Question

我有两个对象数组，我想按照属于每个对象一部分的时间戳顺序将它们组合成一个数组。第一个数组是精确的第二个数组，第二个数组是一个小时的范围。我想将精确的时间戳记在数组中的“湿度”值输入到每小时范围的值中，具体取决于它们在时间上适合的位置。我该如何在打字稿中这样做？这两个数组/对象都按时间排序。

例如：

[{id: 1, timestampUtc: "2019-02-22T08:24:00Z", humidity: 74},
 {id: 2, timestampUtc: "2019-02-24T06:20:00Z", humidity: 39},
 {id: 3, timestampUtc: "2019-02-26T020:03:00Z", humidity: 35}]

和

[{id: 4, starttimestampUtc: "2019-02-22T08:00:00Z", endtimestampUtc: "2019-02-22T09:00:00Z", precipitation: .03},
 {id: 5, starttimestampUtc: "2019-02-24T06:00:00Z", endtimestampUtc: "2019-02-24T07:00:00Z", precipitation: .3},
 {id: 6, starttimestampUtc: "2019-02-26T020:00:00Z",endtimestampUtc: "2019-02-26T021:00:00Z", precipitation: .12}]

外观如下：

[{id: 4, starttimestampUtc: "2019-02-22T08:00:00Z", endtimestampUtc: "2019-02-22T09:00:00Z", precipitation: .03, humidity: 74},
 {id: 5, starttimestampUtc: "2019-02-24T06:00:00Z", endtimestampUtc: "2019-02-24T07:00:00Z", precipitation: .3, humidity: 39},
 {id: 6, starttimestampUtc: "2019-02-26T020:00:00Z", endtimestampUtc: "2019-02-26T021:00:00Z", precipitation: .12, , humidity: 35}]

Answer 1

我能够通过将UTC时间戳转换为日期对象来做到这一点。以下代码似乎对我有用，其中a和b是您的两个数组。我还注意到，每个数组的最后一个元素的时间戳中都有一个额外的0。

import argparse

def dijkstra(graph,src,dest,visited=[],distances={},predecessors={}):
    # a few sanity checks
    if src not in graph:
        raise TypeError('The root of the shortest path tree cannot be found')
    if dest not in graph:
        raise TypeError('The target of the shortest path cannot be found')    
    # ending condition
    if src == dest:
        # We build the shortest path and display it
        path=[]
        pred=dest
        while pred != None:
        path.append(pred)
            pred=predecessors.get(pred,None)
        print('shortest path: '+str(path)+" cost="+str(distances[dest])) 
    else :     
        # if it is the initial  run, initializes the cost
        if not visited: 
            distances[src]=0
        # visit the neighbors
        for neighbor in graph[src] :
            if neighbor not in visited:
                new_distance = distances[src] + graph[src][neighbor]
                if new_distance < distances.get(neighbor,float('inf')):
                    distances[neighbor] = new_distance
                    predecessors[neighbor] = src
        # mark as visited
        visited.append(src)
        # now that all neighbors have been visited:    
    recurse                         
        # select the non visited node with lowest distance 'x'
        # run Dijskstra with src='x'
        unvisited={}
        for k in graph:
            if k not in visited:
                unvisited[k] = distances.get(k,float('inf'))        
        x=min(unvisited, key=unvisited.get)
        dijkstra(graph,x,dest,visited,distances,predecessors)



if __name__ == "__main__":
    parser = argparse.ArgumentParser()
    parser.add_argument('g', help='This is the network')
    # Parse the given arguments
    args = parser.parse_args()
    graph = {'s': {'a': 2, 'b': 1}, 'a': {'s': 3, 'b': 4, 'c':8}, 'b': {'s': 4, 'a': 2, 'd': 2}, 'c': {'a': 2, 'd': 7, 't': 4}, 'd': {'b': 1, 'c': 11, 't': 5}, 't': {'c': 3, 'd': 5}}
    #print args.g
    #print graph
    dijkstra(args.g,'s','t')

如何通过时间戳将项目分类为对象

1 个答案: