统计文字中的单词数量并按出现次数排序(要求：尽量使用正则表达式)
本帖最后由 欧德奈瑞 于 2020-5-21 19:19 编辑现在需要统计若干段文字(英文)中的单词数量，并且还需统计每个单词出现的次数。
注1：单词之间以空格(1个或多个空格)为间隔。
注2：忽略空行或者空格行。
基本版:
统计时，区分字母大小写，且不删除指定标点符号。
进阶版:
1、统计前，需要从文字中删除指定标点符号!.,:*?。 注意：所谓的删除，就是用1个空格替换掉相应字符。
2、统计单词时需要忽略单词的大小写。
输入说明
若干行英文，最后以!!!!!为结束。
输出说明
单词数量
出现次数排名前10的单词（次数按照降序排序，如果次数相同，则按照键值的字母升序排序）及出现次数。
输入样例1
failure is probably the fortification in your pole
it is like a peek your wallet as the thief when you
are thinking how to spend several hard-won lepta
when you are wondering whether new money it has laid
background because of you then at the heart of the
most lax alert and most low awareness and left it
godsend failed
!!!!!
输出样例1
46
the=4
it=3
you=3
and=2
are=2
is=2
most=2
of=2
when=2
your=2
输入样例2
Failure is probably The fortification in your pole!
It is like a peek your wallet as the thief when You
are thinking how to. spend several hard-won lepta.
when yoU are? wondering whether new money it has laid
background Because of: yOu?, then at the heart of the
Tom say: Who is the best? No one dare to say yes.
most lax alert and! most low awareness and* left it
godsend failed
!!!!!
输出样例2
54
the=5
is=3
it=3
you=3
and=2
are=2
most=2
of=2
say=2
to=2 基本版import re
import collections
text = '''failure is probably the fortification in your pole
it is like a peek your wallet as the thief when you
are thinking how to spend several hard-won lepta
when you are wondering whether new money it has laid
background because of you then at the heart of the
most lax alert and most low awareness and left it
godsend failed
!!!!!'''
text=re.sub(r'[^\w\s]','',text)
frequency = collections.Counter(text.split())
l=sorted(list(frequency.items()),key=lambda x:(-x,x))[:10]
print(len(frequency))
for i in l:
print(i,'=',i) 进阶版import re
import collections
text = '''Failure is probably The fortification in your pole!
It is like a peek your wallet as the thief when You
are thinking how to. spend several hard-won lepta.
when yoU are? wondering whether new money it has laid
background Because of: yOu?, then at the heart of the
Tom say: Who is the best? No one dare to say yes.
most lax alert and! most low awareness and* left it
godsend failed
!!!!!'''
text=re.sub(r'[^\w\s]','',text)
print(text)
frequency = collections.Counter(map(str.lower,text.split()))
l=sorted(list(frequency.items()),key=lambda x:(-x,x))[:10]
print(len(frequency))
for i in l:
print(i,'=',i) ouyunfu 发表于 2020-5-21 18:49
进阶版
大佬知道如何接收该题目这样长的输入样例吗？而且您代码的输出等号左右两边带空格，和输出样例不符 欧德奈瑞 发表于 2020-5-21 18:52
大佬知道如何接收该题目这样长的输入样例吗？而且您代码的输出等号左右两边带空格，和输出样例不符
已经按你要求更改
import re
import collections
text = input()
text=re.sub(r'[^\w\s]','',text)
print(text)
frequency = collections.Counter(map(str.lower,text.split()))
l=sorted(list(frequency.items()),key=lambda x:(-x,x))[:10]
print(len(frequency))
for i in l:
print('%s=%d'%(i,i)) ouyunfu 发表于 2020-5-21 18:49
进阶版
虽然大佬的代码还是没法帮我过这道题，但是还是感谢大佬的思路指引。所幸我已经用循环过了这道题了。所以就把大佬的这段代码设为最佳答案吧 ouyunfu 发表于 2020-5-21 19:05
已经按你要求更改
对了，大佬下次接收这种特殊结尾的样本输入的时候可以考虑使用代码text = ''
while True:
strs = input()
if strs != '!!!!!':
text += '\n'
text += strs
else:
break
哦
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