如何替换字符串的多个子串?

问题描述 投票:211回答:20

我想使用.replace函数来替换多个字符串。

我现在有

string.replace("condition1", "")

但是想要有类似的东西

string.replace("condition1", "").replace("condition2", "text")

虽然那感觉不是很好的语法

这样做的正确方法是什么?有点像在grep / regex中你可以做\1\2将字段替换为某些搜索字符串

python text replace
20个回答
222
投票

这是一个简短的例子,应该使用正则表达式:

import re

rep = {"condition1": "", "condition2": "text"} # define desired replacements here

# use these three lines to do the replacement
rep = dict((re.escape(k), v) for k, v in rep.iteritems()) 
#Python 3 renamed dict.iteritems to dict.items so use rep.items() for latest versions
pattern = re.compile("|".join(rep.keys()))
text = pattern.sub(lambda m: rep[re.escape(m.group(0))], text)

例如:

>>> pattern.sub(lambda m: rep[re.escape(m.group(0))], "(condition1) and --condition2--")
'() and --text--'

4
投票

我需要一个解决方案,其中要替换的字符串可以是正则表达式,例如,通过用单个字符替换多个空白字符来帮助规范化长文本。基于其他人的一系列答案,包括MiniQuark和mmj,这就是我想出的:

def multiple_replace(string, reps, re_flags = 0):
    """ Transforms string, replacing keys from re_str_dict with values.
    reps: dictionary, or list of key-value pairs (to enforce ordering;
          earlier items have higher priority).
          Keys are used as regular expressions.
    re_flags: interpretation of regular expressions, such as re.DOTALL
    """
    if isinstance(reps, dict):
        reps = reps.items()
    pattern = re.compile("|".join("(?P<_%d>%s)" % (i, re_str[0])
                                  for i, re_str in enumerate(reps)),
                         re_flags)
    return pattern.sub(lambda x: reps[int(x.lastgroup[1:])][1], string)

它适用于其他答案中给出的示例,例如:

>>> multiple_replace("(condition1) and --condition2--",
...                  {"condition1": "", "condition2": "text"})
'() and --text--'

>>> multiple_replace('hello, world', {'hello' : 'goodbye', 'world' : 'earth'})
'goodbye, earth'

>>> multiple_replace("Do you like cafe? No, I prefer tea.",
...                  {'cafe': 'tea', 'tea': 'cafe', 'like': 'prefer'})
'Do you prefer tea? No, I prefer cafe.'

对我来说最重要的是你也可以使用正则表达式,例如仅替换整个单词,或者规范化空格:

>>> s = "I don't want to change this name:\n  Philip II of Spain"
>>> re_str_dict = {r'\bI\b': 'You', r'[\n\t ]+': ' '}
>>> multiple_replace(s, re_str_dict)
"You don't want to change this name: Philip II of Spain"

如果要将字典键用作普通字符串,则可以在使用例如字符串调用multiple_replace之前对其进行转义。这个功能:

def escape_keys(d):
    """ transform dictionary d by applying re.escape to the keys """
    return dict((re.escape(k), v) for k, v in d.items())

>>> multiple_replace(s, escape_keys(re_str_dict))
"I don't want to change this name:\n  Philip II of Spain"

以下函数可以帮助您在字典键中找到错误的正则表达式(因为来自multiple_replace的错误消息不是很有说服力):

def check_re_list(re_list):
    """ Checks if each regular expression in list is well-formed. """
    for i, e in enumerate(re_list):
        try:
            re.compile(e)
        except (TypeError, re.error):
            print("Invalid regular expression string "
                  "at position {}: '{}'".format(i, e))

>>> check_re_list(re_str_dict.keys())

请注意,它不会链接替换,而是同时执行它们。这样可以在不限制其功能的情况下提高效率。要模仿链接的效果,您可能只需要添加更多字符串替换对并确保对的预期排序:

