我写了一个小程序来计算每个元音在列表中出现的次数,但是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
,它会给我2
,这是正确的,因此我的代码确实更新了aCount变量并正确地存储了它。为什么不打印正确的输出? 最佳答案
问题在于这一行:
vowelCounts = [aCount, eCount, iCount, oCount, uCount] = (0,0,0,0,0)
如果以后开始递增,则不会更新
vowelCounts
。设置
aCount
相当于a = [b, c] = (0, 0)
和a = (0, 0)
。后者相当于设置[b, c] = (0, 0)
和b = 0
。将逻辑重新排序如下,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))