Question

我必须在地址栏中用N S替换北，南等。

如果我有

list = {'NORTH':'N','SOUTH':'S','EAST':'E','WEST':'W'}
address = "123 north anywhere street"

我可以迭代我的字典值来替换我的地址字段吗？

for dir in list[]:
   address.upper().replace(key,value)

我知道我甚至不是很亲密！但是，如果您可以使用这样的字典值，那么任何输入都会受到赞赏。

Answer 1

address = "123 north anywhere street"

for word, initial in {"NORTH":"N", "SOUTH":"S" }.items():
    address = address.replace(word.lower(), initial)
print address

美观，简洁，可读。

Answer 2

你很亲密，实际上：

dictionary = {"NORTH":"N", "SOUTH":"S" } 
for key in dictionary.iterkeys():
    address.upper().replace(key, dictionary[key])

注意：对于Python 3用户，您应该使用.keys()代替.iterkeys()：

dictionary = {"NORTH":"N", "SOUTH":"S" } 
for key in dictionary.keys():
    address.upper().replace(key, dictionary[key])

Answer 3

我认为没有人建议的一个选项是构建一个包含所有键的正则表达式，然后只需对字符串进行一次替换：

>>> import re
>>> l = {'NORTH':'N','SOUTH':'S','EAST':'E','WEST':'W'}
>>> pattern = '|'.join(sorted(re.escape(k) for k in l))
>>> address = "123 north anywhere street"
>>> re.sub(pattern, lambda m: l.get(m.group(0).upper()), address, flags=re.IGNORECASE)
'123 N anywhere street'
>>>

这样做的好处是，正则表达式可以忽略输入字符串的情况而不修改它。

如果您只想对完整的单词进行操作，那么您也可以通过简单修改模式来实现：

>>> pattern = r'\b({})\b'.format('|'.join(sorted(re.escape(k) for k in l)))
>>> address2 = "123 north anywhere southstreet"
>>> re.sub(pattern, lambda m: l.get(m.group(0).upper()), address2, flags=re.IGNORECASE)
'123 N anywhere southstreet'

Answer 4

＆＃34;翻译＆＃34;带字典的字符串是一个非常常见的要求。我提出了一个您可能希望保留在工具箱中的函数：

def translate(text, conversion_dict, before=None):
    """
    Translate words from a text using a conversion dictionary

    Arguments:
        text: the text to be translated
        conversion_dict: the conversion dictionary
        before: a function to transform the input
        (by default it will to a lowercase)
    """
    # if empty:
    if not text: return text
    # preliminary transformation:
    before = before or str.lower
    t = before(text)
    for key, value in conversion_dict.items():
        t = t.replace(key, value)
    return t

然后你可以写：

>>> a = {'hello':'bonjour', 'world':'tout-le-monde'}
>>> translate('hello world', a)
'bonjour tout-le-monde'

Answer 5

您可能正在寻找iteritems()：

d = {'NORTH':'N','SOUTH':'S','EAST':'E','WEST':'W'}
address = "123 north anywhere street"

for k,v in d.iteritems():
    address = address.upper().replace(k, v)

地址现在是'123 N ANYWHERE STREET'

好吧，如果你想保留大小写，空格和嵌套的单词（例如Southstreet不应该转换为Sstreet），请考虑使用这个简单的列表理解：

import re

l = {'NORTH':'N','SOUTH':'S','EAST':'E','WEST':'W'}

address = "North 123 East Anywhere Southstreet    West"

new_address = ''.join(l[p.upper()] if p.upper() in l else p for p in re.split(r'(\W+)', address))

new_address现在是

N 123 E Anywhere Southstreet    W

Answer 6

所有这些答案都很好，但你缺少python字符串替换 - 它简单快捷，但要求你的字符串格式正确。

address = "123 %(direction)s anywhere street"
print(address % {"direction": "N"})

Answer 7

def replace_values_in_string(text, args_dict):
    for key in args_dict.keys():
        text = text.replace(key, str(args_dict[key]))
    return text

Answer 8

尝试，

import re
l = {'NORTH':'N','SOUTH':'S','EAST':'E','WEST':'W'}

address = "123 north anywhere street"

for k, v in l.iteritems():
    t = re.compile(re.escape(k), re.IGNORECASE)
    address = t.sub(v, address)
print(address)

Answer 9

使用replace()和format()都不那么精确：

data =  '{content} {address}'
for k,v in {"{content}":"some {address}", "{address}":"New York" }.items():
    data = data.replace(k,v)
# results: some New York New York

'{ {content} {address}'.format(**{'content':'str1', 'address':'str2'})
# results: ValueError: unexpected '{' in field name

如果您需要精确的地点，最好使用re.sub()进行翻译：

import re
def translate(text, kw, ignore_case=False):
    search_keys = map(lambda x:re.escape(x), kw.keys())
    if ignore_case:
        kw = {k.lower():kw[k] for k in kw}
        regex = re.compile('|'.join(search_keys), re.IGNORECASE)
        res = regex.sub( lambda m:kw[m.group().lower()], text)
    else:
        regex = re.compile('|'.join(search_keys))
        res = regex.sub( lambda m:kw[m.group()], text)

    return res

#'score: 99.5% name:%(name)s' %{'name':'foo'}
res = translate( 'score: 99.5% name:{name}', {'{name}':'foo'})
print(res)

res = translate( 'score: 99.5% name:{NAME}', {'{name}':'foo'}, ignore_case=True)
print(res)

Answer 10

我建议使用正则表达式而不是简单的替换。替换后，您就有替换单词的子部分的风险，这也许不是您想要的。

import json
import re

with open('filePath.txt') as f:
   data = f.read()

with open('filePath.json') as f:
   glossar = json.load(f)

for word, initial in glossar.items():
   data = re.sub(r'\b' + word + r'\b', initial, data)

print(data)

你能用Python中的字典值编写str.replace（）吗？

10 个答案: