计算JSON多段线的中点（Python 2.7.0，无库）

情景：

我有一个向web服务发出请求的服务器

• web服务返回一个JSON对象
• JSON对象包含数组中的多段线顶点

---

JSON对象的一个小片段是：

{
  "objectIdFieldName": "OBJECTID",
  "globalIdFieldName": "",
  "geometryType": "esriGeometryPolyline",
  "spatialReference": {
    "wkid": 476,
    "latestWkid": 476
  },
  "fields": [
    {
      "name": "OBJECTID",
      "alias": "OBJECTID",
      "type": "esriFieldTypeOID"
    }
  ],
  "features": [
    {
      "attributes": {
        "OBJECTID": 3311
      },
      "geometry": {
        "paths": [
          [
            [
              675844.1562959617,
              4861766.9811610579
            ],
            [
              675878.30397594348,
              4861792.5977392439
            ],
            [
              675891.38832408097,
              4861800.4024024364
            ],
            [
              675902.17710777745,
              4861804.9933949765
            ],
            [
              675912.27726199664,
              4861808.2070551421
            ],
            [
              675923.52513550781,
              4861810.2730065044
            ],
            [
              675934.77300901897,
              4861811.1911861338
            ],
            [
              675943.03676202707,
              4861811.1911861338
            ],
            [
              675951.07095439639,
              4861810.502546167
            ],
            [
              675961.17111910321,
              4861808.6661449578
            ],
            [
              675970.35304125212,
              4861806.1411667075
            ],
            [
              675981.51595173683,
              4861800.7007851209
            ],
            [
              675998.03647276573,
              4861792.2469376959
            ]
          ]
        ]
      }
    },

**The JSON object has been cut off.**

可以在此处找到完整的JSON对象：

---

问题：

使用JSON顶点，我想计算多段线的中点（见下面的绿点）：

• 一些行（ObjectID 3716和3385）是多部分的。在这种情况下，只能为直线的最长部分（而不是其他部分）生成中点
• 为了解决这个问题，JSON文本可以保存为文本文件，并加载到python脚本中。在本例中，可以使用Python的JSON库——尽管下面提到了catch

---

输出如下所示（格式可能不同）：

---

捕获物：

这需要在Python2.7.0中完成——因为我的系统使用2.7.0

• 需要注意的是，我无法将任何Python库导入到我正在使用的系统中的Jython实现中。因此，不幸的是，脚本不应该导入任何python库（用于测试的JSON库除外）

---

---

是否可以使用Python 2.7.0计算JSON多段线的中点（无需导入库）？

是的，您可以仅使用内置Python函数/无需导入外部库轻松计算多段线的中点

让我们打破你的要求：

• 在对象的

  features

  字段中迭代该特征
• 能够计算多段线的长度
• 如果有多条要素，请为每个要素拾取最长的多段线
• 找到此多段线的中点（如注释中所建议的，可以通过求和多段线线段的长度沿多段线前进，直到确定位于中点之间的两个点，并使用

因此，我们需要一些辅助函数来计算两点之间的距离，然后是多段线的距离，依此类推：

# Euclidean distance between two points
def get_distance(p1, p2):
    return sum([(x-y)**2 for (x,y) in zip(p1, p2)]) ** (0.5)

# Length of a polyline by summing the length of its segments
def get_distance_line(line):
    total = 0
    for start_index in range(len(line) - 1):
        stop_index = start_index + 1
        total += get_distance(line[start_index], line[stop_index])
    return total

# Get the polyline with the longest distance
# within a list of polyline
def get_longest(li):
    return max(li, key=get_distance_line)

# Compute the target point at `_target_dist`
# of `p1` along the p1-p2 segment
def _get_pt_at_dist(p1, p2, _target_dist):
    # Define the vector from p1 to p2
    vx = p2[0] - p1[0]
    vy = p2[1] - p1[1]
    # Compute the length of the vector
    lv = (vx ** 2 + vy ** 2) ** 0.5
    # Compute the unit vector (the vector with length 1)
    nv = [vx / lv, vy / lv]
    # Compute the target point
    return [
        p1[0] + nv[0] * _target_dist,
        p1[1] + nv[1] * _target_dist,
    ]

