Getting only 1 decimal place

Question

How do I convert 45.34531 to 45.3?

relet · Accepted Answer · 2018-11-21T10:57:15.590

239

Are you trying to represent it with only one digit:

print("{:.1f}".format(number)) # Python3
print "%.1f" % number          # Python2

or actually round off the other decimal places?

round(number,1)

or even round strictly down?

math.floor(number*10)/10

edited Nov 21 '18 at 10:57

answered Aug 03 '10 at 21:42

relet

Is there any chance that your 1st and 3rd solution gives different results? I think they both are exactly same @relet – Devesh Saini Feb 11 '17 at 12:43
1

Try number=-2.55. They also return different types. – relet Feb 11 '17 at 19:10
@DeveshSaini try number 2.36. 1st will give 2.4, 3nd will give 2.3 – Sanyam Jain Jan 22 '18 at 02:51

miku · Answer 2 · 2020-11-15T19:05:03.980

40

>>> "{:.1f}".format(45.34531)
'45.3'

Or use the builtin round:

>>> round(45.34531, 1)
45.299999999999997

edited Nov 15 '20 at 19:05

answered Aug 03 '10 at 21:44

miku

13

Update: Round gives me 45.3 nowdays. – Nathan Mar 30 '12 at 14:44
2

This is answer is correct but the formatting is way to complicated IMO. You should just write `"{:.1f}".format(45.34531)`. – Dave Halter Apr 04 '18 at 22:13
@DaveHalter, writing out `0.1` instead of `.1` is way to complicated, how can anyone even follow that code... – dogmatic69 Jan 15 '19 at 09:38
6

There's also a zero in front that is not necessary. I also think that nobody really understands the format language so keeping it simple is preferred IMO. Now with Python 3.6 I would recommend writing it like this: `f"{number:.1f}"`. – Dave Halter Jan 17 '19 at 13:19

score 21 · Answer 3 · answered Aug 03 '10 at 21:43

21

round(number, 1)

answered Aug 03 '10 at 21:43

DixonD

3 Answers3