How to add two time outputs in bash - bash

Question

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'".

Answer 1

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

Answer 2

For the example i'll use a variable preinitialized:
time="54:32.96";
minutes=$(echo "$time" | cut -d":" -f1)
seconds=$(echo "$time" | cut -d":" -f2 | cut -d"." -f1)
millis=$(echo "$time" | cut -d":" -f2 | cut -d"." -f2)
#Total time in millis
totalMillisOne=$(($millis+$seconds*1000+$minutes*60000))
You do this with every command and you save it in diferent vars, and then you do the average:
let avMillis=$totalMillisOne+$totalMillisTwo
let avMillis=$avMillis/2
And you output it in the same input format:
let avSeconds=$avMillis/1000
let avMillis=$avMillis-$avSeconds*1000;
let avMinutes=$avSeconds/60;
let avSeconds=$avSeconds-$avMinutes*60;
echo "${avMinutes}:${avSeconds}.${avMillis}"