我有一个类似以下元素的 Python 列表:
['Alabama[edit]',
'Auburn (Auburn University)[1]',
'Florence (University of North Alabama)',
'Jacksonville (Jacksonville State University)[2]',
'Livingston (University of West Alabama)[2]',
'Montevallo (University of Montevallo)[2]',
'Troy (Troy University)[2]',
'Tuscaloosa (University of Alabama, Stillman College, Shelton State)[3][4]',
'Tuskegee (Tuskegee University)[5]',
'Alaska[edit]',
'Fairbanks (University of Alaska Fairbanks)[2]',
'Arizona[edit]',
'Flagstaff (Northern Arizona University)[6]',
'Tempe (Arizona State University)',
'Tucson (University of Arizona)',
'Arkansas[edit]',
'Arkadelphia (Henderson State University, Ouachita Baptist University)[2]',
'Conway (Central Baptist College, Hendrix College, University of Central Arkansas)[2]',
'Fayetteville (University of Arkansas)[7]']
该列表并不完整,但足以让您了解其中的内容。
数据结构如下:
有一个美国州名,州名后面是该州的一些城市名。如您所见,州名称以“[edit]”结尾,城市名称以带数字的括号结尾(例如“1”或“[2]”),或者以大学的名称结尾括号内的名称(例如“(北阿拉巴马大学)”)。
(查找此问题的完整引用文件 here )
理想情况下,我想要一个以州名称作为索引的 Python 字典,并且该州的所有城市名称都嵌套在列表中作为该特定索引的值。因此,例如字典应该是这样的:
{'Alabama': ['Auburn', 'Florence', 'Jacksonville'...], 'Arizona': ['Flagstaff', 'Temple', 'Tucson', ....], ......}
现在,我尝试了以下解决方案,以剔除不需要的部分:
import numpy as np
import pandas as pd
def get_list_of_university_towns():
'''
Returns a DataFrame of towns and the states they are in from the
university_towns.txt list. The format of the DataFrame should be:
DataFrame( [ ["Michigan", "Ann Arbor"], ["Michigan", "Yipsilanti"] ],
columns=["State", "RegionName"] )
The following cleaning needs to be done:
1. For "State", removing characters from "[" to the end.
2. For "RegionName", when applicable, removing every character from " (" to the end.
3. Depending on how you read the data, you may need to remove newline character '\n'.
'''
fhandle = open("university_towns.txt")
ftext = fhandle.read().split("\n")
reftext = list()
for item in ftext:
reftext.append(item.split(" ")[0])
#pos = reftext[0].find("[")
#reftext[0] = reftext[0][:pos]
towns = list()
dic = dict()
for item in reftext:
if item == "Alabama[edit]":
state = "Alabama"
elif item.endswith("[edit]"):
dic[state] = towns
towns = list()
pos = item.find("[")
item = item[:pos]
state = item
else:
towns.append(item)
return ftext
get_list_of_university_towns()
我的代码生成的输出片段如下所示:
{'Alabama': ['Auburn',
'Florence',
'Jacksonville',
'Livingston',
'Montevallo',
'Troy',
'Tuscaloosa',
'Tuskegee'],
'Alaska': ['Fairbanks'],
'Arizona': ['Flagstaff', 'Tempe', 'Tucson'],
'Arkansas': ['Arkadelphia',
'Conway',
'Fayetteville',
'Jonesboro',
'Magnolia',
'Monticello',
'Russellville',
'Searcy'],
'California': ['Angwin',
'Arcata',
'Berkeley',
'Chico',
'Claremont',
'Cotati',
'Davis',
'Irvine',
'Isla',
'University',
'Merced',
'Orange',
'Palo',
'Pomona',
'Redlands',
'Riverside',
'Sacramento',
'University',
'San',
'San',
'Santa',
'Santa',
'Turlock',
'Westwood,',
'Whittier'],
'Colorado': ['Alamosa',
'Boulder',
'Durango',
'Fort',
'Golden',
'Grand',
'Greeley',
'Gunnison',
'Pueblo,'],
'Connecticut': ['Fairfield',
'Middletown',
'New',
'New',
'New',
'Storrs',
'Willimantic'],
'Delaware': ['Dover', 'Newark'],
'Florida': ['Ave',
'Boca',
'Coral',
'DeLand',
'Estero',
'Gainesville',
'Orlando',
'Sarasota',
'St.',
'St.',
'Tallahassee',
'Tampa'],
'Georgia': ['Albany',
'Athens',
'Atlanta',
'Carrollton',
'Demorest',
'Fort',
'Kennesaw',
'Milledgeville',
'Mount',
'Oxford',
'Rome',
'Savannah',
'Statesboro',
'Valdosta',
'Waleska',
'Young'],
'Hawaii': ['Manoa'],
但是,输出中存在一个错误:不包括名称中带有空格的州(例如“北卡罗来纳州”)。我知道背后的原因。
我想到了使用正则表达式,但由于我还没有研究它们,所以我不知道如何形成一个。关于使用或不使用正则表达式如何完成它的任何想法?
最佳答案
那就赞美一下正则表达式的强大吧:
states_rx = re.compile(r'''
^
(?P<state>.+?)\[edit\]
(?P<cities>[\s\S]+?)
(?=^.*\[edit\]$|\Z)
''', re.MULTILINE | re.VERBOSE)
cities_rx = re.compile(r'''^[^()\n]+''', re.MULTILINE)
transformed = '\n'.join(lst_)
result = {state.group('state'): [city.group(0).rstrip()
for city in cities_rx.finditer(state.group('cities'))]
for state in states_rx.finditer(transformed)}
print(result)
这产生
{'Alabama': ['Auburn', 'Florence', 'Jacksonville', 'Livingston', 'Montevallo', 'Troy', 'Tuscaloosa', 'Tuskegee'], 'Alaska': ['Fairbanks'], 'Arizona': ['Flagstaff', 'Tempe', 'Tucson'], 'Arkansas': ['Arkadelphia', 'Conway', 'Fayetteville']}
说明:
想法是将任务拆分成几个较小的任务:
- 加入完整列表
\n
- 不同的州
- 独立的城镇
- 对所有找到的项目使用字典理解
第一个子任务
transformed = '\n'.join(your_list)
第二个子任务
^ # match start of the line
(?P<state>.+?)\[edit\] # capture anything in that line up to [edit]
(?P<cities>[\s\S]+?) # afterwards match anything up to
(?=^.*\[edit\]$|\Z) # ... either another state or the very end of the string
第三个子任务
^[^()\n]+ # match start of the line, anything not a newline character or ( or )
参见 another demo on regex101.com .
第四个子任务
result = {state.group('state'): [city.group(0).rstrip() for city in cities_rx.finditer(state.group('cities'))] for state in states_rx.finditer(transformed)}
这大致相当于:
for state in states_rx.finditer(transformed):
# state is in state.group('state')
for city in cities_rx.finditer(state.group('cities')):
# city is in city.group(0), possibly with whitespaces
# hence the rstrip
最后,一些时间问题:
import timeit
print(timeit.timeit(findstatesandcities, number=10**5))
# 12.234304904000965
所以在我的计算机上运行上面的 100.000 次花了我大约 12 秒,所以它应该相当快。
关于python - 用可能的多个单词匹配州和城市,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48049006/