How to Run a Cron Job Randomly Every Hour in Ubuntu

The command in crontab is executed with /bin/sh so you can use arithmetic expansion to calculate whether the current minute modulo 25 equals zero:

*/5 * * * * [ $(( $(date +\%s) / 60 \% 25 )) -eq 0 ] && your_command

cron will run this entire entry every 5 minutes, but only if the current minute (in minutes since the epoch) modulo 25 equals zero will it run your_command.

As others have pointed out, 1 day is not evenly divisible by 25 minutes, so this will not cause your_command to run at the same time every day, but it will run every 25 minutes.

The simplest solution would probably be to run a cronjob more frequently and use a wrapper script to quit without doing anything if not enough time has passed.

To figure out how often you need to run, take the greatest common factor of cron's limits and your desired interval.

So, for "every 30 hours, 30 minutes", that'd be "every 30 minutes" and, for "every 30 hours", that'd be "every 6 hours" (The greatest common factor of 30 and 24)

You can implement the wrapper one of two ways:

First, you could store a timestamp in a file and then check if the time difference between now and the stored timestamp is greater than or equal to 30 hours and 30 minutes.

This seems simple enough, but has two potential gotchas that complicate the code:

1. Failsafe parsing of the saved timestamp file
2. Allowing for some wiggle forward and back when comparing timestamps since other things happening on the system will cause the actual interval to wiggle around.

The second option is to not store a timestamp file at all and, instead, do some math. This is also theoretically faster since the kernel can return the system time without querying the hard drive.

I haven't tested this for typos, but here's Python code for it that's been expanded out for clarity.

import os, time

full_interval = 1830 # (30 hours * 60 minutes) + 30 minutes
cron_interval = 30   #  30 minutes

minutes_since_epoch = time.time() // 60

allowed_back_skew = (cron_interval * 0.1)
sorta_delta = (minutes_since_epoch + allowed_back_skew) % full_interval

if sorta_delta < cron_interval:
    os.execlp('python', 'python', '/root/get_top.py')

Here's the idea behind it:

1. Just as "a stopped clock is right twice a day", the value of minutes_since_epoch % full_interval will only be less than cron_interval once per full_interval.
2. We need to fuzzy-match to account for variations caused by sharing resources with other processes.
3. The easiest way to do this is to use [0, cron_interval) as a window within which a task must fall in order to be executed.
4. To account for jitter in both directions, we slide the starting edge of the window back by 10% of its duration since running too early will be rare while running too late can happen any time the system is so bogged down that the wrapper script is delayed in calling time.time().

If, as I suspect, get_top.py is your own creation, just stick this at the top of it and change the check to

if sorta_delta > cron_interval:
    sys.exit(0)