Python - Rounding by quarter-intervals

谁说胖子不能爱 提交于 2019-11-30 00:54:12

问题


I'm running into the following issue:

Given various numbers like:

10.38

11.12

5.24

9.76

does an already 'built-in' function exists to round them up to the closest 0.25 step like e.g.:

10.38 --> 10.50

11.12 --> 11.00

5.24 --> 5.25

9.76 --> 9-75 ?

Or can I go ahead and hack together a function that performs the desired task?

Thanks in advance and

with best regards

Dan


回答1:


>>> def my_round(x):
...  return round(x*4)/4
... 
>>> 
>>> assert my_round(10.38) == 10.50
>>> assert my_round(11.12) == 11.00
>>> assert my_round(5.24) == 5.25
>>> assert my_round(9.76) == 9.75
>>> 



回答2:


This is a general purpose solution which allows rounding to arbitrary resolutions. For your specific case, you just need to provide 0.25 as the resolution but other values are possible, as shown in the test cases.

def roundPartial (value, resolution):
    return round (value / resolution) * resolution

print "Rounding to quarters"
print roundPartial (10.38, 0.25)
print roundPartial (11.12, 0.25)
print roundPartial (5.24, 0.25)
print roundPartial (9.76, 0.25)

print "Rounding to tenths"
print roundPartial (9.74, 0.1)
print roundPartial (9.75, 0.1)
print roundPartial (9.76, 0.1)

print "Rounding to hundreds"
print roundPartial (987654321, 100)

This outputs:

Rounding to quarters
10.5
11.0
5.25
9.75
Rounding to tenths
9.7
9.8
9.8
Rounding to hundreds
987654300.0



回答3:


There is no builtin, but such a function is trivial to write

def roundQuarter(x):
    return round(x * 4) / 4.0



回答4:


The solution of paxdiablo can be a little bit improved.

def roundPartial (value, resolution):
return round (value /float(resolution)) * resolution

so the function is now: "data-type sensitive".



来源:https://stackoverflow.com/questions/8118679/python-rounding-by-quarter-intervals

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