How to generate 9 digit random number in shell?
I am trying something like this but it only gave numbers below 32768.
#!/bin/bash
mo=$((RANDOM%999999999))
echo "********Random"$mo
Please help
output should be ********Random453351111
In Linux with /dev/urandom:
$ rnd=$(tr -cd "[:digit:]" < /dev/urandom | head -c 9) && echo $rnd
463559879
I think this should make it
shuf -i 99999999-999999999 -n 1
As a work around, we could just simply ask for 1 random integer, for n times:
rand=''
for i in {1..9}; do
rand="${rand}$(( $RANDOM % 10 ))"
done
echo $rand
Try it online!
Note [1]: Since RANDOM's upper limit has a final digit of 7, there's a slightly lesser change for the 'generated' number to contain 8 or 9's.
Because of RANDOM's limited range, it can only be used to retrieve four base-10 digits at a time. Thus, to retrieve 9 digits, you need to call it three times.
If we don't care much about performance (are willing to pay process substitution costs), this may look like:
#!/usr/bin/env bash
get4() {
local newVal=32768
while (( newVal > 29999 )); do # avoid bias because of remainder
newVal=$RANDOM
done
printf '%04d' "$((newVal % 10000))"
}
result="$(get4)$(get4)$(get4)"
result=$(( result % 1000000000 ))
printf '%09d\n' "$result"
If we do care about performance, it may instead look like:
#!/usr/bin/env bash
get4() {
local newVal=32768 outVar=$1
while (( newVal > 29999 )); do # avoid bias because of remainder
newVal=$RANDOM
done
printf -v "$outVar" '%04d' "$((newVal % 10000))"
}
get4 out1; get4 out2; get4 out3
result="${out1}${out2}${out3}"
result=$(( result % 1000000000 ))
printf '%09d\n' "$result"
Use perl, as follows :
perl -e print\ rand | cut -c 3-11
Or
perl -MPOSIX -e 'print floor rand 10**9'
I have a text with repeated data patterns, and grep keeps getting all matches without stop.
for ((count = 1; count !=17; count++)); do # 17 times
xuz1[count]=`grep -e "1 O1" $out_file | cut -c10-29`
xuz2[count]=`grep -e "2 O2" $out_file | cut -c10-29`
xuz3[count]=`grep -e "3 O3" $out_file | cut -c10-29`
echo ${xuz1[count]}
echo ${xuz2[count]}
echo ${xuz3[count]}
done
data looks like:
some text.....
Text....
.....
1 O1 111111 111111 111111
2 O2 222211 222211 222211
3 O3 643653 652346 757686
some text.....
1 O1 111122 111122 111122
2 O2 222222 222222 222222
3 O3 343653 652346 757683
some text.....
1 O1 111333 111333 111333
2 O2 222333 222333 222333
3 O3 343653 652346 757684
.
.
.
And result I'm getting:
xuz1[1] = 111111 111111 111111
xuz2[1] = 222211 222211 222211
xuz3[1] = 643653 652346 757686
xuz1[2] = 111111 111111 111111
xuz2[2] = 222211 222211 222211
xuz3[2] = 643653 652346 757686
...
looking for result like this:
xuz1[1]=111111 111111 111111
xuz2[1]=222211 222211 222211
xuz3[1]=343653 652346 757683
xuz1[2]=111122 111122 111122
xuz2[2]=222222 222222 222222
xuz3[2]=343653 652346 757684
also tried "grep -m 1 -e"
Which way should I go?
for now I ended up with one line
grep -A4 -e "1 O1" $out_file | cut -c10-29
Some text.... Is a huge text part.
A little bash script with a single grep is enough
grep -E '^[0-9]+ +O[0-9]+ +.*'|
while read idx oidx cols; do
if ((idx == 1)); then
let ++i
name=xuz$i
let j=1
fi
echo "$name[$j]=$cols"
let ++j
done
You haven't really described what you want, but I guess something like this.
awk '! /^[1-9][0-9]* O[0-9] / { n++; m=0; if (NR>1) print ""; next }
{ print "xuz" ++m "[" n "]=" substr($0, 10) }' "$out_file"
If the regex doesn't match, we assume we are looking at one of the "some text" pieces, and that this starts a new record. Increment n and reset m. Otherwise, print the output for this item within this record.
If some text could be more than one line, you will need a minor change, but I hope this should be enough at least to send you in the right direction.
You can do this in pure Bash, too, though this is going to be highly inefficient - you would expect a Bash while read loop to be at least a hundred times slower than Awk, and the code is markedly less idiomatic and elegant.
while read -r m x result; do
case $m::$x in
[1-9]::O[1-9])
printf 'xuz%d[%d]=%s\n' $m $n "$result;;
*)
# If n is unset, don't print an empty line
printf '%s' "${n+$'\n'}"
let ((n++));;
esac
done <"$out_file"
I would aggressively challenge any requirement to do this in pure Bash. If it's for homework, the requirement is unrealistic, and a core skill for shell script authors is to understand the limits of the shell and the strengths of the common support tools like Awk. The Awk language is virtually guaranteed to be available wherever you have a shell, in particular a heavy shell like Bash. (In a limited e.g. embedded environment, a limited shell like Dash would make more sense. Then e.g. the let keyword won't be available, though it should not be hard to make this script properly portable.)
The case statement accepts glob patterns, not regular expressions, so the pattern here is slightly less general (we accept one positive digit in the first field).
Thank you all for participating in discussion.
*** this is my home project to help my wife do extract data from research calculations /// speed up is around 400 times **
file used for extracting data from, contains around 2000 lines,
needed data blocks look like this
and they're repeated 10-20 times in the file.
uiyououy COORDINATES
NR ATOM CCCCC X Y Z
1 O1 8.00 0.000000000 0.882236820 -0.789494235
2 O2 8.00 0.000000000 -1.218250722 -1.644061652
3 O3 8.00 0.000000000 1.218328524 0.400260050
4 O4 8.00 0.000000000 -0.882314622 2.033295837
Text text text text
tons of text
to extract 4 lines I used expression below
grep -A4 --no-group-separator -e "1 O1" $from_file | cut -c23-64
>xyz_temp.txt
# grep 4 lines at once to txt
sed -i '/^[ \t]*$/d' xyz_temp.txt
#del empty lines from xyz txt
next is to convert string in to numbers (should use '| bc -l' for arithmetic)
while IFS= read line
do
IFS=' ' read -r -a arr_line <<< "$line"
# break line of xyz into 3 numbers
s1=$(echo "${arr_line[0]}" \* 0.529177249 | bc -l)
# some math convertion
s2=$(echo "${arr_line[1]}" \* 0.529177249 | bc -l)
s3=$(echo "${arr_line[2]}" \* 0.529177249 | bc -l)
#-------to array non sorted ------------
arr[$n]=${n}";"${from_file}";"${gd_}";"${frt[count_4s]}";"${n4}";"${s1}";"${s2}";"${s3}
echo ${arr[n]}
#--------------------------------------------
done <"$from_file_txt"
sort array
IFS=$'\n' sorted=($(sort -t \; -k4 -k5 -g <<<"${arr[*]}"))
# -t separator ';' -k column -g generic * to get new line output
#-k4 -k5 sort by column 4 then5
#printf "%s\n" "${sorted[*]}"
unset IFS
There is Last part which will combine data to result view
echo "$n"
n2=1
n42=1
count_4s2=1
i=0
echo "============================== sorted =============================="
################### loop for empty 4s lines
printf "%s" ";" ";" ";" ";" ";" "${count_4s2}" ";"
printf "%s\n"
printf "%s\n" "${sorted[i]}"
while [ $i -lt $((n-2)) ]
do
i=$((i+1))
if [ "$n42" = "4" ] # 1234
then n42=0
count_4s2=$((count_4s2+1))
printf "%s" ";" ";" ";" ";" ";" "${count_4s2}" ";"
printf "%s\n"
fi
#--------------------------------------------
n2=$((n2+1))
n42=$((n42+1))
printf "%s\n" "${sorted[i]}"
done ############# while
#00000000000000000000000000000000000000
printf "%s\n"
echo ==END===END===END==
Output looks like this
============================== sorted ==============================
;;;;;1;
17;A-13_A1+.out;1.3;0.4;1;0;.221176355474853043;-.523049776514580244
18;A-13_A1+.out;1.3;0.4;2;0;-.550350051428402955;-.734584881824005358
19;A-13_A1+.out;1.3;0.4;3;0;.665269869069959489;.133910683627893251
20;A-13_A1+.out;1.3;0.4;4;0;-.336096173116409577;1.123723974181515102
;;;;;2;
13;A-13_A1+.out;1.3;0.45;1;0;.279265277182782148;-.504490787956469897
14;A-13_A1+.out;1.3;0.45;2;0;-.583907412327951988;-.759310392973448167
15;A-13_A1+.out;1.3;0.45;3;0;.662538493711206290;.146829200993661293
16;A-13_A1+.out;1.3;0.45;4;0;-.357896358566036450;1.116971979936256771
;;;;;3;
9;A-13_A1+.out;1.3;0.5;1;0;.339333719743262501;-.482029749553797105
10;A-13_A1+.out;1.3;0.5;2;0;-.612395507070451545;-.788968880150283253
11;A-13_A1+.out;1.3;0.5;3;0;.658674809217196345;.163289820251690233
12;A-13_A1+.out;1.3;0.5;4;0;-.385613021360830052;1.107708808923212876
==END===END===END==
*note : some code might not shown here
next step is to paste it to excel with ; separator.
I'm trying to write a bash script that calculates the average of numbers by rows and columns. An example of a text file that I'm reading in is:
1 2 3 4 5
4 6 7 8 0
There is an unknown number of rows and unknown number of columns. Currently, I'm just trying to sum each row with a while loop. The desired output is:
1 2 3 4 5 Sum = 15
4 6 7 8 0 Sum = 25
And so on and so forth with each row. Currently this is the code I have:
while read i
do
echo "num: $i"
(( sum=$sum+$i ))
echo "sum: $sum"
done < $2
To call the program it's stats -r test_file. "-r" indicates rows--I haven't started columns quite yet. My current code actually just takes the first number of each column and adds them together and then the rest of the numbers error out as a syntax error. It says the error comes from like 16, which is the (( sum=$sum+$i )) line but I honestly can't figure out what the problem is. I should tell you I'm extremely new to bash scripting and I have googled and searched high and low for the answer for this and can't find it. Any help is greatly appreciated.
You are reading the file line by line, and summing line is not an arithmetic operation. Try this:
while read i
do
sum=0
for num in $i
do
sum=$(($sum + $num))
done
echo "$i Sum: $sum"
done < $2
just split each number from every line using for loop. I hope this helps.
Another non bash way (con: OP asked for bash, pro: does not depend on bashisms, works with floats).
awk '{c=0;for(i=1;i<=NF;++i){c+=$i};print $0, "Sum:", c}'
Another way (not a pure bash):
while read line
do
sum=$(sed 's/[ ]\+/+/g' <<< "$line" | bc -q)
echo "$line Sum = $sum"
done < filename
Using the numsum -r util covers the row addition, but the output format needs a little glue, by inefficiently paste-ing a few utils:
paste "$2" \
<(yes "Sum =" | head -$(wc -l < "$2") ) \
<(numsum -r "$2")
Output:
1 2 3 4 5 Sum = 15
4 6 7 8 0 Sum = 25
Note -- to run the above line on a given file foo, first initialize $2 like so:
set -- "" foo
paste "$2" <(yes "Sum =" | head -$(wc -l < "$2") ) <(numsum -r "$2")
I have a homework assignment that is asking to shift a decimal number by a specified amount of digits. More clearly this bash script will take two input arguments, the first is the number(maximum 9 digits) that the shift will be performed on and the second is the number(-9 to 9) of digits to shift. Another requirement is that when a digit is shifted off the end, it should be attached to the other end of the number. One headache of a requirement is that we cannot use control statements of any kind: no loops, no if, and switch cases.
