遍历字典以替换前导零吗?

时间:2020-06-14 05:23:28

标签: python hyphen

我想遍历此字典并找到具有前导零的任何“ id”(如下面的那个),并替换为不带零的ID。所以'id':'01001'会变成'id':'1001'

这是如何获取我正在使用的数据:

from urllib.request import urlopen
import json
with urlopen('https://raw.githubusercontent.com/plotly/datasets/master/geojson-counties-fips.json') as response:
    counties = json.load(response)

到目前为止,我一次只能获得一个ID,但不确定如何遍历以获取所有ID:

到目前为止我的代码:counties['features'][0]['id']

{ 'type': 'FeatureCollection',
 'features': [{'type': 'Feature',
   'properties': {'GEO_ID': '0500000US01001',
    'STATE': '01',
    'COUNTY': '001',
    'NAME': 'Autauga',
    'LSAD': 'County',
    'CENSUSAREA': 594.436},
   'geometry': {'type': 'Polygon',
    'coordinates': [[[-86.496774, 32.344437],
      [-86.717897, 32.402814],
      [-86.814912, 32.340803],
      [-86.890581, 32.502974],
      [-86.917595, 32.664169],
      [-86.71339, 32.661732],
      [-86.714219, 32.705694],
      [-86.413116, 32.707386],
      [-86.411172, 32.409937],
      [-86.496774, 32.344437]]]},
   'id': '01001'}
    ]
}

3 个答案:

答案 0 :(得分:4)

from urllib.request import urlopen
import json
with urlopen('https://raw.githubusercontent.com/plotly/datasets/master/geojson-counties-fips.json') as response:
    counties = json.load(response)

如果ID与您的JSON结构一起,则遍历列表。并更新ID

counties['features'][0]['id'] = counties['features'][0]['id'].lstrip("0") lstrip将从字符串中删除前导零。

答案 1 :(得分:1)

假设您的词典具有以下数据。您可以使用以下代码:

counties={'type': 'FeatureCollection', 
     'features': [ {'type': 'Feature','properties': {'GEO_ID': '0500000US01001','STATE': '01','COUNTY': '001','NAME': 'Autauga', 'LSAD': 'County','CENSUSAREA': 594.436},
    'geometry': {'type': 'Polygon','coordinates': [[[-86.496774, 32.344437],[-86.717897, 32.402814],[-86.814912, 32.340803],
      [-86.890581, 32.502974],
      [-86.917595, 32.664169],
      [-86.71339, 32.661732],
      [-86.714219, 32.705694],
      [-86.413116, 32.707386],
      [-86.411172, 32.409937],
      [-86.496774, 32.344437] ]] } ,'id': '01001'}, {'type': 'Feature','properties': {'GEO_ID': '0500000US01001','STATE': '01','COUNTY': '001','NAME': 'Autauga', 'LSAD': 'County','CENSUSAREA': 594.436},
    'geometry': {'type': 'Polygon','coordinates': [[[-86.496774, 32.344437],[-86.717897, 32.402814],[-86.814912, 32.340803],
      [-86.890581, 32.502974],
      [-86.917595, 32.664169],
      [-86.71339, 32.661732],
      [-86.714219, 32.705694],
      [-86.413116, 32.707386],
      [-86.411172, 32.409937],
      [-86.496774, 32.344437] ]] } ,'id': '000000000001001'} ]} 

for feature in counties['features']:
    feature ['id']=feature ['id'].lstrip("0")

print(counties)    

答案 2 :(得分:1)

这是使用json对象挂钩实现此目的的更短,更快的方法,

def stripZeroes(d):
    if 'id' in d:
        d['id'] = d['id'].lstrip('0')
        return d
    return d
with urlopen('https://raw.githubusercontent.com/plotly/datasets/master/geojson-counties-fips.json') as response:
    counties = json.load(response, object_hook=stripZeroes)