计算每个元音出现的次数

Question

我编写了一个小程序来计算每个元音出现在列表中的次数，但它没有返回正确的计数，我不明白为什么：

vowels = ['a', 'e', 'i', 'o', 'u']
vowelCounts = [aCount, eCount, iCount, oCount, uCount] = (0,0,0,0,0)
wordlist = ['big', 'cats', 'like', 'really']

for word in wordlist:
    for letter in word:
        if letter == 'a':
            aCount += 1
        if letter == 'e':
            eCount += 1
        if letter == 'i':
            iCount += 1
        if letter == 'o':
            oCount += 1
        if letter == 'u':
            uCount += 1
for vowel, count in zip(vowels, vowelCounts):
    print('"{0}" occurs {1} times.'.format(vowel, count))

输出是

"a" occurs 0 times.
"e" occurs 0 times.
"i" occurs 0 times.
"o" occurs 0 times.
"u" occurs 0 times.

但是，如果我在 Python shell 中键入 aCount，它会得到 2，这是正确的，所以我的代码确实更新了 aCount 变量并正确存储了它。为什么不打印正确的输出？

Answer 1

问题出在这一行：

vowelCounts = [aCount, eCount, iCount, oCount, uCount] = (0,0,0,0,0)

如果您稍后开始增加 aCount，vowelCounts 不会得到更新。

设置a = [b, c] = (0, 0) 等效于a = (0, 0) 和[b, c] = (0, 0)。后者相当于设置b = 0和c = 0。

重新排序你的逻辑如下，它将起作用：

aCount, eCount, iCount, oCount, uCount = (0,0,0,0,0)

for word in wordlist:
    for letter in word:
        # logic 

vowelCounts = [aCount, eCount, iCount, oCount, uCount]

for vowel, count in zip(vowels, vowelCounts):
    print('"{0}" occurs {1} times.'.format(vowel, count))

Answer 2

您还可以使用集合计数器（这是计算事物时自然而然的首选函数，它返回一个字典）：

from collections import Counter

vowels = list('aeiou')
wordlist = ['big', 'cats', 'like', 'really']

lettersum = Counter(''.join(wordlist))

print('\n'.join(['"{}" occurs {} time(s).'.format(i,lettersum.get(i,0)) for i in vowels]))

返回：

"a" occurs 2 time(s).
"e" occurs 2 time(s).
"i" occurs 2 time(s).
"o" occurs 0 time(s).
"u" occurs 0 time(s).

字母：

Counter({'l': 3, 'a': 2, 'e': 2, 'i': 2, 'c': 1, 'b': 1, 
         'g': 1, 'k': 1, 's': 1, 'r': 1, 't': 1, 'y': 1})

Answer 3

您可以使用字典理解：

vowels = ['a', 'e', 'i', 'o', 'u']
wordlist = ['big', 'cats', 'like', 'really']
new_words = ''.join(wordlist)
new_counts = {i:sum(i == a for a in new_words) for i in vowels}

输出：

{'a': 2, 'e': 2, 'i': 2, 'o': 0, 'u': 0}

Answer 4

除了@jpp 所说的 像整数这样的简单数据类型是按值而不是按引用返回的 因此，当您将其分配给某事物并更改该事物时，它不会受到影响

a = 10
b = a #b=a=10
b = 11 #b=11, a=10
print a, b
--> 10 11

我会对此发表评论，但我需要声誉才能做到这一点：D