元音字母出现次数最多的单词的词频统计

Question

问题

问题是这样的，一个词频统计的练习

相关代码

def get_txt():
    txt = open('sentence.txt','rt').read()
    txt = txt.lower()
    return txt
s_txt = get_txt()
wordset = s_txt.split()
yalp = ['a','e','i','o','u']
counts = {}
for word in wordset:
    for alp in word:
        if alp in yalp:
            counts[word] =counts.get(word,0) + 1
        items = list(counts.items())
        items.sort(key = lambda x:x[1],reverse = False)
        for i in range(5):
            word ,counts = items[i]
            print('{0:<10}{1:>5}'.format(word ,counts))

运行结果及报错内容

Traceback (most recent call last):
    word ,counts = items[i]
IndexError: list index out of range
only          1

我想要达到的结果

运行结果却报错了，上网搜了一圈仍然没有解决，可以帮忙看看吗
另外
1.对于下面这行有更优化的方法吗（感觉很蠢）

for word in wordset:
    for alp in word:
        if alp in yalp:

2.这行是抄的，对于为什么lambda要这么用不是很理解

 items.sort(key = lambda x:x[1],reverse = False)

求解答！

Answer 1

要把items排序的部分移到循环外面，不然字典都还没做好呢。而且题目要逆序输出，sort里面的reverse应该为True

wordset = s_txt.split()
yalp = ['a','e','i','o','u']
counts = {}
for word in wordset:
    for alp in word:
        if alp in yalp:
            counts[word] =counts.get(word,0) + 1
items = list(counts.items())
items.sort(key = lambda x:x[1],reverse = True)
for i in range(5):
    word ,counts = items[i]
    print('{0:<10}{1:>5}'.format(word ,counts))

lambda就是为了省事，没什么特别的，不然key关键字后面要跟一个函数，你也可以自定义成这样：

def fun(x):
  return (x[1])
items.sort(key = fun,reverse = True)

看格式也能知道lambda是怎么写的了吧。
至于你说的优化代码部分，可以考虑把所有元音字母转换成数字1，然后统计每个单词里的1的数量

yalp = {ord('a'):ord('1'),ord('e'):ord('1'),ord('i'):ord('1'),ord('o'):ord('1'),ord('u'):ord('1')}
counts = {}
for word in wordset:
    a = word.translate(yalp)
    counts[word] =a.count('1')