# Get a point at a specific distance on a Polyline
# - 1st step to find the two points enclosing the `target_dist
# - 2nd step to calculate the midpoint along the 2 previously selected points
def get_point_at_distance(line, target_dist):
    sum_dist = 0
    for start_index in range(len(line) - 1):
        stop_index = start_index + 1
        n_dist = get_distance(line[start_index], line[stop_index])
        if sum_dist + n_dist > target_dist:
            # We have found the two enclosing points
            p1, p2 = line[start_index], line[stop_index]
            _target_dist = target_dist - sum_dist
            return _get_pt_at_dist(p1, p2, _target_dist)
        else:
            sum_dist += n_dist

    raise ValueError("target distance is greater than the length of the line")

让我们迭代您的数据（我将您的对象命名为

dataset

），并使用这些函数计算中点：

result = {}

for ft in dataset['features']:
    paths = ft['geometry']['paths']

    # Pick the longest path of default to
    # the only existing one:
    if len(paths) == 1:
        p = paths[0]
    else:
        p = get_longest(paths)

    # Compute the distance
    # and the half of the distance
    # for this polyline
    distance_line = get_distance_line(p)
    middle_dist = distance_line / 2

    # Compute the midpoint and save it
    # into a `dict` using the `OBJECTID`
    # attribute as a key
    midpoint = get_point_at_distance(p, middle_dist)
    result[ft['attributes']['OBJECTID']] = midpoint

结果对象：

{3311: [675916.9814634613, 4861809.071098591],
 3385: [676208.6690235228, 4861536.554984818],
 2165: [676163.9343346333, 4861476.37263185],
 2682: [676060.391694662, 4861904.761846619],
 2683: [675743.1665799635, 4861724.081134027],
 2691: [676137.4796253176, 4861557.709372229],
 2375: [676118.1225925689, 4861730.258496471],
 2320: [676142.0016617056, 4861959.660392438],
 3716: [675979.6112380569, 4861724.356721315],
 3546: [676262.2623328466, 4861665.145686949],
 3547: [676167.5737531717, 4861612.6987658115],
 3548: [676021.3677275265, 4861573.140917723],
 3549: [675914.4334252588, 4861669.870444033],
 3550: [675866.6003329497, 4861735.571798388],
 3551: [675800.1231731868, 4861827.48182595],
 3552: [675681.9432478376, 4861918.988687315]}

---

请注意：首先，我选择了节点数最多的路径（而不是距离最长的路径-使用类似

def get_longest（li）：return max（li，key=len）

），我得到的（视觉）结果与提供的结果非常接近，所以，也许这就是为什么你想说

行中最长的部分

，但我不确定

是的，您可以仅使用内置Python函数/轻松计算多段线的中点，而无需导入外部库

让我们打破你的要求：

• 在对象的

  features

  字段中迭代该特征
• 能够计算多段线的长度
• 如果有多条要素，请为每个要素拾取最长的多段线
• 找到此多段线的中点（如注释中所建议的，可以通过求和多段线线段的长度沿多段线前进，直到确定位于中点之间的两个点，并使用

因此，我们需要一些辅助函数来计算两点之间的距离，然后是多段线的距离，依此类推：

# Euclidean distance between two points
def get_distance(p1, p2):
    return sum([(x-y)**2 for (x,y) in zip(p1, p2)]) ** (0.5)

# Length of a polyline by summing the length of its segments
def get_distance_line(line):
    total = 0
    for start_index in range(len(line) - 1):
        stop_index = start_index + 1
        total += get_distance(line[start_index], line[stop_index])
    return total

# Get the polyline with the longest distance
# within a list of polyline
def get_longest(li):
    return max(li, key=get_distance_line)

# Compute the target point at `_target_dist`
# of `p1` along the p1-p2 segment
def _get_pt_at_dist(p1, p2, _target_dist):
    # Define the vector from p1 to p2
    vx = p2[0] - p1[0]
    vy = p2[1] - p1[1]
    # Compute the length of the vector
    lv = (vx ** 2 + vy ** 2) ** 0.5
    # Compute the unit vector (the vector with length 1)
    nv = [vx / lv, vy / lv]
    # Compute the target point
    return [
        p1[0] + nv[0] * _target_dist,
        p1[1] + nv[1] * _target_dist,
    ]

# Get a point at a specific distance on a Polyline
# - 1st step to find the two points enclosing the `target_dist
# - 2nd step to calculate the midpoint along the 2 previously selected points
def get_point_at_distance(line, target_dist):
    sum_dist = 0
    for start_index in range(len(line) - 1):
        stop_index = start_index + 1
        n_dist = get_distance(line[start_index], line[stop_index])
        if sum_dist + n_dist > target_dist:
            # We have found the two enclosing points
            p1, p2 = line[start_index], line[stop_index]
            _target_dist = target_dist - sum_dist
            return _get_pt_at_dist(p1, p2, _target_dist)
        else:
            sum_dist += n_dist

    raise ValueError("target distance is greater than the length of the line")

让我们迭代您的数据（我将您的对象命名为

dataset

），并使用这些函数计算中点：

result = {}

for ft in dataset['features']:
    paths = ft['geometry']['paths']

    # Pick the longest path of default to
    # the only existing one:
    if len(paths) == 1:
        p = paths[0]
    else:
        p = get_longest(paths)

    # Compute the distance
    # and the half of the distance
    # for this polyline
    distance_line = get_distance_line(p)
    middle_dist = distance_line / 2

    # Compute the midpoint and save it
    # into a `dict` using the `OBJECTID`
    # attribute as a key
    midpoint = get_point_at_distance(p, middle_dist)
    result[ft['attributes']['OBJECTID']] = midpoint

结果对象：

{3311: [675916.9814634613, 4861809.071098591],
 3385: [676208.6690235228, 4861536.554984818],
 2165: [676163.9343346333, 4861476.37263185],
 2682: [676060.391694662, 4861904.761846619],
 2683: [675743.1665799635, 4861724.081134027],
 2691: [676137.4796253176, 4861557.709372229],
 2375: [676118.1225925689, 4861730.258496471],
 2320: [676142.0016617056, 4861959.660392438],
 3716: [675979.6112380569, 4861724.356721315],
 3546: [676262.2623328466, 4861665.145686949],
 3547: [676167.5737531717, 4861612.6987658115],
 3548: [676021.3677275265, 4861573.140917723],
 3549: [675914.4334252588, 4861669.870444033],
 3550: [675866.6003329497, 4861735.571798388],
 3551: [675800.1231731868, 4861827.48182595],
 3552: [675681.9432478376, 4861918.988687315]}

---

请注意：首先，我选择了节点数最多的路径（而不是距离最长的路径-使用类似

def get_longest（li）：return max（li，key=len）

），我得到的（视觉）结果与提供的结果非常接近，所以，也许这就是为什么你想说

行中最长的部分

，但我不确定

总长度是每段（或弧，如果可以得到其半径）长度的总和。中点可以通过增加长度来计算，直到达到总长度的一半。你能给我上面的输入多段线的输出样本吗？@Bluespider我提供了一个输出样本（并对问题进行了总体更新）。谢谢。总长度是每段（或弧，如果你能得到它的半径）长度的总和。中点可以通过增加长度来计算，直到达到总长度的一半。你能给我上面的输入多段线的输出样本吗？@Bluespider我提供了一个输出样本（并对问题进行了总体更新）。谢谢，谢谢！关于你在答案末尾的注释：是的，我在寻找长度最长的部分，而不是顶点最多的部分。我最初在问题的建议结果集中包含了错误的部分（最短的部分，但更多的顶点）。对不起；我已经改正了，没问题！希望有帮助！非常感谢。关于你在答案末尾的注释：是的，我在寻找长度最长的部分，而不是顶点最多的部分。我最初在问题的建议结果集中包含了错误的部分（最短的部分，但更多的顶点）。对不起；我已经改正了，没问题！希望有帮助！