Is there a better way to run a command N times in bash?

Question

I occasionally run a bash command line like this:

n=0; while [[ $n -lt 10 ]]; do some_command; n=$((n+1)); done

To run some_command a number of times in a row -- 10 times in this case.

Often some_command is really a chain of commands or a pipeline.

Is there a more concise way to do this?

Answer 1

If your range has a variable, use seq, like this:

count=10
for i in $(seq $count); do
    command
done

Simply:

for run in {1..10}; do
  command
done

Or as a one-liner, for those that want to copy and paste easily:

for run in {1..10}; do command; done

Answer 2

Using a constant:

for ((n=0;n<10;n++)); do
    some_command; 
done

Using a variable (can include math expressions):

x=10; for ((n=0; n < (x / 2); n++)); do some_command; done

Answer 3

Another simple way to hack it:

seq 20 | xargs -Iz echo "Hi there"

run echo 20 times.

---

Notice that seq 20 | xargs -Iz echo "Hi there z" would output:

Hi there 1
Hi there 2
...

Answer 4

If you're using the zsh shell:

repeat 10 { echo 'Hello' }

Where 10 is the number of times the command will be repeated.

Answer 5

Using GNU Parallel you can do:

parallel some_command ::: {1..1000}

If you do not want the number as argument and only run a single job at a time:

parallel -j1 -N0 some_command ::: {1..1000}

Watch the intro video for a quick introduction: https://www.youtube.com/playlist?list=PL284C9FF2488BC6D1

Walk through the tutorial (http://www.gnu.org/software/parallel/parallel_tutorial.html). You command line with love you for it.

Answer 6

A simple function in the bash config file (~/.bashrc often) could work well.

function runx() {
  for ((n=0;n<$1;n++))
    do ${*:2}
  done
}

Call it like this.

$ runx 3 echo 'Hello world'
Hello world
Hello world
Hello world

Answer 7

If you are OK doing it periodically, you could run the following command to run it every 1 sec indefinitely. You can put other custom checks in place to run it n number of times.

watch -n 1 some_command

If you wish to have visual confirmation of changes, append --differences prior to the ls command.

According to the OSX man page, there's also

The --cumulative option makes highlighting "sticky", presenting a running display of all positions that have ever changed. The -t or --no-title option turns off the header showing the interval, command, and current time at the top of the display, as well as the following blank line.

Linux/Unix man page can be found here

Answer 8

for _ in {1..10}; do command; done   

Note the underscore instead of using a variable.

Answer 9

Another form of your example:

n=0; while (( n++ < 10 )); do some_command; done

Answer 10

xargs is fast:

#!/usr/bin/bash
echo "while loop:"
n=0; time while (( n++ < 10000 )); do /usr/bin/true ; done

echo -e "\nfor loop:"
time for ((n=0;n<10000;n++)); do /usr/bin/true ; done

echo -e "\nseq,xargs:"
time seq 10000 | xargs -I{} -P1 -n1 /usr/bin/true

echo -e "\nyes,xargs:"
time yes x | head -n10000 |  xargs -I{} -P1 -n1 /usr/bin/true

echo -e "\nparallel:"
time parallel --will-cite -j1 -N0 /usr/bin/true ::: {1..10000}

On a modern 64-bit Linux, gives:

while loop:

real    0m2.282s
user    0m0.177s
sys     0m0.413s

for loop:

real    0m2.559s
user    0m0.393s
sys     0m0.500s

seq,xargs:

real    0m1.728s
user    0m0.013s
sys     0m0.217s

yes,xargs:

real    0m1.723s
user    0m0.013s
sys     0m0.223s

parallel:

real    0m26.271s
user    0m4.943s
sys     0m3.533s

This makes sense, as the xargs command is a single native process that spawns the /usr/bin/true command multiple time, instead of the for and while loops that are all interpreted in Bash. Of course this only works for a single command; if you need to do multiple commands in each iteration the loop, it will be just as fast, or maybe faster, than passing sh -c 'command1; command2; ...' to xargs

The -P1 could also be changed to, say, -P8 to spawn 8 processes in parallel to get another big boost in speed.

I don't know why GNU parallel is so slow. I would have thought it would be comparable to xargs.

Answer 11

For one, you can wrap it up in a function:

function manytimes {
    n=0
    times=$1
    shift
    while [[ $n -lt $times ]]; do
        $@
        n=$((n+1))
    done
}

Call it like:

$ manytimes 3 echo "test" | tr 'e' 'E'
tEst
tEst
tEst

Answer 12

xargs and seq will help

function __run_times { seq 1 $1| { shift; xargs -i -- "$@"; } }

the view :

abon@abon:~$ __run_times 3  echo hello world
hello world
hello world
hello world

Answer 13

All of the existing answers appear to require bash, and don't work with a standard BSD UNIX /bin/sh (e.g., ksh on OpenBSD).

The below code should work on any BSD:

$ echo {1..4}
{1..4}
$ seq 4
sh: seq: not found
$ for i in $(jot 4); do echo e$i; done
e1
e2
e3
e4
$

Answer 14

I solved with this loop, where repeat is an integer that represents the loops's number

repeat=10
for n in $(seq $repeat); 
    do
        command1
        command2
    done

Answer 15

You can use this command to repeat your command 10 times or more

for i in {1..10}; do **your command**; done

for example

for i in {1..10}; do **speedtest**; done

Answer 16

Yet another answer: Use parameter expansion on empty parameters:

# calls curl 4 times 
curl -s -w "\n" -X GET "http:{,,,}//www.google.com"

Tested on Centos 7 and MacOS.

Answer 17

For loops are probably the right way to do it, but here is a fun alternative:

echo -e {1..10}"\n" |xargs -n1 some_command

If you need the iteration number as a parameter for your invocation, use:

echo -e {1..10}"\n" |xargs -I@ echo now I am running iteration @

Edit: It was rightly commented that the solution given above would work smoothly only with simple command runs (no pipes, etc.). you can always use a sh -c to do more complicated stuff, but not worth it.

Another method I use typically is the following function:

rep() { s=$1;shift;e=$1;shift; for x in `seq $s $e`; do c=${@//@/$x};sh -c "$c"; done;}

now you can call it as:

rep 3 10 echo iteration @

The first two numbers give the range. The @ will get translated to the iteration number. Now you can use this with pipes too:

rep 1 10 "ls R@/|wc -l"

with give you the number of files in directories R1 .. R10.

Answer 18

The script file

bash-3.2$ cat test.sh 
#!/bin/bash

echo "The argument is  arg: $1"

for ((n=0;n<$1;n++));
do
  echo "Hi"
done

and the output below

bash-3.2$  ./test.sh 3
The argument is  arg: 3
Hi
Hi
Hi
bash-3.2$

Answer 19

A little bit naive but this is what I usually remember off the top of my head:

for i in 1 2 3; do
  some commands
done

Very similar to @joe-koberg's answer. His is better especially if you need many repetitions, just harder for me to remember other syntax because in last years I'm not using bash a lot. I mean not for scripting at least.

Answer 20

How about the alternate form of for mentioned in (bashref)Looping Constructs?