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
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)
其他回答
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()
这是我的0.02美元。它基于Andrew Clark的答案,只是更清楚一点,它还涵盖了当一个字符串被替换为另一个字符串的子字符串时的情况(更长的字符串胜出)
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)
这就是这个要点,如果你有任何建议,请随意修改。
或者简单说一下:
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)
你真的不应该这样做,但我觉得这太酷了:
>>> replacements = {'cond1':'text1', 'cond2':'text2'}
>>> cmd = 'answer = s'
>>> for k,v in replacements.iteritems():
>>> cmd += ".replace(%s, %s)" %(k,v)
>>> exec(cmd)
现在,答案是所有替换的结果
再说一次,这是非常俗气的,不是你应该经常使用的东西。但我很高兴知道如果你需要的话,你可以这样做。
下面是另一种使用字典的方法:
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)