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Home/ Questions/Q 6989819
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T19:15:48+00:00 2026-05-27T19:15:48+00:00

I want to calculate in bash the average time spent by several commands. The

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I want to calculate in bash the average time spent by several commands. The output of time command is min:sec.milisec.

I don’t know how to add two outputs of this kind in bash and in final to calculate the average.

I tried to convert the output with date but the output is “date: invalid date `0:01.00′”.

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  1. Editorial Team
    Editorial Team
    2026-05-27T19:15:49+00:00Added an answer on May 27, 2026 at 7:15 pm

    This is a three part answer

    Part one

    First, use the TIMEFORMAT variable to output only seconds elapsed. Then you can add this directly

    From man bash

    TIMEFORMAT
    The value of this parameter is used as a format string specifying how the timing information
    for pipelines prefixed with the time reserved word should be displayed. The % character
    introduces an escape sequence that is expanded to a time value or other information. The
    escape sequences and their meanings are as follows; the braces denote optional portions.

    Here is an example which outputs only seconds with a precision of 0, i.e. no decimal point. Read part three why that’s important.

    TIMEFORMAT='%0R'; time sleep 1
    1
    

    Part two

    Second, how do we capture the output of time? It’s actually a bit tricky, here’s how you do capture the time from the command above

    TIMEFORMAT='%0R'; time1=$( { time sleep 1; } 2>&1 )
    

    Part three

    How do we add the times together and get the average?

    In bash we use the $(( )) construct to do math. Note that bash does not natively support floating point so you will be doing integer division (hence the precision 0.) Here is a script that will capture the time from two commands and output each of the individual times and their average

    #!/bin/bash
    
    TIMEFORMAT='%0R'
    time1=$( { time sleep 1; } 2>&1 )
    time2=$( { time sleep 4; } 2>&1 )
    ave=$(( (time1 + time2) / 2))
    
    echo "time1 is $time1 | time2 is $time2 | average is $ave"
    

    Output

    time1 is 1 | time2 is 4 | average is 2
    

    If integer division is a non-starter for you and you want precision, as long as you don’t mind calling the external binary bc, you can do this quite easily.

    #!/bin/bash
    
    TIMEFORMAT='%3R'
    time1=$( { time sleep 1; } 2>&1 )
    time2=$( { time sleep 4; } 2>&1 )
    ave=$( bc <<<"scale=3; ($time1 + $time2)/2" )
    
    echo "time1 is $time1 | time2 is $time2 | average is $ave"
    

    Output

    time1 is 1.003 | time2 is 4.003 | average is 2.503
    
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