Round a number to a given set of values [duplicate] - python-3.x

Question

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From list of integers, get number closest to a given value
(10 answers)
Closed 5 years ago.
Talking Python 3 here.
I'm looking to round a number to a given set of values which can vary
Assume value_set = [x, y, z] and for the sake of the example x, y, z = 1, 3.12, 4 I'm looking for a function that will round a given float to the closest number
custom_round(0) --> 1
custom_round(2.7) --> 3.12
Notice that it should be generic enough that value_set length will vary also

Answer 1

You can use the min function in order to find the minimum in your list when the key is the absolute value of x-n (x is each item in the list).
value_set = [1, 3.12, 4]
def return_closest(n):
return min(value_set, key=lambda x:abs(x-n))
number_to_check = 3
print (return_closest(number_to_check))
>>> 3.12

Answer 2

You can do this by first sorting the list, and then use binary search:
from bisect import bisect_left
class CustomRound:
def __init__(self,iterable):
self.data = sorted(iterable)
def __call__(self,x):
data = self.data
ndata = len(data)
idx = bisect_left(data,x)
if idx <= 0:
return data[0]
elif idx >= ndata:
return data[ndata-1]
x0 = data[idx-1]
x1 = data[idx]
if abs(x-x0) < abs(x-x1):
return x0
return x1
You can than construct your CustomRound like:
values = [1,3.12,4]
custom_round = CustomRound(values)
and simply call it:
>>> custom_round(0)
1
>>> custom_round(0.5)
1
>>> custom_round(1.5)
1
>>> custom_round(2.5)
3.12
>>> custom_round(3.12)
3.12
>>> custom_round(3.9)
4
>>> custom_round(4.1)
4
>>> custom_round(4.99)
4
This approach will work in O(log n) for rounding and O(n log n) for construction. So you will invest some additional time to construct the custom_round, but if you call it often, it will eventually pay off in rounding individual numbers.