Example: 12345 3 should come out to 345000012 and 12345 -3 should be 12345000
I know that if I mod 12345 by 10^3 I get 345 and then if I divide 12345 by 10^3 I get 12 and then I can just concatenate those two variables together to get 34512. I am not quite sure if that is exactly correct but that is the closest I can get as of now. As far as the -3 shift, I know that 10^-3 is .001 and would work however when I try using 10^-3 in bash I get an error.
I am just lost at this point, any tips would be greatly appreciated.
EDIT: After several hours of bashing (pun intended) my head against this problem, I finally came up with a script that for the most part works. I would post the code right now but I fear another student hopelessly lost might stumble upon it. I will check back and post what I came up with in a week or two. I was able to do it with mods and division. Thank you all for the responses, it really helped me to open up and think about the problem from different angles.
Here's a hint:
echo ${string:0:3}
echo ${#string}
Edit (2011-02-11):
Here's my solution. I added some additional parameters with defaults.
rotate-string ()
{
local s=${1:-1} p=${2:--1} w=${3:-8} c=${4:-0} r l
printf -vr '%0*d' $w 0 # save $w zeros in $r
r=${r//0/$c}$s # change the zeros to the character in $c, append the string
r=${r: -w} # save the last $w characters of $r
l=${r: -p%w} # get the last part of $r ($p mod %w characters)
echo "$l${r::w-${#l}}" # output the Last part on the Left and the Right part which starts at the beginning and goes for ($w minus the_length_of_the_Left_part) characters
}
usage: rotate-string string positions-to-rotate width fill-character
example: rotate-string abc -4 9 =
result: ==abc====
Arguments can be omitted starting from the end and these defaults will be used:
fill-character: "0"
width: 8
positions-to-rotate: -1
string: "1"
More examples:
$ rotate-string
00000010
$ rotate-string 123 4
01230000
Fun stuff:
$ for i in {126..6}; do printf '%s\r' "$(rotate-string Dennis $i 20 .)"; sleep .05; done; printf '\n'
$ while true; do for i in {10..1} {1..10}; do printf '%s\r' "$(rotate-string : $i 10 .)"; sleep .1; done; done
$ while true; do for i in {40..2} {2..40}; do printf '%s\r' "$(rotate-string '/\' $i 40 '_')"; sleep .02; done; done
$ d=0; while true; do for i in {1..10} {10..1}; do printf '%s\r' "$(rotate-string $d $i 10 '_')"; sleep .02; done; ((d=++d%10)); done
$ d=0; while true; do for i in {1..10}; do printf '%s\r' "$(rotate-string $d $i 10 '_')"; sleep .2; ((d=++d%10)); done; done
$ shape='▁▂▃▄▅▆▇█▇▆▅▄▃▂▁'; while true; do for ((i=1; i<=COLUMNS; i++)); do printf '%s\r' "$(rotate-string "$shape" $i $COLUMNS ' ')"; done; done
In the absence of control structures, you need to use recursion, with index values as "choice selections", which is how functional programming often works.
