试图计算字符串中的单词

时间:2013-07-06 23:04:58

标签: python string list function loops

我正在尝试分析字符串的内容。如果在单词中混合了标点符号,我想用空格替换它们。

例如,如果Johnny.Appleseed!是:a * good& farmer作为输入输入,则应该说有6个单词,但我的代码只将其视为0个单词。我不确定如何删除不正确的字符。

仅供参考:我正在使用python 3,我也无法导入任何库

string = input("type something")
stringss = string.split()

    for c in range(len(stringss)):
        for d in stringss[c]:
            if(stringss[c][d].isalnum != True):
                #something that removes stringss[c][d]
                total+=1
print("words: "+ str(total))

7 个答案:

答案 0 :(得分:15)

基于简单循环的解决方案:

strs = "Johnny.Appleseed!is:a*good&farmer"
lis = []
for c in strs:
    if c.isalnum() or c.isspace():
        lis.append(c)
    else:
        lis.append(' ')

new_strs = "".join(lis)
print new_strs           #print 'Johnny Appleseed is a good farmer'
new_strs.split()         #prints ['Johnny', 'Appleseed', 'is', 'a', 'good', 'farmer']

更好的解决方案:

使用regex

>>> import re
>>> from string import punctuation
>>> strs = "Johnny.Appleseed!is:a*good&farmer"
>>> r = re.compile(r'[{}]'.format(punctuation))
>>> new_strs = r.sub(' ',strs)
>>> len(new_strs.split())
6
#using `re.split`:
>>> strs = "Johnny.Appleseed!is:a*good&farmer"
>>> re.split(r'[^0-9A-Za-z]+',strs)
['Johnny', 'Appleseed', 'is', 'a', 'good', 'farmer']

答案 1 :(得分:11)

这是一个不需要导入任何库的单行解决方案 它用空格替换非字母数字字符(如标点符号),然后split为字符串。

灵感来自“Python strings split with multiple separators

>>> s = 'Johnny.Appleseed!is:a*good&farmer'
>>> words = ''.join(c if c.isalnum() else ' ' for c in s).split()
>>> words
['Johnny', 'Appleseed', 'is', 'a', 'good', 'farmer']
>>> len(words)
6

答案 2 :(得分:3)

试试这个:它使用re解析word_list,然后创建word:appearances

字典
import re
word_list = re.findall(r"[\w']+", string)
print {word:word_list.count(word) for word in word_list}

答案 3 :(得分:2)

如何使用收藏品中的Counter?

import re
from collections import Counter

words = re.findall(r'\w+', string)
print (Counter(words))

答案 4 :(得分:1)

for ltr in ('!', '.', ...) # insert rest of punctuation
     stringss = strings.replace(ltr, ' ')
return len(stringss.split(' '))

答案 5 :(得分:1)

我知道这是一个古老的问题但......这个怎么样?

string = "If Johnny.Appleseed!is:a*good&farmer"

a = ["*",":",".","!",",","&"," "]
new_string = ""

for i in string:
   if i not in a:
      new_string += i
   else:
      new_string = new_string  + " "

print(len(new_string.split(" ")))

答案 6 :(得分:0)

#Write a python script to count words in a given string.
 s=str(input("Enter a string: "))
 words=s.split()
 count=0
  for word in words:
      count+=1

  print(f"total number of words in the string is : {count}")