Question

我尝试在python中的两个二维列表（包含时间范围）中找到重叠。我发现了一些相关的问题，但是大多数问题只涉及一维列表。如果您有一个已解决的问题，则非常感谢您的链接。

所以我得到了两个列表，（开始，结束，持续时间都是时间的整数值，以毫秒为单位）

TargetSegments =[[Start,End,Duration],[Start,End,Duration],...]
BackchannelSegments =[[Start,End,Duration],[Start,End,Duration],...]

现在我要获取一个新列表，其中包含[startOverlap，endOverlap，durationOverlap]形式的元素

我的第一种方法是使用for循环，但是它有点复杂，我不确定这是否是最有效的方法

这就是我当前的功能。首先，将列表中包含的字符串转换为整数值（对于ms； ps：str2ms是我自己编写的函数，它只需要一个类似时间的字符串，即“ 1:32:23”并计算相应的时间以毫秒为单位）

def removeBackchannels(self,SegmentsTarget,SegmentsBackchannel):
    #Converts the string for the segments in ms
    c_SegmentsTarget = []
    c_SegmentsBackchannel = []
    for start,end,duration in SegmentsTarget:
        c_SegmentsTarget.append([str2ms(start),str2ms(end),str2ms(duration)])

    for start,end,duration in SegmentsBackchannel:
        c_SegmentsBackchannel.append([str2ms(start),str2ms(end),str2ms(duration)])

    for segmentTar in c_SegmentsTarget:
        for segmentBC in c_SegmentsBackchannel:
            if segmentTar[0] < segmentBC[0] < segmentTar[1]: #if BC Start is before end of the Segmnet
                startOverlap = segmentBC[0]
                if segmentBC[1] > segmentTar[1]:
                    endOverlap = segmentTar[1]
                else:
                    endOverlap = segmentBC[1]
                durationOverlap = endOverlap - startOverlap
                overlaps.append([startOverlap,endOverlap,durationOverlap])
                print "Found Overlap at: %s - %s (%s)" % (startOverlap,endOverlap,durationOverlap)
            elif segmentTar[1] > segmentBC[1] > segmentTar[0]: #if BC End is after the start of the Segment
                startOverlap = segmentTar[0]
                endOverlap = segmentBC[1]
                durationOverlap = endOverlap - startOverlap
                overlaps.append([startOverlap,endOverlap,durationOverlap])
                print "Found Overlap at: %s - %s (%s)" % (startOverlap,endOverlap,durationOverlap)
            if segmentBC[0] > segmentTar[1]: #If start of the back channel is behind the end of the segment of interest
                break
    print len(overlaps)

首先，我不确定我是否能够使用这种方法来解决所有重叠问题，或者我的if陈述是否足够？其次，如果有一种更Python化或更有效的方式来获得重叠，那就太好了。

非常感谢和问候

Python的两个列表以ms为单位的时间范围，确定重叠

0 个答案: