Question

s = str(input("Please enter your line of text: ").lower())
only_vowels = re.sub(r"[^aeiou]", "", s)
c = (Counter(list(only_vowels)))  
print(c)

if len(c) >= 1:
    most = c.most_common()[-1]
    result = (most[0])

    i = -2
    if len(c)>=2:
        while (c.most_common()[i][1]) == most[1]:
            result = ", ".join((result, c.most_common()[i][0]))
            i = i-1


    print("The least common vowel(s) in the inserted sentence is/are",      (result),"and it/they appear(s)",most[1],"times.")

else:
    print("You have not inserted any vowels into this sentence.")

代码跳出这段代码'而（c.most_common（）[i] [1]）==大多数[1]：'

任何想法为什么？

Answer 1

您在列表中有多少个条目？如果i = -2那么你至少需要有2个条目，以便Python可以向后遍历列表。

Answer 2

它打印“计数器”，因为c是Counter类型的对象。

计数器有一种方法可以获得最常见的事件，但它似乎没有一个用于最不常见的事件。怎么样：

from collections import defaultdict

occurrences = defaultdict(list)
vowels = "aeiou"

sentence = input()

for v in vowels:
    if v in sentence:
        occurrences[sentence.count(v)].append(v)

try:
    less_frequent_vowels = occurrences[min(occurrences)]
except ValueError:
    less_frequent_vowels = []

虽然循环列表索引超出范围 - 没有意义？

2 个答案: