如何计算Python 2范围之间重叠的百分比

Question

我想计算两个范围之间重叠的百分比。

例如：我有两个列表 x 和 y x['567.630','592.927'] y ['593.000', '618.297'] 在这种情况下，我希望输出为 0，因为没有重叠。

x['793.843','802.244'] y ['794.843','803.244'] 在这种情况下，我想要 87% 的输出。

x 和 y 的总范围并不总是相同。可能 x 是 10s，y 只是 8s。

难道有一些包可以执行这些计算吗？

先感谢您！

Answer 1

这是更短的解决方案，您可以在其中选择要比较重叠的列表。如果您希望例如[0,1]的重叠和[0,2]为100％，则可以从输出中进行最大（x，y）。我更改了y列表以显示这一点。

x = [793.843, 802.244]

y = [794.843, 900.244] 


def overlap_percentage(xlist,ylist):
    min1 = min(xlist)
    max1 = max(xlist)
    min2 = min(ylist)
    max2 = max(ylist)

    overlap = max(0, min(max1, max2) - max(min1, min2))
    length = max1-min1 + max2-min2
    lengthx = max1-min1
    lengthy = max2-min2

    return 2*overlap/length , overlap/lengthx  , overlap/lengthy 

 average, x,y = overlap_percentage(x,y)
    
 print("average: ", average*100,"%    x: ", x*100,"%    y: ",y*100,"%")

输出

average:  13.0 %    x:  88.0%    y:  7.0 %

Answer 2

无需额外进口。我假设数据是浮点数字列表而不是字符串。百分比计算为与第二个间隔重叠，对于其他情况，请在实现中取消注释该行。

def overlapping_percentage(interval1: list, interval2: list) -> int:
    # min max of each intervals
    m1, M1 = min(interval1), max(interval1)
    m2, M2 = min(interval2), max(interval2)

    # find overlapping boundaries
    lb, ub = 0, 0 # lower and upper bound
    if m1 <= m2 <= M1:
        if M1 >= M2:    # i2 is contained in i1
            lb, ub = m2, M2
        elif M1 < M2:
            lb, ub = m2, M1

    elif m2 <= m1 <= M2:
        if M1 <= M2:    # i1 is contained in i2
            lb, ub = m1, M1
        elif M1 > M2:
            lb, ub = m1, M2

    # percentage
    percent = int((ub - lb) / (interval2[1] - interval2[0]) * 100)
    # percent = int((ub - lb) / (interval1[1] - interval1[0]) * 100) # for the other percentage uncomment this line
    print('overlapping interval', lb, ub)
    print(percent, '%')
    
    return percent


# intervals
#i1, i2 = [567.630, 592.927], [593.000, 618.297]
i1, i2 = [793.843, 802.244], [794.843, 803.244]

overlapping_percentage(i1, i2)
# overlapping interval 794.843 - 802.244
# 88 %

重叠部分可以进一步压缩如下

    ...
    # find overlapping boundaries
    lb, ub = 0, 0 # lower and upper bound
    if m1 <= m2 <= M1:
        lb, ub = m2, min(M1, M2)
    elif m2 <= m1 <= M2:
        lb, ub = m1, min(M1, M2)
    ...