Python words.txt文件中有多少个元音按顺序排列？

函数

使用

将始终返回True。如果我做对了，就只有12个字了

如何使

uses

返回False

def count_vowels():
    fin = open ('words.txt')
    count = 0
    vowels = ['aeiou']
    for line in fin:
        word = line.strip()
        if uses(word, vowels):
            count = count + 1
            print "There are", count ," words with a vowel."

def uses(word, vowels):
    for vowels in word:
        if vowels in vowels:
            return True
    return False

def count_words():                                
    fin = open ('words.txt')
    vowels = ['aeiou']
    for line in fin:
            word = line.strip()
            if uses(word, vowels):
                print word

count_vowels()
count_words()

#This is what returns in the shell when I run my code:

>>>There are 1 words with a vowel.
There are 2 words with a vowel.
There are 3 words with a vowel.
There are 4 words with a vowel.
There are 5 words with a vowel.
....
#and so on...

# However, if I use the following code to replace the middle function in my original code:

def uses(word, vowels):
    found = ''
    for l in word:
        if l not in found:
            if l in vowels:
                found += l
        else:
            return False
    return found == vowels

# When I run the code the shell returns this:

>>>
>>>

---

如@void所说，使用

re

模块

import re
s = 'hello world'
words = re.findall(r'\w+', s)
print 'words'''

---

您只需使用

all

功能即可！让魔法发挥作用：

def uses(word, vowels):
    return all(letter in word for letter in vowels)

演示：

这样，您就可以将

元音作为
“aiueo”
字符串传递，而不是
['aiueo']

但是，如果您希望按顺序找到元音，请使用以下命令：

def uses(word, vowels):
    found = ''
    for l in word:
        if l not in found:
            if l in vowels:
                found += l
        else:
            return False
    return found == vowels

这里，有点令人失望，但我只给你完整的固定代码：

def count_vowels():               
    with open('words.txt') as fin:
        count = 0
        vowels = 'aeiou'
        for line in fin:
            words = line.split()
            for word in words:
                if uses(word, vowels):
                    count = count + 1
                    print "There are", count ," words with a vowel."

def uses(word, vowels):
    return all(letter in word for letter in vowels)

def count_words():                                
    with open('words.txt') as fin:
        vowels = 'aeiou'
        for line in fin:
                words = line.split()
                for word in words:
                    if uses(word, vowels):
                        print word

count_vowels()
count_words()

如果您只查找所有字母都按该顺序排列的单词（但可能中间有其他字母），则需要检查第一个元音，直到找到为止，然后检查第二个元音，然后是第三个元音，直到找到所有字母（或到达单词末尾）。下面是一些代码：

def uses(word, vowels)
    vowel_index = 0
    for letter in word:
        if letter == vowels[vowel_index]:
            vowel_index += 1
            if vowel_index == len(vowels):
                return True
    return False

我没有删除发现的元音，而是保留一个整数索引，告诉我当前测试的元音

您需要使用
元音的可下标值调用此函数，可以是一个包含一个字符串作为其内容的列表，也可以是一个包含所有元音的字符串。
如果我做对了，我将实现一个
uses
-函数，如下所示：

def uses(word, vowels):
    # remove every letter, that's not a vowel, make the word lower case
    reduced_word = filter( lambda character: character in vowels, word.lower() )
    # check, whether the vowels are in order
    return vowels in reduced_word

vowels = "aeiou"

test_words = [ "atetitotu", "Atetitotu", "Haemodiutory", "FgeAihou" ]

for word in test_words: print word, uses(word, vowels)

输出：

atetitotu True
Atetitotu True
Haemodiutory False
FgeAihou False

但是，如果您想查找文件中的所有这些单词，我建议您使用
re
-module:

import re
file_content = open('file.txt').read()
vowels = "aeiou"
# the pattern starts with word boundary and there can be any kind of letters at the beginning
pattern = r"\b\w*"
# than between every vowel there can be any letter, except of a vowel
pattern += (r"[^%s]*?" % vowels ).join(vowels)
# and then, after the last vowel there can be any letter before the word ends
pattern += r"\w*\b"
# here the result: pattern = r"\b\w*a[^aeiou]*?e[^aeiou]*?i[^aeiou]*?o[^aeiou]*?u\w*\b"
# now we look for this pattern all over the file, ignoring case
re.findall(pattern, file_content, flags = re.I)

输出是这样的

['atetitotu', 'Atetitotu']

目前的结果是什么？也，你能给我们一个输入样本吗？
'atetitotu'
是否按顺序计算？为什么你把元音都列为一个字符串？这可能是因为你正在从
元音中删除字母，但从来没有把它们放回下一个单词中…一个单词中元音的顺序必须是
aeiou
？元音
aeiou
是否有序？不一定？你什么意思？啊，我明白了。。等一下，OP还没有澄清。我想我有点匆忙：）它应该返回12次：“有”，计数，“带元音的单词。”
['atetitotu', 'Atetitotu']

def uses_only(wrd_trnsv, str_chars):
    for chr_trnsv in wrd_trnsv:
        if chr_trnsv not in str_chars:
            return False
    return True

wrd_list = open('C:\\Temp\\Complete_Words_list.txt')
def uses_all_txt(wrd_list, req_str_chrs):
    count = 0
    for each_wrd in wrd_list :
        if uses_only(req_str_chrs,each_wrd):
            count = count + 1
     return count