在字符串中分离城市和州?
一个字符串,citystate,可能是这样格式的:
"Greenwich, CT"
"Greenwich,CT"
"Greenwich CT"
我该如何把这个字符串 citystate 拆分成两个独立的字符串,分别是城市和州(用 Python)呢?
(这样 city 就是 Greenwich,state 就是 CT)
谢谢大家的帮助!
相关问题:
4 个回答
0
def cityState(s):
# split on last comma
res = s.rsplit(',', 1)
if len(res)==2:
return res[0].strip(), res[1].strip()
# otherwise split on last non-trailing whitespace
res = s.rsplit(None, 1)
if len(res)==2:
return res[0].lstrip(), res[1]
# otherwise split failed
raise ValueError("Could not split {0} into city, state".format(s))
for city_state in ["Greenwich, CT", "Greenwich,CT", "Greenwich CT", "New York, NY"]:
print "City '{0}' is in '{1}'".format(*cityState(city_state))
City 'Greenwich' is in 'CT'
City 'Greenwich' is in 'CT'
City 'Greenwich' is in 'CT'
City 'New York' is in 'NY'
结果是
1
做一些假设:
(1) 一个州的代码由两个字符组成,你可以通过查找州代码表来验证分割的结果。
(2) 你的输入数据已经转换成了 unicode(Python 2.X)或 str(Python 3.x)。
(3) 你的数据可能经历了很多不同的输入错误,或者是一些初学者的错误。
代码:
tests = [
u"Foo YZ", u"Foo YZ\n", u"Foo\xa0YZ", u"Foo YZ",
u"Foo, YZ", u"Foo, YZ\n", u"Foo,\xa0YZ",
u"Foo,YZ", u"Foo,YZ\n", u"Foo,,YZ",
u"FooYZ", u" Foo \t,\xa0 YZ \n",
]
import re
def machin(strg):
strg = strg.strip()
state = strg[-2:]
city = u" ".join(strg[:-2].split()) # normalise whitespace
city = city.rstrip(", ")
return city, state
def kettler(strg):
split_loc = re.split(r' ?,? ?', strg)
city = ' '.join(split_loc[:-1]) # Everything up to last element joined by space
state = split_loc[-1]
return city, state
def bothwell(strg):
# split on last comma
res = strg.rsplit(',', 1)
if len(res)==2:
return res[0].strip(), res[1].strip()
# otherwise split on last non-trailing whitespace
res = strg.rsplit(None, 1)
if len(res)==2:
return res[0].lstrip(), res[1]
# otherwise split failed
# raise ValueError("Could not split {0} into city, state".format(s))
return None, None
def paulger_v2(strg):
return strg[:-3].strip(", "), strg[-2:]
funcs = (kettler, bothwell, paulger_v2, machin)
for func in funcs:
print "\n===", func.__name__, "==="
nok = 0
for test in tests:
city, state = func(test)
ok = city == u"Foo" and state == u"YZ"; nok += ok
print [ok, test, city, state]
test = test.replace(u"Foo", u"Foo Baa")
city, state = func(test)
ok = city == u"Foo Baa" and state == u"YZ"; nok += ok
print [ok, test, city, state]
print "\n+++ %s: %d OK +++" % (func.__name__, nok)
输出:
=== kettler ===
[True, u'Foo YZ', u'Foo', u'YZ']
[True, u'Foo Baa YZ', u'Foo Baa', u'YZ']
[False, u'Foo YZ\n', u'Foo', u'YZ\n']
[False, u'Foo Baa YZ\n', u'Foo Baa', u'YZ\n']
[False, u'Foo\xa0YZ', '', u'Foo\xa0YZ']
[False, u'Foo Baa\xa0YZ', u'Foo', u'Baa\xa0YZ']
[True, u'Foo YZ', u'Foo', u'YZ']
[True, u'Foo Baa YZ', u'Foo Baa', u'YZ']
[True, u'Foo, YZ', u'Foo', u'YZ']
[True, u'Foo Baa, YZ', u'Foo Baa', u'YZ']
[False, u'Foo, YZ\n', u'Foo', u'YZ\n']
[False, u'Foo Baa, YZ\n', u'Foo Baa', u'YZ\n']
[False, u'Foo,\xa0YZ', u'Foo', u'\xa0YZ']
[False, u'Foo Baa,\xa0YZ', u'Foo Baa', u'\xa0YZ']
[True, u'Foo,YZ', u'Foo', u'YZ']
[True, u'Foo Baa,YZ', u'Foo Baa', u'YZ']
[False, u'Foo,YZ\n', u'Foo', u'YZ\n']
[False, u'Foo Baa,YZ\n', u'Foo Baa', u'YZ\n']
[False, u'Foo,,YZ', u'Foo ', u'YZ']
[False, u'Foo Baa,,YZ', u'Foo Baa ', u'YZ']
[False, u'FooYZ', '', u'FooYZ']
[False, u'Foo BaaYZ', u'Foo', u'BaaYZ']
[False, u' Foo \t,\xa0 YZ \n', u' Foo \t \xa0 YZ', u'\n']
[False, u' Foo Baa \t,\xa0 YZ \n', u' Foo Baa \t \xa0 YZ', u'\n']
+++ kettler: 8 OK +++
=== bothwell ===
[True, u'Foo YZ', u'Foo', u'YZ']
[False, u'Foo Baa YZ', u'Foo Baa', u'YZ']
[True, u'Foo YZ\n', u'Foo', u'YZ']
[False, u'Foo Baa YZ\n', u'Foo Baa', u'YZ']
[True, u'Foo\xa0YZ', u'Foo', u'YZ']
[False, u'Foo Baa\xa0YZ', u'Foo Baa', u'YZ']
[True, u'Foo YZ', u'Foo', u'YZ']
[False, u'Foo Baa YZ', u'Foo Baa', u'YZ']
[True, u'Foo, YZ', u'Foo', u'YZ']
[False, u'Foo Baa, YZ', u'Foo Baa', u'YZ']
[True, u'Foo, YZ\n', u'Foo', u'YZ']
[False, u'Foo Baa, YZ\n', u'Foo Baa', u'YZ']
[True, u'Foo,\xa0YZ', u'Foo', u'YZ']
[False, u'Foo Baa,\xa0YZ', u'Foo Baa', u'YZ']
[True, u'Foo,YZ', u'Foo', u'YZ']
[False, u'Foo Baa,YZ', u'Foo Baa', u'YZ']
[True, u'Foo,YZ\n', u'Foo', u'YZ']
[False, u'Foo Baa,YZ\n', u'Foo Baa', u'YZ']
[False, u'Foo,,YZ', u'Foo,', u'YZ']
[False, u'Foo Baa,,YZ', u'Foo Baa,', u'YZ']
[False, u'FooYZ', None, None]
[False, u'Foo BaaYZ', u'Foo', u'BaaYZ']
[True, u' Foo \t,\xa0 YZ \n', u'Foo', u'YZ']
[False, u' Foo Baa \t,\xa0 YZ \n', u'Foo Baa', u'YZ']
+++ bothwell: 10 OK +++
=== paulger_v2 ===
[True, u'Foo YZ', u'Foo', u'YZ']
[False, u'Foo Baa YZ', u'Foo Baa', u'YZ']
[False, u'Foo YZ\n', u'Foo', u'Z\n']
[False, u'Foo Baa YZ\n', u'Foo Baa', u'Z\n']
[True, u'Foo\xa0YZ', u'Foo', u'YZ']
[False, u'Foo Baa\xa0YZ', u'Foo Baa', u'YZ']
[True, u'Foo YZ', u'Foo', u'YZ']
[False, u'Foo Baa YZ', u'Foo Baa', u'YZ']
[True, u'Foo, YZ', u'Foo', u'YZ']
[False, u'Foo Baa, YZ', u'Foo Baa', u'YZ']
[False, u'Foo, YZ\n', u'Foo', u'Z\n']
[False, u'Foo Baa, YZ\n', u'Foo Baa', u'Z\n']
[True, u'Foo,\xa0YZ', u'Foo', u'YZ']
[False, u'Foo Baa,\xa0YZ', u'Foo Baa', u'YZ']
[True, u'Foo,YZ', u'Foo', u'YZ']
[False, u'Foo Baa,YZ', u'Foo Baa', u'YZ']
[False, u'Foo,YZ\n', u'Foo', u'Z\n']
[False, u'Foo Baa,YZ\n', u'Foo Baa', u'Z\n']
[True, u'Foo,,YZ', u'Foo', u'YZ']
[False, u'Foo Baa,,YZ', u'Foo Baa', u'YZ']
[False, u'FooYZ', u'Fo', u'YZ']
[False, u'Foo BaaYZ', u'Foo Ba', u'YZ']
[False, u' Foo \t,\xa0 YZ \n', u'Foo \t,\xa0 Y', u' \n']
[False, u' Foo Baa \t,\xa0 YZ \n', u'Foo Baa \t,\xa0 Y', u' \n']
+++ paulger_v2: 7 OK +++
=== machin ===
[True, u'Foo YZ', u'Foo', u'YZ']
[True, u'Foo Baa YZ', u'Foo Baa', u'YZ']
[True, u'Foo YZ\n', u'Foo', u'YZ']
[True, u'Foo Baa YZ\n', u'Foo Baa', u'YZ']
[True, u'Foo\xa0YZ', u'Foo', u'YZ']
[True, u'Foo Baa\xa0YZ', u'Foo Baa', u'YZ']
[True, u'Foo YZ', u'Foo', u'YZ']
[True, u'Foo Baa YZ', u'Foo Baa', u'YZ']
[True, u'Foo, YZ', u'Foo', u'YZ']
[True, u'Foo Baa, YZ', u'Foo Baa', u'YZ']
[True, u'Foo, YZ\n', u'Foo', u'YZ']
[True, u'Foo Baa, YZ\n', u'Foo Baa', u'YZ']
[True, u'Foo,\xa0YZ', u'Foo', u'YZ']
[True, u'Foo Baa,\xa0YZ', u'Foo Baa', u'YZ']
[True, u'Foo,YZ', u'Foo', u'YZ']
[True, u'Foo Baa,YZ', u'Foo Baa', u'YZ']
[True, u'Foo,YZ\n', u'Foo', u'YZ']
[True, u'Foo Baa,YZ\n', u'Foo Baa', u'YZ']
[True, u'Foo,,YZ', u'Foo', u'YZ']
[True, u'Foo Baa,,YZ', u'Foo Baa', u'YZ']
[True, u'FooYZ', u'Foo', u'YZ']
[True, u'Foo BaaYZ', u'Foo Baa', u'YZ']
[True, u' Foo \t,\xa0 YZ \n', u'Foo', u'YZ']
[True, u' Foo Baa \t,\xa0 YZ \n', u'Foo Baa', u'YZ']
+++ machin: 24 OK +++
4
import re
city, state = re.split(r' ?,? ?', location)
要解决纽约这个问题,可以使用切片的方法:
split_loc = re.split(r' ?,? ?', location)
city = ' '.join(split_loc[:-1]) # Everything up to last element joined by space
state = split_loc[-1]