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
下面是一个简短的例子,应该做的技巧与正则表达式:
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--'
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。
对于只替换一个字符,使用翻译和str.maketrans是我最喜欢的方法。
Tl;dr > result_string = your_string.translate(str.maketrans(dict_mapping))
demo
my_string = 'This is a test string.'
dict_mapping = {'i': 's', 's': 'S'}
result_good = my_string.translate(str.maketrans(dict_mapping))
result_bad = my_string
for x, y in dict_mapping.items():
result_bad = result_bad.replace(x, y)
print(result_good) # ThsS sS a teSt Strsng.
print(result_bad) # ThSS SS a teSt StrSng.
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
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()
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)