python - 在 Python 2.7 中计算和存储函数输出？

Question

我有这样的代码:

nList = [[[0,0,0],[100420,0,623400]],\
[[]],\
[[100043,1324000,123240]],\
[[0,0,543],[3002340,443000,34300],[334000,4043400,7342],[0,0,134020]]
import math
prevX, prevY, prevT = 0, 0, 0

#Entry point
for traceIndex in range(0, len(nList)):
    print 'trace: ' + str(traceIndex+1)
    trace = nList[traceIndex]
    for pointIndex in range(0, len(trace)):
        point = trace[pointIndex]
        if len(point)>0:
            tempX, tempY, tempT = point[0], point[1], point[2]
            if pointIndex != 0:
           #calculate time difference here
                timeDiff = calculateTime (tempT,prevT)

基本上，nList每个 \ 之前都有子列表称为迹线，每条迹线都有三个元素的点。例如，nList[0][0]生成迹线 1，点 1= [0,0,0] 。 point=[x-coordinate, y-coordinate, time] 。我已经计算了每条轨迹中每个点的 timeDiff 。现在我需要总结不同跟踪的 timeDiff 并打印它们，以便:

trace: 1
623400
trace: 2
trace: 3
trace: 4
187393

nList 由称为“trace”的子列表组成，每个“trace”都有一个或多个带有 3 个元素 [x, y, t] 的点。例如，trace1 有 2 个点，trace1point1 = [0,0,0] 且trace1point2=[100420,0,623400]。 timeDiff 计算 t2 和 t1 之间的差。对于trace1，这将是(623400-0)。与迹线 1 相比，迹线 4 有更多的点，并且 timeDiff 对于单个迹线 4pointN 而言为 1=<N=<4 、(34300-543)、(7342-34300)和(134020-7342)。我想编写一个程序，获取每个跟踪中的所有 timeDiff，并以产生上述输出的方式对它们进行求和。

Answer 1

使用 zip 并直接迭代元素以避免需要在变量中存储太多内容，可以更轻松地解决此问题。根据您的示例输出，您需要每个时间点之间的绝对差异:

traces = [[[0,0,0],[100420,0,623400]],\
[[]],\
[[100043,1324000,123240]],\
[[0,0,543],[3002340,443000,34300],[334000,4043400,7342],[0,0,134020]]]
TIME_INDEX = 2  
traceCounter = 1
for trace in traces:
    print "trace:", traceCounter
    traceCounter += 1

    if len(trace[0]) < 2:
       #no coordinate in first element of trace, nothing to do
       continue

    #Zip takes several lists as arguments and returns list of lists with every 0th element in the 0th list, every 1st element in the 1st list etc. 
    timeStamps = zip(*trace)[TIME_INDEX]


    sumOfTimeDiffs = sum([abs(y-x) for x, y in zip(timeStamps[:-1], timeStamps[1:])] )

    if sumOfTimeDiffs > 0:
       print sumOfTimeDiffs

输出:

trace: 1
623400
trace: 2
trace: 3
trace: 4
187393