我有一個數據框，如下所示： col 1 col 20 59 538 Walton Avenue, Chester, FY6 7NP1 62 42 Chesterton Road, Peterborough, FR7 2NY2 179 3 Wallbridge Street, Essex, 4HG 3HT3 180 6 Stevenage Avenue, Coventry, 7PY 9NP列表類似于：[Stevenage, Essex, Coventry, Chester]按照此處的解決方案：How to check if Pandas rows contain any full string or substring of a list? 是這樣的：city_list = list(cities["name"])df["col3"] = np.where(df["col2"].str.contains('|'.join(city_list)), df["col2"], '')我發現 col 2 中的一些與列表中的字符串匹配，但 col3 與 col2 相同。我希望 col3 成為列表中的值，而不是與 col3 相同。這將是： col 1 col 2 col30 59 538 Walton Avenue, Chester, FY6 7NP Chester 1 62 42 Chesterton Road, Peterborough, FR7 2NY 2 179 3 Wallbridge Street, Essex, 4HG 3HT Essex3 180 6 Stevenage Avenue, Coventry, 7PY 9NP Coventry我試過了：pat = "|".join(cities.name)df.insert(0, "name", df["col2"].str.extract('(' + pat + ')', expand = False))但這返回了一個錯誤，說在期望 1 時有 456 個輸入。還：df["col2"] = df["col2"].apply(lambda x: difflib.get_close_matches(x, cities["name"])[0])df.merge(cities)但這返回時錯誤列表索引超出范圍。有沒有辦法做到這一點？df1 大約有 160,000 個條目，col2 中的每個地址來自不同的國家，因此沒有標準的書寫方式，而城市列表大約有 170,000 個條目
4 回答
嗶嗶one
TA貢獻1854條經驗 獲得超8個贊
你可以這樣做：
city_list = ["Stevenage", "Essex", "Coventry", "Chester"]
def get_match(row):
col_2 = row["col 2"].replace(",", " ").split() # Here you should process the string as you want
for c in city_list:
if difflib.get_close_matches(col_2, c)
return c
return ""
df["col 3"] = df.apply(lambda row: get_match(row), axis=1)
慕的地10843
TA貢獻1785條經驗 獲得超8個贊
查看str.contains測試模式是否匹配系列的函數：
df = pd.DataFrame([[59, '538 Walton Avenue, Chester,', 'FY6 7NP'],
[62, '42 Chesterton Road, Peterborough', '4HG 3HT'],
[179, '3 Wallbridge Street, Essex', '4HG 3HT'],
[180, '6 Stevenage Avenue, Coventry', '7PY 9NP']])
city_list = ["Stevenage", "Essex", "Coventry", "Chester"]
for city in city_list:
df.loc[df[1].str.contains(city), 'match'] = city
慕慕森
TA貢獻1856條經驗 獲得超17個贊
試試這個
def aux_func(address):
aux_list = ['Stevenage', 'Essex', 'Coventry', 'Chester']
# remove commas
address = address.split(',')
# avoide matches with the first part of the address
if len(address)>1:
# remove the first element of the address
address = address[1:]
for v in aux_list:
for chunk in address:
if v in chunk:
return v
return ""
df['col 3'] = [aux_func(address) for address in df['col 2']]
波斯汪
TA貢獻1811條經驗 獲得超4個贊
依靠這樣的輔助功能：
df = pd.DataFrame({'col 1': [59, 62, 179, 180],
'col 2': ['538 Walton Avenue, Chester, FY6 7NP',
'42 Chesterton Road, Peterborough, FR7 2NY',
'3 Wallbridge Street, Essex, 4HG 3HT',
'6 Stevenage Avenue, Coventry, 7PY 9NP'
]})
def aux_func(x):
# split by comma and select the interesting part ([1])
x = x.split(',')
x = x[1]
aux_list = ['Stevenage', 'Essex', 'Coventry', 'Chester']
for v in aux_list:
if v in x:
return v
return ""
df['col 3'] = [aux_func(name) for name in df['col 2']]
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