Python字符串元音计数器

Question

我正在尝试创建一个程序来计算给定句子中元音的数量并返回最常见的元音和它（它们）出现的次数以及最不常见的元音（ s) 同时忽略那些根本不发生的。 这是我当前的代码

import collections, string

print("""This program will take a sentence input by the user and work out
the least common vowel in the sentence, vowels being A, E, I, O and U.
""")

sent = None

while sent == None or "":
    try:
        sent = input("Please enter a sentence: ").lower()
    except ValueError:
        print("That wasn't a valid string, please try again")
        continue

punct = str(set(string.punctuation))
words = sent.split()
words = [''.join(c for c in s if c not in string.punctuation) for s in words]

a = 0
e = 0
i = 0
o = 0
u = 0

for c in sent:
    if c is "a":
        a = a + 1
    if c is "e":
        e = e + 1
    if c is "i":
        i = i + 1
    if c is "o":
        o = o + 1
    if c is "u":
        u = u + 1

aeiou = {"a":a, "e":e, "i":i, "o":o, "u":u}
print("The most common occuring vowel(s) was: ", max(aeiou, key=aeiou.get))
print("The least common occuring vowel(s) was: ", min(aeiou, key=aeiou.get))

ender = input("Please press enter to end")

目前，它打印出出现次数最多和最少的元音，而不是全部，它也不会打印出现次数，也不会忽略那些根本不出现的元音。任何有关我将如何去做的帮助将不胜感激。

谢谢

Answer 1

collections.Counter 在这里会很棒！

vowels = set('aeiou') 
counter = collections.Counter(v for v in sentence.lower() if v in vowels)
ranking = counter.most_common()
print ranking[0]  # most common
print ranking[-1]  # least common

---

关于您的代码的一些注释。

• 不要使用a = a + 1，使用a += 1。
• 不要使用is比较字符串，使用==：if c == a: a += 1。

最后，要获得最大值，您需要整个项目，而不仅仅是价值。这意味着（不幸的是）您将需要一个比aeiou.get 更复杂的“关键”功能。

# This list comprehension filters out the non-seen vowels.
items = [item for item in aeiou.items() if item[1] > 0]

# Note that items looks something like this:  [('a', 3), ('b', 2), ...]
# so an item that gets passed to the key function looks like
# ('a', 3) or ('b', 2)
most_common = max(items, key=lambda item: item[1])
least_common = min(items, key=lambda item: item[1])

---

lambda 在您刚看到它的前几次可能会很棘手。请注意：

function = lambda x: expression_here

等同于：

def function(x):
    return expression_here

Answer 2

你可以使用字典：

>>> a = "hello how are you"
>>> vowel_count = { x:a.count(x) for x in 'aeiou' }
>>> vowel_count 
{'a': 1, 'i': 0, 'e': 2, 'u': 1, 'o': 3}
>>> keys = sorted(vowel_count,key=vowel_count.get)
>>> print "max -> " + keys[-1] + ": " + str(vowel_count[keys[-1]])
max -> o: 3
>>> print "min -> " + keys[0] + ": " + str(vowel_count[keys[0]])
min -> i: 0

count统计元素出现的次数

你也可以使用列表推导：

>>> vowel_count = [ [x,a.count(x)] for x in 'aeiou' ]
>>> vowel_count
[['a', 1], ['e', 2], ['i', 0], ['o', 3], ['u', 1]]
>>> sorted(vowel_count,key=lambda x:x[1])
[['i', 0], ['a', 1], ['u', 1], ['e', 2], ['o', 3]]

Answer 3

class VowelCounter:
    def __init__(self, wrd):
        self.word = wrd
        self.found = 0
        self.vowels = "aeiouAEIOU"
    def getNumberVowels(self):
        for i in range(0, len(self.word)):
            if self.word[i] in self.vowels:
                self.found += 1
        return self.found

    def __str__(self):
        return "There are " + str(self.getNumberVowels()) + " vowels in the String you entered."
def Vowelcounter():
    print("Welcome to Vowel Counter.")
    print("In this program you can count how many vowel there are in a String.")
    string = input("What is your String: \n")
    obj = VowelCounter(string)
    vowels = obj.getNumberVowels()
    if vowels > 1:
        print("There are " + str(vowels) + " vowels in the string you entered.")
    if vowels == 0:
        print("There are no vowels in the string you entered.")
    if vowels == 1:
        print("There is only 1 vowel in the string you entered.")
    recalc = input("Would you like to count how many vowels there are in a different string? \n")
    if recalc == "Yes" or recalc == "yes" or recalc == "YES" or recalc == "y" or recalc == "Y":
        Vowelcounter()
    else:
        print("Thank you")
Vowelcounter()

Answer 4

从我的角度来看，为了简单起见，我建议您按以下方式进行操作：

 def vowel_detector():
        vowels_n = 0
        index= 0
        text = str(input("Please enter any sentence: "))
        stop = len(text)
        while index != stop:
            for char in text:
                if char == "a" or char == "e" or char == "i" or char=="o" or char == "u":
                    vowels_n += 1
                    index += 1
                else: 
                    index +=1




          print("The number of vowels in your sentence is " + str(vowels_n))
        return

    vowel_detector()