英文:
Extracting two letter state abbreviation from string in column output to new column
问题
我需要从列full_address中提取最后两个州的缩写。
states = ['AK', 'AL', 'AR', 'AZ', 'CA', 'CO', 'CT', 'DC', 'DE', 'FL', 'GA','HI', 'IA', 'ID', 'IL', 'IN', 'KS', 'KY', 'LA', 'MA', 'MD', 'ME','MI', 'MN', 'MO', 'MS', 'MT', 'NC', 'ND', 'NE', 'NH', 'NJ', 'NM','NV', 'NY', 'OH', 'OK', 'OR', 'PA', 'RI', 'SC', 'SD', 'TN', 'TX','UT', 'VA', 'VT', 'WA', 'WI', 'WV', 'WY']
但是我在列state中得到了N/A。
如何才能获得列state中正确的州缩写值？
我希望得到：
state
0 AL1 AL2 AL3 AL4 AL5 AL
英文:
I need to extract the last two state abbreviation from the column full_address
id position name score \0 1 19 PJ Fresh (224 Daniel Payne Drive) NaN1 2 9 J' ti`'z Smoothie-N-Coffee Bar NaN2 3 6 Philly Fresh Cheesesteaks (541-B Graymont Ave) NaN3 4 17 Papa Murphy's (1580 Montgomery Highway) NaN4 5 162 Nelson Brothers Cafe (17th St N) 4.7ratings category price_range \0 NaN Burgers, American, Sandwiches $1 NaN Coffee and Tea, Breakfast and Brunch, Bubble Tea NaN2 NaN American, Cheesesteak, Sandwiches, Alcohol $3 NaN Pizza $4 22.0 Breakfast and Brunch, Burgers, Sandwiches NaNfull_address zip_code lat \0 224 Daniel Payne Drive, Birmingham, AL, 35207 35207 33.5623651 1521 Pinson Valley Parkway, Birmingham, AL, 35217 35217 33.5836402 541-B Graymont Ave, Birmingham, AL, 35204 35204 33.5098003 1580 Montgomery Highway, Hoover, AL, 35226 35226 33.4044394 314 17th St N, Birmingham, AL, 35203 35203 33.514730
states = [ 'AK', 'AL', 'AR', 'AZ', 'CA', 'CO', 'CT', 'DC', 'DE', 'FL', 'GA','HI', 'IA', 'ID', 'IL', 'IN', 'KS', 'KY', 'LA', 'MA', 'MD', 'ME','MI', 'MN', 'MO', 'MS', 'MT', 'NC', 'ND', 'NE', 'NH', 'NJ', 'NM','NV', 'NY', 'OH', 'OK', 'OR', 'PA', 'RI', 'SC', 'SD', 'TN', 'TX','UT', 'VA', 'VT', 'WA', 'WI', 'WV', 'WY']
df['state']=df['full_address'].apply(lambda x: x if x in states else 'N/A')
But I get N/A in the column state
state0 N/A1 N/A2 N/A3 N/A4 N/A5 N/A6 N/A7 N/A8 N/A9 N/A
How do I get the correct values of the state abbreviation in the column state?
I'm aiming for:
state
0 AL1 AL2 AL3 AL4 AL5 AL
答案1
得分: 0
你可以只是这样做，无需使用正则表达式：
df["State"] = df["full_address"].str.split(", ").str[-2]
或者这样：
df["full_address"].str.extract(r"(?<=, )(\w{2})(?=, )")
英文:
You could just do this, no need for regex:
df["State"] = df["full_address"].str.split(", ").str[-2]
Or this:
df["full_address"].str.extract(r"(?<=, )(\w{2})(?=, )")
答案2
得分: 0
鉴于它看起来像是完整地址的自由文本，我会避免尝试解析它，因为由于不规则的输入，肯定会出现一些错误。
我会使用类似于 https://pypi.org/project/uszipcode/ 的工具，并利用邮政编码字段来确定州的缩写。
英文:
given that it looks like the full address is free form text i would steer clear of attempting to parse it because some errors are sure to come up just because of irregular inputs.
i would use something like https://pypi.org/project/uszipcode/ and leverage the zipcode field to determine the state abbreviation instead
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