Question

我设法编写了这个代码：

def vowels(s) :
    result = 0
    n = 0
    while n<len(s):
        if s[n] in "AEIOUaeiou" : result = result+1
        n = n+1
    return result

line = input("Type a line of text: ")
print("The line contains", vowels(line), "vowels")

这给了我总共有多少个元音符号.. 但我想知道如何更改它以便输出最常发生的元音和发生的次数

Answer 1

您可以使用collections.Counter来获取文本中每个元音的出现次数。

>>> from collections import Counter

>>> def vowels(s):
...     return Counter(c for c in s if c in "AEIOUaeiou")

>>> counter = vowels("Lorem ipsum lorem")
>>> print counter
Counter({'e': 2, 'o': 2, 'i': 1, 'u': 1})
>>> print sum(counter.values())
6
>>> print counter.most_common()
[('e', 2), ('o', 2), ('i', 1), ('u', 1)]

Answer 2

使用collections.Counter及其most_common方法

from collections import Counter

def vowels(s) :
    vow_found = [i for i in s.lower() if i in 'aeiou']
    c = Counter(vow_found)
    return len(vow_found), c.most_common()

line = input("Type a line of text: ")
numvow, most_com = vowels(line)
print("The line contains", numvow, "vowels")
print("and the most common are", most_com)

，输入

hello I am your friend today

产生

The line contains 10 vowels
and the most common are [('o', 3), ('i', 2), ('a', 2), ('e', 2), ('u', 1)]

Answer 3

您可以使用collections.Counter执行此操作：

import collections

vowels = lambda s: collections.Counter(i for i in s if i in 'AEIOUaeiou').most_common(1)[0]

line = input("Type a line of text: ")
v = vowels(line)

print("The most frequently vowel is", v[0], 'the times is', v[1])

Answer 4

我的做法是什么：

使用lower()将输入设为小写。
计算输入和商店中每个元音的频率。
对列表进行反向排序，以便最大频率出现。
选择排序列表的第一个元素。

希望这会有所帮助：

line = input("Type a line of text: ").lower()
print(sorted([(i,line.count(i)) for i in "aeiou"], key=lambda x:x[1], reverse=True)[0])

Answer 5

def vowels(s) :
    vowel_letters = "AEIOUaeiou"
    result = []
    for vowel in vowel_letters:
        result.append((vowel,s.count(vowel)))
    result.sort(key=lambda tup: tup[1])
    return result[-1]

line = raw_input("Type a line of text: ")
print vowels(line)

Answer 6

相同

的更简单的简单代码

from collections import Counter
def most_common_vowel(input_string):
    return Counter(
            filter(lambda x: x in "aeiou", input_string.lower())
        ).most_common(1)

使用filter函数从字符串中删除非元音，并使用Counter获取最常用的元素

如何输出最常见的元音

6 个答案: