根据输入提取字符串的特定部分
我有一个非常大的.json
转换为string
,其中包含许多城市/国家。根据输入提取字符串的特定部分
我想根据用户的国家选择提取城市信息(伦敦只是一个示例)。
例如,如果Country
用户inputed是:UK
,以下信息会从字符串中提取:
我不是我怎么可能做到这一点,由于我没有经验不太清楚,但我知道这需要一个if语句。到目前为止,我的进步:
Country = raw_input('Country: ')
if 'UK' in string:
???
最初的反应是不是很大,因为我们几个人忽略了生JSON。然而,你提供将来会更好一些,因为你显示的代码片段中有一个更完整(和有效)的代码片段。
这就是说,我将数据加载到一个字典,这样做:
import json
json_string = """{
"response": {
"version":"0.1",
"termsofService":"http://www.wunderground.com/weather/api/d/terms.html",
"features": {
"conditions": 1
}
, "results": [
{
"name": "London",
"city": "London",
"state": "AR",
"country": "US",
"country_iso3166":"US",
"country_name":"USA",
"zmw": "72847.1.99999",
"l": "https://*.com/q/zmw:72847.1.99999"
}
,
{
"name": "London",
"city": "London",
"state": "KY",
"country": "US",
"country_iso3166":"US",
"country_name":"USA",
"zmw": "40741.1.99999",
"l": "https://*.com/q/zmw:40741.1.99999"
}
,
{
"name": "London",
"city": "London",
"state": "MN",
"country": "US",
"country_iso3166":"US",
"country_name":"USA",
"zmw": "56036.3.99999",
"l": "https://*.com/q/zmw:56036.3.99999"
}
,
{
"name": "London",
"city": "London",
"state": "OH",
"country": "US",
"country_iso3166":"US",
"country_name":"USA",
"zmw": "43140.1.99999",
"l": "https://*.com/q/zmw:43140.1.99999"
}
,
{
"name": "London",
"city": "London",
"state": "ON",
"country": "CA",
"country_iso3166":"CA",
"country_name":"Canada",
"zmw": "00000.1.71623",
"l": "https://*.com/q/zmw:00000.1.71623"
}
,
{
"name": "London",
"city": "London",
"state": "TX",
"country": "US",
"country_iso3166":"US",
"country_name":"USA",
"zmw": "76854.1.99999",
"l": "https://*.com/q/zmw:76854.1.99999"
}
,
{
"name": "London",
"city": "London",
"state": "",
"country": "UK",
"country_iso3166":"GB",
"country_name":"United Kingdom",
"zmw": "00000.1.03772",
"l": "https://*.com/q/zmw:00000.1.03772"
}
,
{
"name": "London",
"city": "London",
"state": "WV",
"country": "US",
"country_iso3166":"US",
"country_name":"USA",
"zmw": "25126.1.99999",
"l": "https://*.com/q/zmw:25126.1.99999"
}
]
}
}"""
json_object = json.loads(json_string)
world_dict = {}
for item in json_object['response']['results']:
item_country = item['country']
in_dict = world_dict.get(item_country)
if in_dict:
world_dict[item_country].extend([item])
else:
world_dict[item_country] = [item]
country = raw_input('Country: ')
response = world_dict.get(country)
if response:
for item in response:
print item
else:
print "Not a valid country"
编辑: 基于评论时使用的网址,而不是一个JSON字符串。
import requests
url = 'http://api.wunderground.com/api/a8c3e5ce8970ae66/conditions/q/London.json'
data = requests.get(url).json()
world_dict = {}
for item in data['response']['results']:
item_country = item['country']
in_dict = world_dict.get(item_country)
if in_dict:
world_dict[item_country].extend([item])
else:
world_dict[item_country] = [item]
country = raw_input('Country: ')
response = world_dict.get(country)
if response:
for item in response:
print item
else:
print "Not a valid country"
老实说,谢谢你!谢谢你的耐心!代码很好用! – ThatOnePythonNoob
@ThatOnePythonNoob你是非常欢迎:)我更新,以便它会拒绝不正确的国家名称。不要因以前的困难而感到灰心;就像我说的那样,您发布了一个精确的json结构,只是非常容易让底部显示正确,并在顶部显示无效的代码片段。如果此代码已解决您的问题,我将非常感谢您将其标记为正确(单击答案旁边的勾号),以便其他人不认为它仍未回答。 – roganjosh
是否可以将'json_string'更改为API链接(http://api.wunderground.com/api/a8c3e5ce8970ae66/conditions/q/London.json),然后将其转换为字符串?该计划是否仍然有效? – ThatOnePythonNoob
import json
country = raw_input('Country: ')
jsondata = "the large json string mentioned in your post"
info = json.loads(jsondata)
for item in info:
if item['country'] == country:
print item
是不是'信息['results']'中的项目? – roganjosh
@John Gordon,字面上打印10000行 – ThatOnePythonNoob
然后,限制它只是你想要的信息:'print item ['name'],item ['city'],item ['state']等... ' –
你可以试试这个。可能仍然想在代码中考虑一些用户输入错误。例如,str.strip()和大写敏感。
import json
input_country = raw_input('Please enter country:')
with open('London.json') as fp:
london_json = fp.read()
london = json.loads(london_json)
for item in london["response"]["results"]:
if item['country'] == input_country:
print json.dumps(item, indent = 2)
感谢您的建议!运行时,我目前得到的错误:IOError:[Errno 2]没有这样的文件或目录:'http://api.wunderground.com/api/a8c3e5ce8970ae66/conditions/q/London.json'有什么想法? – ThatOnePythonNoob
'open()'只适用于本地文件,而不适用于网址。 –
你知道如何使用'json'模块解析JSON吗? – FamousJameous
@FamousJameous嗯,不,我会看看我猜。 – ThatOnePythonNoob
以下是供将来参考的链接:https://docs.python.org/2.7/library/json.html – FamousJameous