>>> multiple_replace("button", {"but": "mut", "mutton": "lamb"})
'mutton'
>>> multiple_replace("button", [("button", "lamb"),
...                             ("but", "mut"), ("mutton", "lamb")])
'lamb'

4
投票

启动Python 3.8,并引入assignment expressions (PEP 572):=运算符),我们可以在列表理解中应用替换:

# text = "The quick brown fox jumps over the lazy dog"
# replacements = [("brown", "red"), ("lazy", "quick")]
[text := text.replace(a, b) for a, b in replacements]
# text = 'The quick red fox jumps over the quick dog'

2
投票

我建议代码应该是,例如:

z = "My name is Ahmed, and I like coding "
print(z.replace(" Ahmed", " Dauda").replace(" like", " Love" ))

它将按要求打印出所有更改。


1
投票

你真的不应该这样做,但我觉得它太酷了:

>>> replacements = {'cond1':'text1', 'cond2':'text2'}
>>> cmd = 'answer = s'
>>> for k,v in replacements.iteritems():
>>>     cmd += ".replace(%s, %s)" %(k,v)
>>> exec(cmd)

现在,answer是所有替补的结果

再次,这是非常hacky,并不是你应该经常使用的东西。但是如果你需要的话,知道你可以做这样的事情真是太好了。


1
投票

这是一个在长字符串上更有效的示例,有许多小的替换。

source = "Here is foo, it does moo!"

replacements = {
    'is': 'was', # replace 'is' with 'was'
    'does': 'did',
    '!': '?'
}

def replace(source, replacements):
    finder = re.compile("|".join(re.escape(k) for k in replacements.keys())) # matches every string we want replaced
    result = []
    pos = 0
    while True:
        match = finder.search(source, pos)
        if match:
            # cut off the part up until match
            result.append(source[pos : match.start()])
            # cut off the matched part and replace it in place
            result.append(replacements[source[match.start() : match.end()]])
            pos = match.end()
        else:
            # the rest after the last match
            result.append(source[pos:])
            break
    return "".join(result)

print replace(source, replacements)

重点是避免很多长串的连接。我们将源字符串剪切为片段,在我们构成列表时替换一些片段,然后将整个事物连接回字符串。


1
投票

我不知道速度,但这是我的工作日快速修复:

reduce(lambda a, b: a.replace(*b)
    , [('o','W'), ('t','X')] #iterable of pairs: (oldval, newval)
    , 'tomato' #The string from which to replace values
    )

...但我喜欢上面的#1正则表达式答案。注意 - 如果一个新值是另一个的子字符串,则该操作不可交换。


0
投票

或者只是为了快速入侵:

for line in to_read:
    read_buffer = line              
    stripped_buffer1 = read_buffer.replace("term1", " ")
    stripped_buffer2 = stripped_buffer1.replace("term2", " ")
    write_to_file = to_write.write(stripped_buffer2)

0
投票

以下是使用字典执行此操作的另一种方法:

listA="The cat jumped over the house".split()
modify = {word:word for number,word in enumerate(listA)}
modify["cat"],modify["jumped"]="dog","walked"
print " ".join(modify[x] for x in listA)

0
投票

从Andrew的宝贵答案开始,我开发了一个脚本,该脚本从文件加载字典并详细说明打开的文件夹上的所有文件以进行替换。该脚本从外部文件加载映射,您可以在其中设置分隔符。我是初学者,但我发现这个脚本在多个文件中进行多次替换时非常有用。它在几秒钟内加载了一个包含1000多个条目的字典。它不优雅,但它对我有用

import glob
import re

mapfile = input("Enter map file name with extension eg. codifica.txt: ")
sep = input("Enter map file column separator eg. |: ")
mask = input("Enter search mask with extension eg. 2010*txt for all files to be processed: ")
suff = input("Enter suffix with extension eg. _NEW.txt for newly generated files: ")

rep = {} # creation of empy dictionary

with open(mapfile) as temprep: # loading of definitions in the dictionary using input file, separator is prompted
    for line in temprep:
        (key, val) = line.strip('\n').split(sep)
        rep[key] = val

for filename in glob.iglob(mask): # recursion on all the files with the mask prompted

    with open (filename, "r") as textfile: # load each file in the variable text
        text = textfile.read()

        # start replacement
        #rep = dict((re.escape(k), v) for k, v in rep.items()) commented to enable the use in the mapping of re reserved characters
        pattern = re.compile("|".join(rep.keys()))
        text = pattern.sub(lambda m: rep[m.group(0)], text)

        #write of te output files with the prompted suffice
        target = open(filename[:-4]+"_NEW.txt", "w")
        target.write(text)
        target.close()

0
投票

这是我解决问题的方法。我在聊天机器人中使用它来一次替换不同的单词。

def mass_replace(text, dct):
    new_string = ""
    old_string = text
    while len(old_string) > 0:
        s = ""
        sk = ""
        for k in dct.keys():
            if old_string.startswith(k):
                s = dct[k]
                sk = k
        if s:
            new_string+=s
            old_string = old_string[len(sk):]
        else:
            new_string+=old_string[0]
            old_string = old_string[1:]
    return new_string

print mass_replace("The dog hunts the cat", {"dog":"cat", "cat":"dog"})

这将成为The cat hunts the dog


104
投票

你可以做一个很好的小循环功能。

def replace_all(text, dic):
    for i, j in dic.iteritems():
        text = text.replace(i, j)
    return text

其中text是完整的字符串,dic是一个字典 - 每个定义都是一个字符串,它将替换该术语的匹配。

注意:在Python 3中,iteritems()已被items()取代


小心:Python字典没有可靠的迭代顺序。此解决方案仅解决您的问题,如果:

  • 替换顺序无关紧要
  • 替换可以改变以前替换的结果

例如:

d = { "cat": "dog", "dog": "pig"}
mySentence = "This is my cat and this is my dog."
replace_all(mySentence, d)
print(mySentence)

可能的输出#1:

"This is my pig and this is my pig."

可能的输出#2

"This is my dog and this is my pig."

一种可能的解决方法是使用OrderedDict。

from collections import OrderedDict
def replace_all(text, dic):
    for i, j in dic.items():
        text = text.replace(i, j)
    return text
od = OrderedDict([("cat", "dog"), ("dog", "pig")])
mySentence = "This is my cat and this is my dog."
replace_all(mySentence, od)
print(mySentence)

输出:

"This is my pig and this is my pig."

小心#2:如果你的text字符串太大或字典中有很多对,效率低下。


0
投票

另一个例子:输入列表

error_list = ['[br]', '[ex]', 'Something']
words = ['how', 'much[ex]', 'is[br]', 'the', 'fish[br]', 'noSomething', 'really']

期望的输出将是

words = ['how', 'much', 'is', 'the', 'fish', 'no', 'really']

代码:

[n[0][0] if len(n[0]) else n[1] for n in [[[w.replace(e,"") for e in error_list if e in w],w] for w in words]] 

82
投票

以下是使用reduce的第一个解决方案的变体,以备您正常使用。 :)

repls = {'hello' : 'goodbye', 'world' : 'earth'}
s = 'hello, world'
reduce(lambda a, kv: a.replace(*kv), repls.iteritems(), s)

martineau甚至更好的版本:

repls = ('hello', 'goodbye'), ('world', 'earth')
s = 'hello, world'
reduce(lambda a, kv: a.replace(*kv), repls, s)

67
投票

为什么不这样的解决方案呢?

s = "The quick brown fox jumps over the lazy dog"
for r in (("brown", "red"), ("lazy", "quick")):
    s = s.replace(*r)

#output will be:  The quick red fox jumps over the quick dog

30
投票

这只是对F.J和MiniQuark的一个更简洁的回顾。您需要实现多个同时串替换的功能如下:

def multiple_replace(string, rep_dict):
    pattern = re.compile("|".join([re.escape(k) for k in sorted(rep_dict,key=len,reverse=True)]), flags=re.DOTALL)
    return pattern.sub(lambda x: rep_dict[x.group(0)], string)

用法:

>>>multiple_replace("Do you like cafe? No, I prefer tea.", {'cafe':'tea', 'tea':'cafe', 'like':'prefer'})
'Do you prefer tea? No, I prefer cafe.'

如果您愿意,您可以从这个更简单的更换功能开始。


28
投票

我在F.J.s上建立了这个很好的答案:

import re

def multiple_replacer(*key_values):
    replace_dict = dict(key_values)
    replacement_function = lambda match: replace_dict[match.group(0)]
    pattern = re.compile("|".join([re.escape(k) for k, v in key_values]), re.M)
    return lambda string: pattern.sub(replacement_function, string)

def multiple_replace(string, *key_values):
    return multiple_replacer(*key_values)(string)

一次性用法:

>>> replacements = (u"café", u"tea"), (u"tea", u"café"), (u"like", u"love")
>>> print multiple_replace(u"Do you like café? No, I prefer tea.", *replacements)
Do you love tea? No, I prefer café.

请注意,由于替换仅在一次通过中完成,“café”更改为“tea”,但它不会更改回“café”。

如果您需要多次进行相同的更换,您可以轻松创建替换功能:

>>> my_escaper = multiple_replacer(('"','\\"'), ('\t', '\\t'))
>>> many_many_strings = (u'This text will be escaped by "my_escaper"',
                       u'Does this work?\tYes it does',
                       u'And can we span\nmultiple lines?\t"Yes\twe\tcan!"')
>>> for line in many_many_strings:
...     print my_escaper(line)
... 
This text will be escaped by \"my_escaper\"
Does this work?\tYes it does
And can we span
multiple lines?\t\"Yes\twe\tcan!\"

改进:

  • 将代码转换为函数
  • 增加了多线支持
  • 修复了转义中的错误
  • 易于为特定的多次替换创建功能

请享用! :-)


20
投票

我想提出字符串模板的用法。只需将要替换的字符串放在字典中即可完成所有设置!来自docs.python.org的例子

>>> from string import Template
>>> s = Template('$who likes $what')
>>> s.substitute(who='tim', what='kung pao')
'tim likes kung pao'
>>> d = dict(who='tim')
>>> Template('Give $who $100').substitute(d)
Traceback (most recent call last):
[...]
ValueError: Invalid placeholder in string: line 1, col 10
>>> Template('$who likes $what').substitute(d)
Traceback (most recent call last):
[...]
KeyError: 'what'
>>> Template('$who likes $what').safe_substitute(d)
'tim likes $what'

11
投票

在我的情况下,我需要用名称简单替换唯一键,所以我想到了这一点:

a = 'This is a test string.'
b = {'i': 'I', 's': 'S'}
for x,y in b.items():
    a = a.replace(x, y)
>>> a
'ThIS IS a teSt StrIng.'

8
投票

我的0.02美元。它基于Andrew Clark的答案,只是更清楚一点,它还涵盖了当要替换的字符串是要替换的另一个字符串的子字符串时(更长的字符串获胜)

def multireplace(string, replacements):
    """
    Given a string and a replacement map, it returns the replaced string.

    :param str string: string to execute replacements on
    :param dict replacements: replacement dictionary {value to find: value to replace}
    :rtype: str

    """
    # Place longer ones first to keep shorter substrings from matching
    # where the longer ones should take place
    # For instance given the replacements {'ab': 'AB', 'abc': 'ABC'} against 
    # the string 'hey abc', it should produce 'hey ABC' and not 'hey ABc'
    substrs = sorted(replacements, key=len, reverse=True)

    # Create a big OR regex that matches any of the substrings to replace
    regexp = re.compile('|'.join(map(re.escape, substrs)))

    # For each match, look up the new string in the replacements
    return regexp.sub(lambda match: replacements[match.group(0)], string)

在这个this gist,如果您有任何建议,请随时修改它。