我想使用.replace函数替换多个字符串。

我目前有

string.replace("condition1", "")

但想要一些像

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

尽管这样的语法感觉不太好

正确的做法是什么?有点像在grep/regex中,你可以用\1和\2来替换某些搜索字符串的字段


当前回答

这是我对这个问题的解决办法。我把它用在聊天机器人上,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"})

这就成了猫捉狗

其他回答

注意:测试你的案例,见注释。

这里有一个例子,它在长弦上更有效,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)

关键是要避免长字符串的多次连接。我们将源字符串切成片段,在我们形成列表时替换一些片段,然后将整个字符串连接回字符串。

sentence='its some sentence with a something text'

def replaceAll(f,Array1,Array2):
    if len(Array1)==len(Array2):
        for x in range(len(Array1)):
            return f.replace(Array1[x],Array2[x])

newSentence=replaceAll(sentence,['a','sentence','something'],['another','sentence','something something'])

print(newSentence)

从安德鲁的宝贵答案开始,我开发了一个脚本,从一个文件加载字典,并详细说明所有文件上打开的文件夹做替换。脚本从一个外部文件加载映射,您可以在该文件中设置分隔符。我是一个初学者,但我发现这个脚本在多个文件中做多个替换时非常有用。它在几秒钟内加载了一个包含1000多个条目的字典。这并不优雅,但对我来说很管用

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()

我把这句话建立在fj的精彩回答上:

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)

一针用法:

>>> 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.

注意,由于替换只在一次传递中完成,“café”会变成“tea”,但不会变回“café”。

如果你需要做相同的替换多次,你可以很容易地创建一个替换函数:

>>> 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 修正了逃跑的错误 容易创建一个函数,用于特定的多个替换

享受吧!: -)

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]]