Convert a number range to another range, maintaining ratio

I'm trying to convert one range of numbers to another, maintaining ratio. Maths is not my strong point.

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).

Solution 1:

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

Or a little more readable:

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

Or if you want to protect for the case where the old range is 0 (OldMin = OldMax):

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

Note that in this case we're forced to pick one of the possible new range values arbitrarily. Depending on context, sensible choices could be: NewMin (see sample), NewMax or (NewMin + NewMax) / 2

Solution 2:

That's a simple linear conversion.

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

So converting 10000 on the scale of -16000 to 16000 to a new scale of 0 to 100 yields:

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

Solution 3:

Actually there are some cases that above answers would break. Such as wrongly input value, wrongly input range, negative input/output ranges.

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

Solution 4:

There is a condition, when all of the values that you are checking are the same, where @jerryjvl's code would return NaN.

if (OldMin != OldMax && NewMin != NewMax):
    return (((OldValue - OldMin) * (NewMax - NewMin)) / (OldMax - OldMin)) + NewMin
else:
    return (NewMax + NewMin) / 2

Convert a number range to another range, maintaining ratio

Solution 1:

Solution 2:

Solution 3:

Solution 4:

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