Question

我想将几个Json对象与LineStrings

组合成一个GeoJSON特征集合

例如，我有以下格式错误的json对象：

 {"lat":16.0269337,"lon":40.073042,"score":1,"ID":"13800006252028","TYPES":"Regional","N2C":"2","NAME":"Strada Statale della Val Sinni"}
{"lat":16.0316522,"lon":41.6808931,"score":1,"ID":"13800006260148","TYPES":"Scenic","N2C":"2"}
{"lat":16.0316523,"lon":41.6808932,"score":1,"ID":"13800006260148","TYPES":"Scenic","N2C":"2","NAME":"Giro del Gargano"}
{"lat":16.0316522,"lon":41.6808931,"score":1,"ID":"13800006203620","TYPES":"Scenic","N2C":"2"}
{"lat":16.0316522,"lon":41.6808931,"score":1,"ID":"13800006203620","TYPES":"Regional","N2C":"2","NAME":"Giro del Gargano"}
{"lat":16.032322,"lon":40.6097567,"score":1,"ID":"13800005963252","TYPES":"Scenic","N2C":"2"}
{"lat":16.032322,"lon":40.6097567,"score":1,"ID":"13800005963252","TYPES":"Regional","N2C":"2","NAME":"Via Appia"}

我的目标是做两件事

使用ID创建GeoJson线串，如果ID和＆amp;相同的类型是相同的
根据得分属性添加匹配的json文件的得分。

Answer 1

我不是专家，但这是第一次削减。 This post有助于获得基础知识。对于每个“Geo”，“JSON”和pandas操作部件，可能有比我更优雅的方法。

目前还不能确定您的数据的确切格式，但是如果它真的是使用pd.DataFrame的字典列表，或者直接从JSON文件中使用{{1}，您将能够直接导入它。 }}。我只是和前者一起工作，但这很容易改变。

pd.from_json

然后使用pandas来操作数据并使用geojson包创建LineString对象

json_objs = [{"lat":16.0269337,"lon":40.073042,"score":1,"ID":"13800006252028","TYPES":"Regional","N2C":"2","NAME":"Strada Statale della Val Sinni"},
{"lat":16.0316522,"lon":41.6808931,"score":1,"ID":"13800006260148","TYPES":"Scenic","N2C":"2"},
{"lat":16.0316523,"lon":41.6808932,"score":1,"ID":"13800006260148","TYPES":"Scenic","N2C":"2","NAME":"Giro del Gargano"},
{"lat":16.0316522,"lon":41.6808931,"score":1,"ID":"13800006203620","TYPES":"Scenic","N2C":"2"},
{"lat":16.0316522,"lon":41.6808931,"score":1,"ID":"13800006203620","TYPES":"Regional","N2C":"2","NAME":"Giro del Gargano"},
{"lat":16.032322,"lon":40.6097567,"score":1,"ID":"13800005963252","TYPES":"Scenic","N2C":"2"},
{"lat":16.032322,"lon":40.6097567,"score":1,"ID":"13800005963252","TYPES":"Regional","N2C":"2","NAME":"Via Appia"}

输出：

import geojson
import pandas as pd

df = pd.DataFrame(json_objs)

grouped = df.groupby(["ID","TYPES"])
features = []

for name,indices in grouped.groups.items():
    group = df.iloc[indices]

    score = group["score"].sum()   
    points = list(zip(group["lat"].values,group["lon"].values))    
    properties = {"ID": name[0], "record_type": name[1], "score": score}

    ls = geojson.LineString(coordinates = points, properties = properties)
    print(ls)

    features.append(ls)

然后将{"coordinates": [[16.032322, 40.6097567]], "properties": {"ID": "13800005963252", "record_type": "Scenic", "score": 1}, "type": "LineString"} {"coordinates": [[16.0269337, 40.073042]], "properties": {"ID": "13800006252028", "record_type": "Regional", "score": 1}, "type": "LineString"} {"coordinates": [[16.032322, 40.6097567]], "properties": {"ID": "13800005963252", "record_type": "Regional", "score": 1}, "type": "LineString"} {"coordinates": [[16.0316522, 41.6808931], [16.0316523, 41.6808932]], "properties": {"ID": "13800006260148", "record_type": "Scenic", "score": 2}, "type": "LineString"} {"coordinates": [[16.0316522, 41.6808931]], "properties": {"ID": "13800006203620", "record_type": "Scenic", "score": 1}, "type": "LineString"} {"coordinates": [[16.0316522, 41.6808931]], "properties": {"ID": "13800006203620", "record_type": "Regional", "score": 1}, "type": "LineString"}转换为LineStrings：

FeatureCollection

显然你也可以手动构建这种格式，但我希望feature = geojson.FeatureCollection(features)为你提供额外的灵活性。显然还有一个我无法安装的geojson包。

使用属性从JSON对象创建新的GeoJSON LineString

1 个答案: