Shell command to sum integers, one per line? - shell

Question

I am looking for a command that will accept (as input) multiple lines of text, each line containing a single integer, and output the sum of these integers.
As a bit of background, I have a log file which includes timing measurements. Through grepping for the relevant lines and a bit of sed reformatting I can list all of the timings in that file. I would like to work out the total. I can pipe this intermediate output to any command in order to do the final sum. I have always used expr in the past, but unless it runs in RPN mode I do not think it is going to cope with this (and even then it would be tricky).
How can I get the summation of integers?

Answer 1

Bit of awk should do it?
awk '{s+=$1} END {print s}' mydatafile
Note: some versions of awk have some odd behaviours if you are going to be adding anything exceeding 2^31 (2147483647). See comments for more background. One suggestion is to use printf rather than print:
awk '{s+=$1} END {printf "%.0f", s}' mydatafile

Answer 2

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

Answer 3

The one-liner version in Python:
$ python -c "import sys; print(sum(int(l) for l in sys.stdin))"

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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:
Define "add_top_of_stack" as:
Remove the two top values off the stack and add the result back
If the stack has two or more values, run "add_top_of_stack" recursively
If the stack has two or more values, run "add_top_of_stack"
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()

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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).

Answer 12

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

Answer 13

Plain bash one liner
$ cat > /tmp/test
1
2
3
4
5
^D
$ echo $(( $(cat /tmp/test | tr "\n" "+" ) 0 ))

Answer 14

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

Answer 15

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)

Answer 16

The following works in bash:
I=0
for N in `cat numbers.txt`
do
I=`expr $I + $N`
done
echo $I

Answer 17

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.

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

Alternative pure Perl, fairly readable, no packages or options required:
perl -e "map {$x += $_} <> and print $x" < infile.txt

Answer 22

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

Answer 23

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

Answer 24

You can do it in python, if you feel comfortable:
Not tested, just typed:
out = open("filename").read();
lines = out.split('\n')
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()))"

Answer 25

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

Answer 26

$ 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.

Answer 27

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

Answer 28

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)

Answer 29

My version:
seq -5 10 | xargs printf "- - %s" | xargs | bc

Answer 30

C++ (simplified):
echo {1..10} | scc 'WRL n+=$0; n'
SCC project - http://volnitsky.com/project/scc/
SCC is C++ snippets evaluator at shell prompt