Shell command to sum integers, one per line?

Paste typically merges lines of multiple files, but it can also be used to convert individual lines of a file into a single line. The delimiter flag allows you to pass a x+x type equation to bc.

paste -s -d+ infile | bc

Alternatively, when piping from stdin,

<commands> | paste -s -d+ - | bc

The one-liner version in Python:

$ python -c "import sys; print(sum(int(l) for l in sys.stdin))"

or, in a functional formulation rather than using a loop:

$ python -c "import sys; print(sum(map(int, sys.stdin)))"

I would put a big WARNING on the commonly approved solution:

awk '{s+=$1} END {print s}' mydatafile # DO NOT USE THIS!!

that is because in this form awk uses a 32 bit signed integer representation: it will overflow for sums that exceed 2147483647 (i.e., 2^31).

A more general answer (for summing integers) would be:

awk '{s+=$1} END {printf "%.0f\n", s}' mydatafile # USE THIS INSTEAD

With jq:

seq 10 | jq -s 'add' # 'add' is equivalent to 'reduce .[] as $item (0; . + $item)'

Plain bash:

$ cat numbers.txt 
1
2
3
4
5
6
7
8
9
10
$ sum=0; while read num; do ((sum += num)); done < numbers.txt; echo $sum
55

dc -f infile -e '[+z1<r]srz1<rp'

Note that negative numbers prefixed with minus sign should be translated for dc, since it uses _ prefix rather than - prefix for that. For example, via tr '-' '_' | dc -f- -e '...'.

Edit: Since this answer got so many votes "for obscurity", here is a detailed explanation:

The expression [+z1<r]srz1<rp does the following:

[   interpret everything to the next ] as a string
  +   push two values off the stack, add them and push the result
  z   push the current stack depth
  1   push one
  <r  pop two values and execute register r if the original top-of-stack (1)
      is smaller
]   end of the string, will push the whole thing to the stack
sr  pop a value (the string above) and store it in register r
z   push the current stack depth again
1   push 1
<r  pop two values and execute register r if the original top-of-stack (1)
    is smaller
p   print the current top-of-stack

As pseudo-code:

1. Define "add_top_of_stack" as:
   1. Remove the two top values off the stack and add the result back
   2. If the stack has two or more values, run "add_top_of_stack" recursively
2. If the stack has two or more values, run "add_top_of_stack"
3. Print the result, now the only item left in the stack

To really understand the simplicity and power of dc, here is a working Python script that implements some of the commands from dc and executes a Python version of the above command:

### Implement some commands from dc
registers = {'r': None}
stack = []
def add():
    stack.append(stack.pop() + stack.pop())
def z():
    stack.append(len(stack))
def less(reg):
    if stack.pop() < stack.pop():
        registers[reg]()
def store(reg):
    registers[reg] = stack.pop()
def p():
    print stack[-1]

### Python version of the dc command above

# The equivalent to -f: read a file and push every line to the stack
import fileinput
for line in fileinput.input():
    stack.append(int(line.strip()))

def cmd():
    add()
    z()
    stack.append(1)
    less('r')

stack.append(cmd)
store('r')
z()
stack.append(1)
less('r')
p()

Pure and short bash.

f=$(cat numbers.txt)
echo $(( ${f//$'\n'/+} ))

perl -lne '$x += $_; END { print $x; }' < infile.txt

I've done a quick benchmark on the existing answers which

• use only standard tools (sorry for stuff like lua or rocket),
• are real one-liners,
• are capable of adding huge amounts of numbers (100 million), and
• are fast (I ignored the ones which took longer than a minute).

I always added the numbers of 1 to 100 million which was doable on my machine in less than a minute for several solutions.

Here are the results:

Python

:; seq 100000000 | python -c 'import sys; print sum(map(int, sys.stdin))'
5000000050000000
# 30s
:; seq 100000000 | python -c 'import sys; print sum(int(s) for s in sys.stdin)'
5000000050000000
# 38s
:; seq 100000000 | python3 -c 'import sys; print(sum(int(s) for s in sys.stdin))'
5000000050000000
# 27s
:; seq 100000000 | python3 -c 'import sys; print(sum(map(int, sys.stdin)))'
5000000050000000
# 22s
:; seq 100000000 | pypy -c 'import sys; print(sum(map(int, sys.stdin)))'
5000000050000000
# 11s
:; seq 100000000 | pypy -c 'import sys; print(sum(int(s) for s in sys.stdin))'
5000000050000000
# 11s

Awk

:; seq 100000000 | awk '{s+=$1} END {print s}'
5000000050000000
# 22s

Paste & Bc

This ran out of memory on my machine. It worked for half the size of the input (50 million numbers):

:; seq 50000000 | paste -s -d+ - | bc
1250000025000000
# 17s
:; seq 50000001 100000000 | paste -s -d+ - | bc
3750000025000000
# 18s

So I guess it would have taken ~35s for the 100 million numbers.

Perl

:; seq 100000000 | perl -lne '$x += $_; END { print $x; }'
5000000050000000
# 15s
:; seq 100000000 | perl -e 'map {$x += $_} <> and print $x'
5000000050000000
# 48s

Ruby

:; seq 100000000 | ruby -e "puts ARGF.map(&:to_i).inject(&:+)"
5000000050000000
# 30s

C

Just for comparison's sake I compiled the C version and tested this also, just to have an idea how much slower the tool-based solutions are.

#include <stdio.h>
int main(int argc, char** argv) {
    long sum = 0;
    long i = 0;
    while(scanf("%ld", &i) == 1) {
        sum = sum + i;
    }
    printf("%ld\n", sum);
    return 0;
}

:; seq 100000000 | ./a.out 
5000000050000000
# 8s

Conclusion

C is of course fastest with 8s, but the Pypy solution only adds a very little overhead of about 30% to 11s. But, to be fair, Pypy isn't exactly standard. Most people only have CPython installed which is significantly slower (22s), exactly as fast as the popular Awk solution.

The fastest solution based on standard tools is Perl (15s).

My fifteen cents:

$ cat file.txt | xargs  | sed -e 's/\ /+/g' | bc

Example:

$ cat text
1
2
3
3
4
5
6
78
9
0
1
2
3
4
576
7
4444
$ cat text | xargs  | sed -e 's/\ /+/g' | bc 
5148

Using the GNU datamash util:

seq 10 | datamash sum 1

Output:

55

If the input data is irregular, with spaces and tabs at odd places, this may confuse datamash, then either use the -W switch:

<commands...> | datamash -W sum 1

...or use tr to clean up the whitespace:

<commands...> | tr -d '[[:blank:]]' | datamash sum 1

If the input is large enough, the output will be in scientific notation.

seq 100000000 | datamash sum 1

Output:

5.00000005e+15

To convert that to decimal, use the the --format option:

seq 100000000 | datamash  --format '%.0f' sum 1

Output:

5000000050000000

BASH solution, if you want to make this a command (e.g. if you need to do this frequently):

addnums () {
  local total=0
  while read val; do
    (( total += val ))
  done
  echo $total
}

Then usage:

addnums < /tmp/nums

Plain bash one liner

$ cat > /tmp/test
1 
2 
3 
4 
5
^D

$ echo $(( $(cat /tmp/test | tr "\n" "+" ) 0 ))

You can using num-utils, although it may be overkill for what you need. This is a set of programs for manipulating numbers in the shell, and can do several nifty things, including of course, adding them up. It's a bit out of date, but they still work and can be useful if you need to do something more.

https://suso.suso.org/programs/num-utils/index.phtml

It's really simple to use:

$ seq 10 | numsum
55

But runs out of memory for large inputs.

$ seq 100000000 | numsum
Terminado (killed)

Cannot avoid submitting this, it is the most generic approach to this Question, please check:

jot 1000000 | sed '2,$s/$/+/;$s/$/p/' | dc

It is to be found over here, I was the OP and the answer came from the audience:

• Most elegant unix shell one-liner to sum list of numbers of arbitrary precision?

And here are its special advantages over awk, bc, perl, GNU's datamash and friends:

• it uses standards utilities common in any unix environment
• it does not depend on buffering and thus it does not choke with really long inputs.
• it implies no particular precision limits -or integer size for that matter-, hello AWK friends!
• no need for different code, if floating point numbers need to be added, instead.
• it theoretically runs unhindered in the minimal of environments

I realize this is an old question, but I like this solution enough to share it.

% cat > numbers.txt
1 
2 
3 
4 
5
^D
% cat numbers.txt | perl -lpe '$c+=$_}{$_=$c'
15

If there is interest, I'll explain how it works.

sed 's/^/.+/' infile | bc | tail -1

The following works in bash:

I=0

for N in `cat numbers.txt`
do
    I=`expr $I + $N`
done

echo $I

Pure bash and in a one-liner :-)

$ cat numbers.txt
1
2
3
4
5
6
7
8
9
10


$ I=0; for N in $(cat numbers.txt); do I=$(($I + $N)); done; echo $I
55

Here's a nice and clean Raku (formerly known as Perl 6) one-liner:

say [+] slurp.lines

We can use it like so:

% seq 10 | raku -e "say [+] slurp.lines"
55

It works like this:

slurp without any arguments reads from standard input by default; it returns a string. Calling the lines method on a string returns a list of lines of the string.

The brackets around + turn + into a reduction meta operator which reduces the list to a single value: the sum of the values in the list. say then prints it to standard output with a newline.

One thing to note is that we never explicitly convert the lines to numbers—Raku is smart enough to do that for us. However, this means our code breaks on input that definitely isn't a number:

% echo "1\n2\nnot a number" | raku -e "say [+] slurp.lines"
Cannot convert string to number: base-10 number must begin with valid digits or '.' in '⏏not a number' (indicated by ⏏)
  in block <unit> at -e line 1

Alternative pure Perl, fairly readable, no packages or options required:

perl -e "map {$x += $_} <> and print $x" < infile.txt

For Ruby Lovers

ruby -e "puts ARGF.map(&:to_i).inject(&:+)" numbers.txt

You can do it in Python:

lines = input();
ints = map(int, lines);
s = sum(ints);
print s;

Sebastian pointed out a one liner script:

cat filename | python -c"from fileinput import input; print sum(map(int, input()))"

A benefit of using fileinput.input is its support both for using the standard input stream as the input, and alternatively for specifying one or more input files as arguments.

The following should work (assuming your number is the second field on each line).

awk 'BEGIN {sum=0} \
 {sum=sum + $2} \
END {print "tot:", sum}' Yourinputfile.txt

C (not simplified)

seq 1 10 | tcc -run <(cat << EOF
#include <stdio.h>
int main(int argc, char** argv) {
    int sum = 0;
    int i = 0;
    while(scanf("%d", &i) == 1) {
        sum = sum + i;
    }
    printf("%d\n", sum);
    return 0;
}
EOF)

$ cat n
2
4
2
7
8
9

$ perl -MList::Util -le 'print List::Util::sum(<>)' < n
32

Or, you can type in the numbers on the command line:

$ perl -MList::Util -le 'print List::Util::sum(<>)'
1
3
5
^D
9

However, this one slurps the file so it is not a good idea to use on large files. See j_random_hacker's answer which avoids slurping.

One-liner in Racket:

racket -e '(define (g) (define i (read)) (if (eof-object? i) empty (cons i (g)))) (foldr + 0 (g))' < numlist.txt

My version:

seq -5 10 | xargs printf "- - %s" | xargs  | bc

Real-time summing to let you monitor progress of some number-crunching task.

$ cat numbers.txt 
1
2
3
4
5
6
7
8
9
10

$ cat numbers.txt | while read new; do total=$(($total + $new)); echo $total; done
1
3
6
10
15
21
28
36
45
55

(There is no need to set $total to zero in this case. Neither you can access $total after the finish.)