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
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
我需要一个解决方案,其中字符串可以被替换为正则表达式, 例如,通过将多个空格字符替换为一个空格字符来帮助规范化长文本。根据其他人(包括MiniQuark和mmj)的一系列答案,我得出了以下结论:
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例子,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,例如只替换整个单词,或规范化空白:
>>> 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"
如果你想使用字典键作为普通字符串, 你可以在调用multiple_replace之前转义这些,例如使用下面的函数:
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"
下面的函数可以帮助在你的字典键中找到错误的正则表达式(因为来自multiple_replace的错误消息不是很明显):
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,它没有链接替换,而是同时执行它们。这使得它更有效率,而不会限制它能做什么。为了模仿链接的效果,你可能只需要添加更多的字符串替换对,并确保这些对的预期顺序:
>>> multiple_replace("button", {"but": "mut", "mutton": "lamb"})
'mutton'
>>> multiple_replace("button", [("button", "lamb"),
... ("but", "mut"), ("mutton", "lamb")])
'lamb'
其他回答
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)
我不知道速度如何,但这是我日常的快速解决方法:
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值的子字符串,那么该操作是不可交换的。
在我的情况下,我需要一个简单的唯一键替换名称,所以我想到了这个:
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.'
我的方法是首先将字符串标记化,然后决定每个标记是否包含它。
潜在地,如果我们可以假设一个hashmap/set的O(1)查找,可能会更好:
remove_words = {"we", "this"}
target_sent = "we should modify this string"
target_sent_words = target_sent.split()
filtered_sent = " ".join(list(filter(lambda word: word not in remove_words, target_sent_words)))
Filtered_sent现在是'应该修改字符串'
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()