Question

我写了一个小程序来计算每个元音在列表中出现的次数，但它没有返回正确的计数，我看不出原因：

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

"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，它会给我2，这是正确的，所以我的代码确实更新了aCount变量并正确存储它。为什么不打印正确的输出？

Answer 1

问题在于这一行：

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

如果您稍后开始递增vowelCounts，则

aCount不会更新。

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

重新排列您的逻辑，如下所示：

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

你也可以使用集合计数器（在计算事物时它是自然的首选函数，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).

lettersum：

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

您可以使用词典理解：

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通过引用返回的所以当你把它分配给某个东西并改变它不受影响的东西时

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

我已经对他做了评论，但我需要声誉：D

计算每个元音出现的次数

4 个答案: