将一个数字范围转换为另一个范围，保持比率

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

捷径/简化方案

 NewRange/OldRange = Handy multiplicand or HM
 Convert OldValue in OldRange to NewValue in NewRange = 
 (OldValue - OldMin x HM) + NewMin

韦恩

2015-04-08 04:30:54

其他回答

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

2019-08-20 10:32:21

我个人使用支持泛型的helper类(Swift 3,4)。x兼容)

struct Rescale<Type : BinaryFloatingPoint> {
    typealias RescaleDomain = (lowerBound: Type, upperBound: Type)

    var fromDomain: RescaleDomain
    var toDomain: RescaleDomain

    init(from: RescaleDomain, to: RescaleDomain) {
        self.fromDomain = from
        self.toDomain = to
    }

    func interpolate(_ x: Type ) -> Type {
        return self.toDomain.lowerBound * (1 - x) + self.toDomain.upperBound * x;
    }

    func uninterpolate(_ x: Type) -> Type {
        let b = (self.fromDomain.upperBound - self.fromDomain.lowerBound) != 0 ? self.fromDomain.upperBound - self.fromDomain.lowerBound : 1 / self.fromDomain.upperBound;
        return (x - self.fromDomain.lowerBound) / b
    }

    func rescale(_ x: Type )  -> Type {
        return interpolate( uninterpolate(x) )
    }
}

Ex:

   let rescaler = Rescale<Float>(from: (-1, 1), to: (0, 100))
    
   print(rescaler.rescale(0)) // OUTPUT: 50

2017-04-20 16:38:35

使用Numpy和interp函数，你可以将你的值从旧范围转换为新范围:

>>> import numpy as np
>>> np.interp(0, [-16000,16000], [0,100])
50.0

你也可以尝试映射一个值列表:

>>> np.interp([-16000,0,12000] ,[-16000,16000], [0,100])
array([ 0. , 50. , 87.5])

2021-09-21 04:53:17

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

2009-05-30 05:50:45

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

2017-04-23 04:27:08

将一个数字范围转换为另一个范围，保持比率

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