Python: Rounding output of the time module

Question

This is the code I have for measuring how long it took the user to complete the program:

start_time = time.time() # Sits at the top of my code print("%f seconds" % (time.time() - start_time)) # Goes at the bottom of my code

My question is, how do I round the output of this to one decimal place? For example, if my output was 3.859639 seconds how would I present this like: 3.8 Secounds?

So, you want to round down all the time? Instead of rounding 3.859... to 3.9? — Andy
– Andy ♦, Commented Oct 12, 2015 at 15:31
If you need the value rounded, the look into the round() function: docs.python.org/2/library/functions.html#round. To round down you could subtract 0.5 and then round() — AChampion
– AChampion, Commented Oct 12, 2015 at 15:31

ilyakhov · Accepted Answer · 2015-10-12 15:42:10Z

5

It looks like you've forgotten ".1" before "f". Try this:

print("%.1f seconds" % (time.time() - start_time))

edited Oct 12, 2015 at 15:42

answered Oct 12, 2015 at 15:31

ilyakhov

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1 Comment

Csarg Over a year ago

This works for me, having 3.9 instead of 3.8 won't kill my program so it's fine.

Anurag Verma · Accepted Answer · 2015-10-12 15:45:13Z

Use round function with second argument as 1

start_time = time.time() # Sits at the top of my code x=round((time.time() - start_time),1) print x

Jon Clements · Accepted Answer · 2015-10-12 15:48:34Z

Primitive way... multiply out, truncate the rest, then divide again:

diff = time.time() - start_time rounded_down = int(diff * 10) / 10 # = 3.8

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