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
您可以使用pandas库和replace函数,它既支持精确匹配,也支持正则表达式替换。例如:
df = pd.DataFrame({'text': ['Billy is going to visit Rome in November', 'I was born in 10/10/2010', 'I will be there at 20:00']})
to_replace=['Billy','Rome','January|February|March|April|May|June|July|August|September|October|November|December', '\d{2}:\d{2}', '\d{2}/\d{2}/\d{4}']
replace_with=['name','city','month','time', 'date']
print(df.text.replace(to_replace, replace_with, regex=True))
修改后的文本为:
0 name is going to visit city in month
1 I was born in date
2 I will be there at time
你可以在这里找到一个例子。请注意,文本上的替换是按照它们在列表中出现的顺序进行的
其他回答
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--'
为什么没有这样的解决方案呢?
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
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'
您可以使用pandas库和replace函数,它既支持精确匹配,也支持正则表达式替换。例如:
df = pd.DataFrame({'text': ['Billy is going to visit Rome in November', 'I was born in 10/10/2010', 'I will be there at 20:00']})
to_replace=['Billy','Rome','January|February|March|April|May|June|July|August|September|October|November|December', '\d{2}:\d{2}', '\d{2}/\d{2}/\d{4}']
replace_with=['name','city','month','time', 'date']
print(df.text.replace(to_replace, replace_with, regex=True))
修改后的文本为:
0 name is going to visit city in month
1 I was born in date
2 I will be there at time
你可以在这里找到一个例子。请注意,文本上的替换是按照它们在列表中出现的顺序进行的