我试图将一个范围的数字转换为另一个,保持比率。数学不是我的强项。

I have an image file where point values may range from -16000.00 to 16000.00 though the typical range may be much less. What I want to do is compress these values into the integer range 0-100, where 0 is the value of the smallest point, and 100 is the value of the largest. All points in between should keep a relative ratio even though some precision is being lost I'd like to do this in python but even a general algorithm should suffice. I'd prefer an algorithm where the min/max or either range can be adjusted (ie, the second range could be -50 to 800 instead of 0 to 100).


当前回答

NewValue = (((OldValue - OldMin) * (NewMax - NewMin)) / (OldMax - OldMin)) + NewMin

或者更容易读懂:

OldRange = (OldMax - OldMin)  
NewRange = (NewMax - NewMin)  
NewValue = (((OldValue - OldMin) * NewRange) / OldRange) + NewMin

或者如果你想保护旧范围为0的情况(OldMin = OldMax):

OldRange = (OldMax - OldMin)
if (OldRange == 0)
    NewValue = NewMin
else
{
    NewRange = (NewMax - NewMin)  
    NewValue = (((OldValue - OldMin) * NewRange) / OldRange) + NewMin
}

注意,在这种情况下,我们被迫任意选择一个可能的新范围值。根据上下文,明智的选择可能是:NewMin(见示例),NewMax或(NewMin + NewMax) / 2

其他回答

实际上,在某些情况下,上述答案会失效。 如错误的输入值,错误的输入范围,负输入/输出范围。

def remap( x, oMin, oMax, nMin, nMax ):

    #range check
    if oMin == oMax:
        print "Warning: Zero input range"
        return None

    if nMin == nMax:
        print "Warning: Zero output range"
        return None

    #check reversed input range
    reverseInput = False
    oldMin = min( oMin, oMax )
    oldMax = max( oMin, oMax )
    if not oldMin == oMin:
        reverseInput = True

    #check reversed output range
    reverseOutput = False   
    newMin = min( nMin, nMax )
    newMax = max( nMin, nMax )
    if not newMin == nMin :
        reverseOutput = True

    portion = (x-oldMin)*(newMax-newMin)/(oldMax-oldMin)
    if reverseInput:
        portion = (oldMax-x)*(newMax-newMin)/(oldMax-oldMin)

    result = portion + newMin
    if reverseOutput:
        result = newMax - portion

    return result

#test cases
print remap( 25.0, 0.0, 100.0, 1.0, -1.0 ), "==", 0.5
print remap( 25.0, 100.0, -100.0, -1.0, 1.0 ), "==", -0.25
print remap( -125.0, -100.0, -200.0, 1.0, -1.0 ), "==", 0.5
print remap( -125.0, -200.0, -100.0, -1.0, 1.0 ), "==", 0.5
#even when value is out of bound
print remap( -20.0, 0.0, 100.0, 0.0, 1.0 ), "==", -0.2

Java版本

不管你喂它什么,它都能工作!

我把所有内容都展开了,这样便于学习。当然,最后舍入是可选的。

    private long remap(long p, long Amin, long Amax, long Bmin, long Bmax ) {

    double deltaA = Amax - Amin;
    double deltaB = Bmax - Bmin;
    double scale  = deltaB / deltaA;
    double negA   = -1 * Amin;
    double offset = (negA * scale) + Bmin;
    double q      = (p * scale) + offset;
    return Math.round(q);

}

这是一个简单的线性变换。

new_value = ( (old_value - old_min) / (old_max - old_min) ) * (new_max - new_min) + new_min

因此,将10000在-16000到16000的范围内转换为0到100的新范围会得到:

old_value = 10000
old_min = -16000
old_max = 16000
new_min = 0
new_max = 100

new_value = ( ( 10000 - -16000 ) / (16000 - -16000) ) * (100 - 0) + 0
          = 81.25

我没有为此挖掘BNF,但Arduino文档有一个很好的函数示例,它是分解的。我可以在Python中通过简单地添加一个def重命名到remap(因为map是内置的)并删除类型强制转换和花括号(即删除所有的'long')来使用它。

原始

long map(long x, long in_min, long in_max, long out_min, long out_max)
{
  return (x - in_min) * (out_max - out_min) / (in_max - in_min) + out_min;
}

Python

def remap(x, in_min, in_max, out_min, out_max):
  return (x - in_min) * (out_max - out_min) / (in_max - in_min) + out_min

https://www.arduino.cc/en/reference/map

列出理解一行的解决方案

color_array_new = [int((((x - min(node_sizes)) * 99) / (max(node_sizes) - min(node_sizes))) + 1) for x in node_sizes]

完整版

def colour_specter(waste_amount):
color_array = []
OldRange = max(waste_amount) - min(waste_amount)
NewRange = 99
for number_value in waste_amount:
    NewValue = int((((number_value - min(waste_amount)) * NewRange) / OldRange) + 1)
    color_array.append(NewValue)
print(color_array)
return color_array