我有一个最接近 8 的浮点数7.999999985666533
,我使用它math.isclose
math.isclose(a, b)
但是对于像 1.5999999991220535
这样的浮点值,最接近的整数是 2,但如果我将其乘以 10 (10 ** 1)
,我得到 16 作为最接近的整数,这也是结果 isclose
再比如:
1.2799999997163347
乘以 128 后应得到 100 (10 ** 2)
有没有办法做到这一点?
连续分数非常强大。也许这里有一个小的矫枉过正,但它有效。
import numpy as np
def get_cont_fraction(x, depth=0, maxdepth=10, precision=1e-6):
if depth > maxdepth:
out = []
else:
assert x >= 0
out=[]
if not depth:
out += [ int(x) ] + get_cont_fraction( x - int( x ), depth=depth + 1, maxdepth=maxdepth, precision=precision)
elif x < precision :
out=[]
else:
out += [ int(1./ x) ] + get_cont_fraction(1. / x - int( 1. / x ), depth=depth + 1, maxdepth=maxdepth, precision=precision)
return out
def get_fraction(inList):
num = inList[-1]
den = 1
testList = inList[:-1:]
testList = testList[::-1]
for a in testList:
num , den = a * num + den, num
return ( num, den )
if __name__ == "__main__":
a = get_fraction( get_cont_fraction( 1.5999999991220535 ) )
print a
print a[0]*1./a[1]
a = get_fraction( get_cont_fraction( 1.2799999997163347 ) )
print a
print a[0]*1./a[1]
给:
>> (8, 5)
>> 1.6
>> (32, 25)
>> 1.28
由于使用字符串操作的解决方案似乎没问题,这里有一个不错的简短解决方案:
def nearest( x, Max9 = 2 ):
s = str(x).replace('.','')
splitter = Max9 * '9'
sOut = s.split( splitter )[0]
return int( sOut ) + 1
a = 7.999999985666533
b = 1.5999999991220535
c = 1.2799999997163347
print nearest( a )
print nearest( b )
print nearest( c )
只是提供:
>> 8
>> 16
>> 128
编辑
正如@gc7__正确指出的那样,上述解决方案忽略了值稍大的情况。这使得代码有点复杂,但仍然很好。
import re
def nearest( x, Max09 = 2, digits=25 ):
s = ('{val:.{dig}f}'.format( dig=digits, val=x ) ).split('.')
rnd = 0
if len(s) < 2 or s[1] == '0':## input is integer xyz or float of type xyz.
out = int( x )
else:
s0, s9 = Max09*'0', Max09*'9'
splitter = '{}|{}'.format( s0, s9)
body = s[0]
p0, p9 = s[1].find(s0), s[1].find(s9) ### returns -1 if nothing is found
tail = re.split( splitter, s[1] )[0]
out = int( body + tail )
if p9 > -1 and ( p9 < p0 or p0 < 0 ):
rnd = 1
return out + rnd
a = 7.999998560066533
b = 1.5999999991220535
c = 1.2799999997163347
d = 1233
e = 19935
f = 1.6000000000123
g = 10006.6000000000123
h = 500001.0
print nearest( a )
print nearest( b )
print nearest( c )
print nearest( d )
print nearest( e )
print nearest( f )
print nearest( g )
print nearest( h )
提供:
>> 8
>> 16
>> 128
>> 1233
>> 19935
>> 16
>> 100066
>> 500001
math.isclose
告诉您给定参数 rel_tol
和 abs_tol
的情况下,两个floats
是否接近。 默认设置是
rel_tol=1e-09, abs_tol=0.0
round
将跳到下一个整数,这可能远远超出这些容差。
将abs_tol
设置为<0.5
的内容将使您round
使用过的内容isclose
True
:
from math import isclose
f = 1.5999999991220535
r = round(f)
print(r) # 2
print(isclose(f, r)) # False
print(isclose(f, r, abs_tol=0.5-0.00001)) # True