Essential notes5.py - #problem5_AirBnB-TX 001 e0 \u4ece\u4e00\u4e2a LIST \u4e2d\uff0c\u53bb\u9664\u6389$\uff0c \u4e14\u5c06 string \u8f6c\u6210

Essential notes5.py - #problem5_AirBnB-TX 001 e0...

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#problem5_AirBnB-TX """ 001 e0 从一个 LIST 中,去除掉 $ , 且将 string 转成 integer """ [['$27', ''], ['$149', ''], ['$59', ''], ['$60', ' %20TX'], ['$75', '']] prices = ['$27' , '$149' , '$59' , '$60' , '$75'] convertint(prices) == [27 , 149 , 59 , 60 , 75] def convertint(prices): return [int(i.split("$")[1]) for i in prices] [i.split("$") for i in prices] #--> [['', '27'], ['', '149'], ['', '59'], ['', '60'], ['', '75']] [i.split("$")[1] for i in prices] #--> ['27', '149', '59', '60', '75'] """ 002 e1 从一个包含多个 URL LIST 中,找出 CITY """ urls = ['', '', '', ' %20TX'] parseurls(urls) == ['Cleveland','Cibolo','Beach City','College Station'] import re def parseurl(urls): raw_locs = get_raw_locations(urls) cities = clean_raw_locations(raw_locs) return cities def get_raw_locations(urls): locs = [] for url in urls: start = re.search('location=', url).span()[1] #without[1], then (38, 47) (38, 47) (38, 47) (38, 47) stop = re.search('%2C', url).span()[0] #without[0], then (56, 59) (53, 56) (59, 62) (64, 67) locs.append(url[start:stop]) return locs #['Cleveland', 'Cibolo', 'Beach%20City', 'College%20Station'] def clean_raw_locations(locs): cleaned = [] for raw_loc in locs: cleaned.append(' '.join(raw_loc.split('%20'))) return cleaned #['Cleveland', 'Cibolo', 'Beach City', 'College Station'] """ 003 e2 给出一个数值范围,查找出 LIST OF LIST 中对应的 LIST """ [[27, 'Cleveland'], [149, 'Cibolo'],
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[59, 'Beach City'], [60, 'College Station'], [75, 'Colleyville']] filterdata(data, 50, 100) would return [[59, 'Beach City'], [60, 'College Station'], [75, 'Colleyville']] def filterdata(data, low, high): return [i for i in data if i[0] >= low and i[0] <= high] #i[0] 27 149 59 60 75 """ 004 e3 找出 LIST 中每个值的发生次数,输出为字典 DICT """ findcount(['Cleveland','Austin','Dallas','Austin','Cleveland']) == {'Cleveland':2, 'Austin':2, 'Dallas':1} def findcount(s): #solution 1 from collections import defaultdict items = defaultdict(int) for i in s: items[i] += 1 return dict(items) def findcount__version1(s): #solution 2 from collections import Counter return dict(Counter(s)) """ 005 e4 按价格范围,以字典的形式,排出搜索排前三名的城市 """ result2 = {'Price 0-50': ['Coppell', 'Colleyville', 'Carrollton'], 'Price 51-100': ['Bellaire', 'Alvin', 'Alamo Heights'], 'Price 101-200': ['Bayou Vista', 'Center Point', 'Corpus Christi'], 'Price 201-10000': ['Baffin Bay', 'Burnet', 'Buchanan Dam']} TopCities = {} # zerofifty = filterdata(Master_list,0,50) fiftyhundred = filterdata(Master_list,51,100) hundredtwohundred = filterdata(Master_list,101,200) twohundredplus = filterdata(Master_list,201,10000) def citiescounts(data): cities = [i[1] for i in data] cities_counts = findcount(cities) return cities_counts zerofifty_citycounts = citiescounts(zerofifty) fiftyhundred_citycounts = citiescounts(fiftyhundred) hundredtwohundred_citycounts = citiescounts(hundredtwohundred) twohundredplus_citycounts = citiescounts(twohundredplus) def topthree(d): return sorted(d, key=d.get, reverse=True)[0:3]
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top3_0_50 = topthree(zerofifty_citycounts) top3_51_100 = topthree(fiftyhundred_citycounts)
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  • Fall '19
  • susan lee

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