Python 计算坐标之间距离的程序

在这个程序中，我将包含key-'id'和value-'coord'的json数据解析到python字典中。我是python新手，很抱歉代码不是python的方式。我的问题在z=myDict.getn行。它总是返回相同的值，但我希望每49次迭代就更改一个next键，因为我有49个坐标，这是我的json看起来像：

 {"status" : "200", 
    "result" : [{"id":10001,"coord":"45.781231 78.784321"},{"id":10002,"coord":"45.481684 78.743410},{"id":10003,"coord":"45.411934 78.543113}, etc the same 50 dicts.

如果你能帮助我，我将非常感激

from math import sin, cos, sqrt, atan2, radians
import requests 
import itertools 


def data():
    response = requests.get('')
    json_data = response.json()
    global myDict
    myDict = {}
    for each in json_data['result']:
    myDict.update ({each['id']: (each['coord'])})


    myDict[10252] = '43.238454 76.949185'
    myDict = {str(k):(v) for k,v in myDict.items()}
    keys, values = zip(*myDict.items())
    for v in itertools.product(*values):
        experiment = dict(zip(keys, v))    
        for key,value in experiment.items():
        x = myDict[key]
        for n,m in experiment.items():
            z = myDict.get(n)
        var1,var2 = x.split()
        var3,var4 = z.split()                                                                                                   
        R = 6373.0 
        lat1 = radians(float(var1)),radians(float(var2))
        lat2 = radians(float(var3)),radians(float(var4))
        dlon = lat2[1]-lat1[1]
        dlat = lat2[0]-lat1[0]
        a = sin(dlat / 2)**2 + cos(lat1[0]) * cos(lat2[0]) * sin(dlon / 2)**2
        c = 2 * atan2(sqrt(a), sqrt(1 - a))
        distance = int((R * c)*1000)
        print ("Result:", distance)
        print(x,z)
 g = data()
 print(g)

我给出的输出是： 正如您所看到的，每49次迭代，数字都是相等的。它将每个id值与唯一的id:10001进行比较。我想把它和id:10002100310004等进行比较

Result: 0
Result: 55
Result: 1987
Result: 1345
Result: 3290
Result: 4380
Result: 2058
Result: 4715
Result: 910
Result: 2501
Result: 1176
Result: 3958
Result: 3622
Result: 534
Result: 1176
Result: 922
Result: 2154
Result: 3440
Result: 2749
Result: 1530
Result: 2139
Result: 919
Result: 1264
Result: 3331
Result: 1559
Result: 3542
Result: 3752
Result: 4684
Result: 3262
Result: 5653
Result: 6239
Result: 5827
Result: 6245
Result: 1444
Result: 754
Result: 1755
Result: 453
Result: 5459
Result: 7287
Result: 1642
Result: 4506
Result: 4266
Result: 493
Result: 1029
Result: 4944
Result: 4744
Result: 5206
Result: 2683
Result: 0
Result: 55
Result: 1987
Result: 1345
Result: 3290
Result: 4380
Result: 2058
Result: 4715
Result: 910
Result: 2501
Result: 1176
Result: 3958
Result: 3622
Result: 534
Result: 1176
Result: 922
Result: 2154
Result: 3440
Result: 2749
Result: 1530
Result: 2139
Result: 919
Result: 1264
Result: 3331
Result: 1559
Result: 3542
Result: 3752
Result: 4684
Result: 3262
Result: 5653
Result: 6239
Result: 5827
Result: 6245
Result: 1444
Result: 754
Result: 1755
Result: 453
Result: 5459
Result: 7287
Result: 1642
Result: 4506
Result: 4266
Result: 493
Result: 1029
Result: 4944
Result: 4744
Result: 5206
Result: 2683
Result: 0
Result: 55
Result: 1987
Result: 1345
Result: 3290
Result: 4380
Result: 2058
Result: 4715
Result: 910
Result: 2501
Result: 1176
Result: 3958
Result: 3622
Result: 534
Result: 1176
Result: 922
Result: 2154
Result: 3440
Result: 2749
Result: 1530
Result: 2139
Result: 919
Result: 1264
Result: 3331
Result: 1559
Result: 3542
Result: 3752
Result: 4684
Result: 3262
Result: 5653
Result: 6239
Result: 5827
Result: 6245
Result: 1444
Result: 754
Result: 1755
Result: 453
Result: 5459
Result: 7287
Result: 1642
Result: 4506
Result: 4266
Result: 493
Result: 1029
Result: 4944
Result: 4744
Result: 5206
Result: 2683

---

我认为您应该编辑第二个for循环并重新对齐，以便将每个id与其他id进行比较

for v in itertools.product(*values):
        experiment = dict(zip(keys, v))    
        for key,value in experiment.items():
         x = myDict[key]
         z = myDict.get(key)
         var1,var2 = x.split()
         var3,var4 = z.split()                                                                                                   
         R = 6373.0 
         lat1 = radians(float(var1)),radians(float(var2))
         lat2 = radians(float(var3)),radians(float(var4))
         dlon = lat2[1]-lat1[1]
         dlat = lat2[0]-lat1[0]
         a = sin(dlat / 2)**2 + cos(lat1[0]) * cos(lat2[0]) * sin(dlon / 2)**2
         c = 2 * atan2(sqrt(a), sqrt(1 - a))
         distance = int((R * c)*1000)
         print ("Result:", distance)
         print(x,z)

可以使用geopy中的vincenty计算距离

from geopy.distance import vincenty
dist = vincenty(p1, p2)

如果希望结果以公里为单位：

dist = vincenty(p1, p2).kilometers

---

p1和p2应该是元组

您是否尝试打印n以便查看值的变化？您的程序太复杂了。为什么要构建一个dict，然后将其用作一个列表？并放置提供的示例。哦，我尝试打印，发现问题出在x=myDict[key]中，因为我必须将字符串更改为float。哦，不，问题不在这里。如果您提供预期输出和给定输出，我可以编辑我的答案，使其更有帮助。