.421s
user    0m0.359s
sys     0m0.063s

vs using $0

$ time awk '{ sum += $0 } END { print sum }' random_numbers
16388542582

real    0m0.302s
user    0m0.234s
sys     0m0.063s
glenn jackman
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35

Another option is to use jq:

$ seq 10|jq -s add
55

-s (--slurp) reads the input lines into an array.

nisetama
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  • Wonderful solution. I had a tab delimited file where I wanted to sum column 6. Did that with the following command: `awk '{ print $6 }' myfile.log | jq -s add` – Lo-Tan Sep 13 '21 at 06:52
11

This is straight Bash:

sum=0
while read -r line
do
    (( sum += line ))
done < file
echo $sum
Dennis Williamson
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    And it's probably one of the slowest solutions and therefore not so suitable for large amounts of numbers. – David Apr 23 '19 at 16:06
10

I prefer to use GNU datamash for such tasks because it's more succinct and legible than perl or awk. For example

datamash sum 1 < myfile

where 1 denotes the first column of data.

hertzsprung
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8

I prefer to use R for this:

$ R -e 'sum(scan("filename"))'
fedorn
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  • I'm a fan of R for other applications but it's not good for performance in this way. File I/O is a major issue. I've tested passing args to a script which can be sped up using the vroom package. I'll post more details when I've benchmarked some other scripts on the same server. – Tom Kelly Jun 24 '19 at 05:54
7

Here's another one-liner

( echo 0 ; sed 's/$/ +/' foo ; echo p ) | dc

This assumes the numbers are integers. If you need decimals, try

( echo 0 2k ; sed 's/$/ +/' foo ; echo p ) | dc

Adjust 2 to the number of decimals needed.

lhf
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7

Perl 6

say sum lines
~$ perl6 -e '.say for 0..1000000' > test.in

~$ perl6 -e 'say sum lines' < test.in
500000500000
Brad Gilbert
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6
$ perl -MList::Util=sum -le 'print sum <>' nums.txt
Zaid
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5

More succinct:

# Ruby
ruby -e 'puts open("random_numbers").map(&:to_i).reduce(:+)'

# Python
python -c 'print(sum(int(l) for l in open("random_numbers")))'
Vidul
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  • Converting to float seems to be about twice as fast on my system (320 vs 640 ms). `time python -c "print(sum([float(s) for s in open('random_numbers','r')]))"` – user12719 May 16 '20 at 19:06
5

I couldn't just pass by... Here's my Haskell one-liner. It's actually quite readable:

sum <$> (read <$>) <$> lines <$> getContents

Unfortunately there's no ghci -e to just run it, so it needs the main function, print and compilation.

main = (sum <$> (read <$>) <$> lines <$> getContents) >>= print

To clarify, we read entire input (getContents), split it by lines, read as numbers and sum. <$> is fmap operator - we use it instead of usual function application because sure this all happens in IO. read needs an additional fmap, because it is also in the list.

$ ghc sum.hs
[1 of 1] Compiling Main             ( sum.hs, sum.o )
Linking sum ...
$ ./sum 
1
2
4
^D
7

Here's a strange upgrade to make it work with floats:

main = ((0.0 + ) <$> sum <$> (read <$>) <$> lines <$> getContents) >>= print
$ ./sum 
1.3
2.1
4.2
^D
7.6000000000000005
Peter K
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4
cat nums | perl -ne '$sum += $_ } { print $sum'

(same as brian d foy's answer, without 'END')

edibleEnergy
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  • I like this, but could you explain the curly brackets? It's weird to see } without { and vice versa. – drumfire Jun 17 '20 at 21:40
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    @drumfire see @brian d foy's answer above with `perl -MO=Deparse` to see how perl parses the program. or the docs for perlrun: https://perldoc.perl.org/perlrun.html (search for -n). perl wraps your code with { } if you use -n so it becomes a complete program. – edibleEnergy Jun 18 '20 at 23:12
4

Just for fun, lets do it with PDL, Perl's array math engine!

perl -MPDL -E 'say rcols(shift)->sum' datafile

rcols reads columns into a matrix (1D in this case) and sum (surprise) sums all the element of the matrix.

Joel Berger
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  • How fix Can't locate PDL.pm in @INC (you may need to install the PDL module) (@INC contains: /etc/perl /usr/local/lib/x86_64-linux-gnu/perl/5.22.1 ?)) for fun of course=) – Fortran Apr 29 '17 at 15:01
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    You have to install PDL first, it isn't a Perl native module. – Joel Berger May 01 '17 at 20:06
3

Here is a solution using python with a generator expression. Tested with a million numbers on my old cruddy laptop.

time python -c "import sys; print sum((float(l) for l in sys.stdin))" < file

real    0m0.619s
user    0m0.512s
sys     0m0.028s
dwurf
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    A simple list comprehension with a named function is a nice use-case for `map()`: `map(float, sys.stdin)` – sevko May 01 '15 at 19:50
3

C++ "one-liner":

#include <iostream>
#include <iterator>
#include <numeric>
using namespace std;

int main() {
    cout << accumulate(istream_iterator<int>(cin), istream_iterator<int>(), 0) << endl;
}
Peter K
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2

C always wins for speed:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char **argv) {
    ssize_t read;
    char *line = NULL;
    size_t len = 0;
    double sum = 0.0;

    while (read = getline(&line, &len, stdin) != -1) {
        sum += atof(line);
    }

    printf("%f", sum);
    return 0;
}

Timing for 1M numbers (same machine/input as my python answer):

$ gcc sum.c -o sum && time ./sum < numbers 
5003371677.000000
real    0m0.188s
user    0m0.180s
sys     0m0.000s
dwurf
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2
sed ':a;N;s/\n/+/;ta' file|bc
ghostdog74
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2

Running R scripts

I've written an R script to take arguments of a file name and sum the lines.

#! /usr/local/bin/R
file=commandArgs(trailingOnly=TRUE)[1]
sum(as.numeric(readLines(file)))

This can be sped up with the "data.table" or "vroom" package as follows:

#! /usr/local/bin/R
file=commandArgs(trailingOnly=TRUE)[1]
sum(data.table::fread(file))
#! /usr/local/bin/R
file=commandArgs(trailingOnly=TRUE)[1]
sum(vroom::vroom(file))

Benchmarking

Same benchmarking data as @glenn jackman.

for ((i=0; i<1000000; i++)) ; do echo $RANDOM; done > random_numbers

In comparison to the R call above, running R 3.5.0 as a script is comparable to other methods (on the same Linux Debian server).

$ time R -e 'sum(scan("random_numbers"))'  
 0.37s user
 0.04s system
 86% cpu
 0.478 total

R script with readLines

$ time Rscript sum.R random_numbers
  0.53s user
  0.04s system
  84% cpu
  0.679 total

R script with data.table

$ time Rscript sum.R random_numbers     
 0.30s user
 0.05s system
 77% cpu
 0.453 total

R script with vroom

$ time Rscript sum.R random_numbers     
  0.54s user 
  0.11s system
  93% cpu
  0.696 total

Comparison with other languages

For reference here as some other methods suggested on the same hardware

Python 2 (2.7.13)

$ time python2 -c "import sys; print sum((float(l) for l in sys.stdin))" < random_numbers 
 0.27s user 0.00s system 89% cpu 0.298 total

Python 3 (3.6.8)

$ time python3 -c "import sys; print(sum((float(l) for l in sys.stdin)))" < random_number
0.37s user 0.02s system 98% cpu 0.393 total

Ruby (2.3.3)

$  time ruby -e 'sum = 0; File.foreach(ARGV.shift) {|line| sum+=line.to_i}; puts sum' random_numbers
 0.42s user
 0.03s system
 72% cpu
 0.625 total

Perl (5.24.1)

$ time perl -nle '$sum += $_ } END { print $sum' random_numbers
 0.24s user
 0.01s system
 99% cpu
 0.249 total

Awk (4.1.4)

$ time awk '{ sum += $0 } END { print sum }' random_numbers
 0.26s user
 0.01s system
 99% cpu
 0.265 total
$ time awk '{ sum += $1 } END { print sum }' random_numbers
 0.34s user
 0.01s system
 99% cpu
 0.354 total

C (clang version 3.3; gcc (Debian 6.3.0-18) 6.3.0 )

 $ gcc sum.c -o sum && time ./sum < random_numbers   
 0.10s user
 0.00s system
 96% cpu
 0.108 total

Update with additional languages

Lua (5.3.5)

$ time lua -e 'sum=0; for line in io.lines() do sum=sum+line end; print(sum)' < random_numbers 
 0.30s user 
 0.01s system
 98% cpu
 0.312 total

tr (8.26) must be timed in bash, not compatible with zsh

$time { { tr "\n" + < random_numbers ; echo 0; } | bc; }
real    0m0.494s
user    0m0.488s
sys 0m0.044s

sed (4.4) must be timed in bash, not compatible with zsh

$  time { head -n 10000 random_numbers | sed ':a;N;s/\n/+/;ta' |bc; }
real    0m0.631s
user    0m0.628s
sys     0m0.008s
$  time { head -n 100000 random_numbers | sed ':a;N;s/\n/+/;ta' |bc; }
real    1m2.593s
user    1m2.588s
sys     0m0.012s

note: sed calls seem to work faster on systems with more memory available (note smaller datasets used for benchmarking sed)

Julia (0.5.0)

$ time julia -e 'print(sum(readdlm("random_numbers")))'
 3.00s user 
 1.39s system 
 136% cpu 
 3.204 total
$  time julia -e 'print(sum(readtable("random_numbers")))'
 0.63s user 
 0.96s system 
 248% cpu 
 0.638 total

Notice that as in R, file I/O methods have different performance.

Tom Kelly
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2

Bash variant

raw=$(cat file)
echo $(( ${raw//$'\n'/+} ))

$ wc -l file
10000 file

$ time ./test
323390

real    0m3,096s
user    0m3,095s
sys     0m0,000s

What is happening here? Read the content of a file into $raw var. Then create math statement from this var by changing all new lines into '+'

Ivan
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1

Another for fun

sum=0;for i in $(cat file);do sum=$((sum+$i));done;echo $sum

or another bash only

s=0;while read l; do s=$((s+$l));done<file;echo $s

But awk solution is probably best as it's most compact.

nickjb
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1

With Ruby:

ruby -e "File.read('file.txt').split.inject(0){|mem, obj| mem += obj.to_f}"
sites
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1

In Go:

package main

import (
    "bufio"
    "fmt"
    "os"
    "strconv"
)

func main() {
    scanner := bufio.NewScanner(os.Stdin)
    sum := int64(0)
    for scanner.Scan() {
        v, err := strconv.ParseInt(scanner.Text(), 10, 64)
        if err != nil {
            fmt.Fprintf(os.Stderr, "Not an integer: '%s'\n", scanner.Text())
            os.Exit(1)
        }
        sum += v
    }
    fmt.Println(sum)
}
dwurf
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  • What is "64"? "10" I suppose is base? – Peter K Mar 17 '20 at 00:37
  • Yes, 10 is the base. 64 is the number of bits, if the resulting int can't be represented with that many bits then an error is returned. See https://golang.org/pkg/strconv/#ParseInt – dwurf Mar 18 '20 at 01:30
0

Here's another:

open(FIL, "a.txt");

my $sum = 0;
foreach( <FIL> ) {chomp; $sum += $_;}

close(FIL);

print "Sum = $sum\n";
ruben2020
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0

I don't know if you can get a lot better than this, considering you need to read through the whole file.

$sum = 0;
while(<>){
   $sum += $_;
}
print $sum;
Stefan Kendall
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    Very readable. For perl. But yeah, it's going to have to be something like that... – dmckee --- ex-moderator kitten Apr 23 '10 at 23:40
  • `$_` is the default variable. The line input operator, `<>`, puts it's result in there by default when you use `<>` in `while`. – brian d foy Apr 23 '10 at 23:53
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    @Mark, `$_` is the topic variable--it works like the 'it'. In this case `<>` assigns each line to it. It gets used in a number of places to reduce code clutter and help with writing one-liners. The script says "Set the sum to 0, read each line and add it to the sum, then print the sum." – daotoad Apr 23 '10 at 23:59
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    @Stefan, with warnings and strictures off, you can skip declaring and initializing `$sum`. Since this is so simple, you can even use a statement modifier `while`: `$sum += $_ while <>; print $sum;` – daotoad Apr 24 '10 at 00:00
  • for the rest of us who can't easily, how about you indicate which language this is in? `PHP`? `Perl`? – Dut A. Oct 20 '21 at 02:29
0

I have not tested this but it should work:

cat f | tr "\n" "+" | sed 's/+$/\n/' | bc

You might have to add "\n" to the string before bc (like via echo) if bc doesn't treat EOF and EOL...

DVK
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0

You can do it with Alacon - command-line utility for Alasql database.

It works with Node.js, so you need to install Node.js and then Alasql package:

To calculate sum from TXT file you can use the following command:

> node alacon "SELECT VALUE SUM([0]) FROM TXT('mydata.txt')"
agershun
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0

It is not easier to replace all new lines by +, add a 0 and send it to the Ruby interpreter?

(sed -e "s/$/+/" file; echo 0)|irb

If you do not have irb, you can send it to bc, but you have to remove all newlines except the last one (of echo). It is better to use tr for this, unless you have a PhD in sed .

(sed -e "s/$/+/" file|tr -d "\n"; echo 0)|bc
Daniel Porumbel
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0

In shell using awk, I have used below script to do so:

    #!/bin/bash


total=0;

for i in $( awk '{ print $1; }' <myfile> )
do
 total=$(echo $total+$i | bc )
 ((count++))
done
echo "scale=2; $total " | bc
Channa
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0

One in tcl:

#!/usr/bin/env tclsh
set sum 0
while {[gets stdin num] >= 0} { incr sum $num }
puts $sum
Shawn
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0

GNU Parallel can presumably be used to improve many of the above answers by spreading the workload across multiple cores.

In the example below we send chunks of 500 numbers (--max-lines=500) to bc processes which are executed in parallel 4 at a time (-j 4). The results are then aggregated by a final bc.

time parallel --max-lines=500 -j 4 --pipe "paste -sd+ - | bc" < random_numbers | paste -sd+ - | bc

The optimal choice of work size and number of parallel processes depends on the machine and problem. Note that this solution only really shines when there's a large number of parallel processes with substantial work each.

user12719
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0

in awk, just getting a 2nd column with cumulative sum is far less syntax than saving it towards the end :

jot 20 61111111889 - 799973766543 | 

mawk '$2=_+=$1'        # skips rows with zero(0) as its value
gawk '($2=_+=$1)_'     # no rows left behind

61111111889 61111111889
861084878432 922195990321
1661058644975 2583254635296
2461032411518 5044287046814

3261006178061 8305293224875
4060979944604 12366273169479
4860953711147 17227226880626
5660927477690 22888154358316

6460901244233 29349055602549
7260875010776 36609930613325
8060848777319 44670779390644
8860822543862 53531601934506

9660796310405 63192398244911
10460770076948 73653168321859
11260743843491 84913912165350
12060717610034 96974629775384

12860691376577 109835321151961
13660665143120 123495986295081
14460638909663 137956625204744
15260612676206 153217237880950

for all practical purposes, perl5 python3 and mawk2 are tied for speed summing up from 1 to 99,999,999 ::


(echo '99999999' | mawk2 '$++NF = (__=+$++_)*++__/++_'

99999999 4999999950000000

(all input digits were re-generated on the fly and piped in to eliminate any potential cache access advantage) :

      in0:  847MiB 0:00:10 [81.1MiB/s] [81.1MiB/s] [ <=> ]
     1  4999999950000000

( python3 -c 'import sys; print(int(sum((float(_) for _ in sys.stdin))))')  
      19.14s user 0.55s system 188% cpu 10.473 total
gcat -b  0.00s user 0.00s system 0% cpu 10.473 total

      in0:  847MiB 0:00:10 [81.0MiB/s] [81.0MiB/s] [ <=> ]
     1  4999999950000000

( perl536 -nle '$sum += $_ } END { print $sum' )  
          19.37s user 0.55s system 190% cpu 10.472 total
gcat -b      0.00s user 0.00s system 0% cpu 10.472 total

      in0:  847MiB 0:00:10 [81.1MiB/s] [81.1MiB/s] [ <=>]
     1  4999999950000000

( mawk1996 '{ _+=$__ } END { print _ }' )  

      17.51s user 0.57s system 172% cpu 10.463 total
gcat -b  0.00s user 0.00s system 0% cpu 10.463 total

However, once you eliminate the pipe and hashing speed factors and ask them to sum it among itself, perl5.36 is some 52% slower :

 ( time (

 mawk2 'BEGIN { for(___=_-=_=__=((_+=++_)+(_*=_+_))^_; ++_<__;)___+=_
        print ___ }' 
                     ) | gcat -b ) | lgp3 ;

 ( time ( 

perl5 -e '$y = $x = 0; $z = 10**8; while(++$x < $z) { $y += $x } print $y' 
         ) | gcat -b ) | lgp3 ;

     1  4999999950000000

( mawk2 ; )  1.97s user 0.01s system 99% cpu 1.981 total
gcat -b  0.00s user 0.00s system 0% cpu 1.979 total

( perl5 -e '$y = $x = 0; $z = 10**8; while(++$x < $z) { $y += $x } print $y';   2.98s user 0.03s system 99% cpu 3.015 total
gcat -b  0.00s user 0.00s system 0% cpu 3.014 total
     1  4999999950000000

as for gnu-parallel, they're more than half an order of magnitude slower

  • 36 concurrent jobs with 5,000,000 rows per job and very generous 100 MB size upper cap running on M1 Max with 64 GB ram and it still took nearly 53 seconds compare to about 10.5 secs for the other 3.

  ( time ( mawk2 'BEGIN { for(_-=_=__=((_+=++_)+(_*=_+_))^_; ++_ < __; ) print _ }' | pvE0 | 

   parallel --block 100M -N 5000000 -j 36 --pipe "gpaste -sd+ - | bc" | gpaste -sd+ - | bc 


   ) |  gcat -b ) | lgp3 | lgp3 -1;


   in0:  847MiB 0:00:47 [17.8MiB/s] [17.8MiB/s] [  <=> ]
   1    4999999950000000

   0.00s user 0.00s system 0% cpu 52.895 total
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