这是我的任务
journey = """Just a small tone girl
Leaving in a lonely whirl
She took the midnight tray going anywhere
Just a seedy boy
Bored and raised in South Detroit or something
He took the midnight tray going anywhere"""
太棒了好的,因此,对于本练习,您的工作是使用Python的字符串替换方法来修复此字符串并将新版本输出到控制台。
这就是我所做的
journey = """ just a small tone girl
Leaving in a lonely whirl
she took a midnight tray going anywhere
Just a seedy boy
bored and raised in south detroit or something
He took the midnight tray going anywhere"""
journeyEdit = journey.replace("tone" ,
"town").replace("tray","train").replace("seedy","city").replace("Leaving",
"living").replace("bored","born").replace("whirl","world").replace("or
something", " ")
print (journeyEdit)
答案 0 :(得分:4)
这里是从文本替换单词的示例方法。您可以使用 python re 软件包。
请找到以下代码作为指导。
import re
journey = """ just a small tone girl Leaving in a lonely whirl she took a
midnight tray going anywhere Just a seedy boy bored and raised in south
detroit or something He took the midnight tray going anywhere"""
# define desired replacements here
journeydict = {"tone" : "town",
"tray":"train",
"seedy":"city",
"Leaving": "living",
"bored":"born",
"whirl":"world"
}
# use these given three lines to do the replacement
rep = dict((re.escape(k), v) for k, v in journeydict.items())
#Python 3 renamed dict.iteritems to dict.items so use rep.items() for latest
versions
pattern = re.compile("|".join(journeydict.keys()))
text = pattern.sub(lambda m: journeydict[re.escape(m.group(0))], journey)
print(journey)
print(text)
答案 1 :(得分:2)
可能比您指定的;-)更长。
如How to replace multiple substrings of a string?所示:
import re
journey = """ just a small tone girl Leaving in a lonely whirl she took a
midnight tray going anywhere Just a seedy boy bored and raised in south
detroit or something He took the midnight tray going anywhere"""
rep = {"tone": "town",
"tray": "train",
"seedy":"city",
"Leaving": "living",
"bored":"born",
"whirl":"world",
"or something": " "}
# 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()))
journeyEdit = pattern.sub(lambda m: rep[re.escape(m.group(0))], journey)
print(journeyEdit)