#!/bin/sh
#
# cshift NUMBER N
cshift() {
let num=10#$1
num=`printf '%09d' $num`
lshift="${num:1:8}${num:0:1}"
rshift="${num:8:1}${num:0:8}"
next=( "cshift $lshift $(($2 + 1))" "echo $num" "cshift $rshift $(( $2 - 1 ))" )
x=$(( $2 == 0 ? 1 : $2 < 0 ? 0 : 2 ))
eval "${next[x]}"
}
cshift $1 $2
and, the testing:
$ for ((i=-9;i<=9;i++)); do cshift 12345 $i ; done
000012345
500001234
450000123
345000012
234500001
123450000
012345000
001234500
000123450
000012345
500001234
450000123
345000012
234500001
123450000
012345000
001234500
000123450
000012345
You can also do some math on the indexes and avoid the recursion, but I don't mind making the computer work harder so I don't have to. It's easy to think of how to do the shift by one in either direction, and then I use an evaluated choice that is selected by the signum of the shift value, outputting a value and stopping when the shift value is zero.
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?
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
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))"
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
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
With jq:
seq 10 | jq -s 'add' # 'add' is equivalent to 'reduce .[] as $item (0; . + $item)'
dc -f infile -e '[+z1<r]srz1<rp'
Note that negative numbers prefixed with minus sign should be translated for dc, since it uses _ prefix rather than - prefix for that. For example, via tr '-' '_' | dc -f- -e '...'.
Edit: Since this answer got so many votes "for obscurity", here is a detailed explanation:
The expression [+z1<r]srz1<rp does the following:
[ interpret everything to the next ] as a string
+ push two values off the stack, add them and push the result
z push the current stack depth
1 push one
<r pop two values and execute register r if the original top-of-stack (1)
is smaller
] end of the string, will push the whole thing to the stack
sr pop a value (the string above) and store it in register r
z push the current stack depth again
1 push 1
<r pop two values and execute register r if the original top-of-stack (1)
is smaller
p print the current top-of-stack
As pseudo-code:
Define "add_top_of_stack" as:
Remove the two top values off the stack and add the result back
If the stack has two or more values, run "add_top_of_stack" recursively
If the stack has two or more values, run "add_top_of_stack"
Print the result, now the only item left in the stack
To really understand the simplicity and power of dc, here is a working Python script that implements some of the commands from dc and executes a Python version of the above command:
### Implement some commands from dc
registers = {'r': None}
stack = []
def add():
stack.append(stack.pop() + stack.pop())
def z():
stack.append(len(stack))
def less(reg):
if stack.pop() < stack.pop():
registers[reg]()
def store(reg):
registers[reg] = stack.pop()
def p():
print stack[-1]
### Python version of the dc command above
# The equivalent to -f: read a file and push every line to the stack
import fileinput
for line in fileinput.input():
stack.append(int(line.strip()))
def cmd():
add()
z()
stack.append(1)
less('r')
stack.append(cmd)
store('r')
z()
stack.append(1)
less('r')
p()
Pure and short bash.
f=$(cat numbers.txt)
echo $(( ${f//$'\n'/+} ))
perl -lne '$x += $_; END { print $x; }' < infile.txt
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
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).
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
Plain bash one liner
$ cat > /tmp/test
1
2
3
4
5
^D
$ echo $(( $(cat /tmp/test | tr "\n" "+" ) 0 ))
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
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)
The following works in bash:
I=0
for N in `cat numbers.txt`
do
I=`expr $I + $N`
done
echo $I
I realize this is an old question, but I like this solution enough to share it.
% cat > numbers.txt
1
2
3
4
5
^D
% cat numbers.txt | perl -lpe '$c+=$_}{$_=$c'
15
If there is interest, I'll explain how it works.
Cannot avoid submitting this, it is the most generic approach to this Question, please check:
jot 1000000 | sed '2,$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
sed 's/^/.+/' infile | bc | tail -1
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 $I
55
Alternative pure Perl, fairly readable, no packages or options required:
perl -e "map {$x += $_} <> and print $x" < infile.txt
For Ruby Lovers
ruby -e "puts ARGF.map(&:to_i).inject(&:+)" numbers.txt
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
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()))"
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
$ 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
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)
My version:
seq -5 10 | xargs printf "- - %s" | xargs | bc
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