How to Have an Algorithm Round in a "Smart" Way Automatically
I Would Like to Round Number in a Code but in a Way That It Adapts to Each Values. for Example I Would Like a Rounding Algorithm to Return: 0.999999 It Should...
I would like to round number in a code but in a way that it adapts to each values.
For example i would like a rounding algorithm to return :
- 0.999999 it should return 1
- 0.0749999 it should return 0.075
- 0.006599 it should return 0.0066 and so on ...
I don't know in advance the number of digits (which is kinda my problem)
I was thinking to use strings to find where are the 9s (or count the 0s) but it is quite a lot of effort for that i was thinking ?
If you know any way to do that (if possible without advanced libraries) i would appreciate.
Thanks.
3 Answers
It's some complicated. but, It works. Please make sure the result is what you want. I think you can understand how to round the number from code.
def clear9(numstr):
liststr = list(numstr)
for index in range(len(liststr)-1,-1,-1):
if liststr[index] == '.': continue
if liststr[index] == '9':
liststr[index] = '0'
if index == 0:
liststr.insert(0, '1')
else:
if index != len(liststr)-1:
liststr[index] = str(int(liststr[index])+1)
break
numstr = ''
for item in liststr:
numstr += item
return numstr
def myround(num):
numstr = str(num)
numstr = clear9(numstr)
return float(numstr)
print (myround(9.05))
print (myround(9.999999))
print (myround(0.999999))
print (myround(0.0749999))
print (myround(0.006599))
print (myround(0.00659923))
print (myround(0.09659923))
print (myround(-0.00659923))
9.05
10.0
1.0
0.075
0.0066
0.00659923
0.09659923
-0.00659923
import math
def round_(number):
dist = int(math.log10(abs(number))) #number of zeros after decimal point
return (round(number, abs(dist) + 2) if dist != 0 else round(number))
print(round_(0.999999))
print(round_(0.0749999))
print(round_(0.006599))
print(round_(-0.00043565))
output:
1
0.075
0.0066
-0.00044
Dealing with floating point numbers is tricky. You want to do a kind of round-off in base 10, but floating point numbers are base 2.
So I propose to use the decimal module, which can represent real numbers exactly, as opposed to base-2 floating point.:
from decimal import Decimal
def myround(num):
dec = Decimal(num)
adj = abs(dec.adjusted())+1
return round(num, adj)
Look at the documentation for Decimal.adjusted() to understand how this works.
A test:
In [1]: from decimal import Decimal
In [2]: def myround(num):
...: dec = Decimal(num)
...: adj = abs(dec.adjusted())+1
...: return round(num, adj)
...:
In [3]: myround(0.999999)
Out[3]: 1.0
In [4]: myround(0.006599)
Out[4]: 0.0066
In [5]: myround(0.0749999)
Out[5]: 0.075
Sarah Jenkins is a veteran tech journalist with over 12 years of experience covering artificial intelligence, mobile innovations, and digital ethics. Her insights have appeared in leading technology publications worldwide.
