Python：创建单词分析

Question

我正在尝试创建一个分析用户输入单词的工具。它说明了字母的数量（完成），元音的数量（需要帮助），大写字母的数量（已完成）以及最常见的字母（尚未担心）。我也做了下面的代码：

word = raw_input("Please enter a word:") 
print word 


if len(word) == 1: 
    print word + " has " + str(len(word)) + " letter." 
else: 
    print word + " has " + str(len(word)) + " letters." 


if sum(1 for v in word if v ==["a", "e", "i", "o", "u", "A", "E", "I", "O", "U"]) == 1: 
    print "It also has ", sum(1 for v in word if v == ["a", "e", "i", "o", "u", "A", "E", "I", "O", "U"]), " vowel." 
else: 
    print "It also has ", sum(1 for v in word if v == ["a", "e", "i", "o", "u", "A", "E", "I", "O", "U"]), " vowels." 


if sum(1 for c in word if c.isupper()) == 1: 
    print "It has ", sum(1 for c in word if c.isupper()), " capital letter." 
else: 
    print "It has ", sum(1 for c in word if c.isupper()), " capital letters." 

使用例如字“你好”，它返回如下：

HeLLo 
HeLLo has 5 letters. 
It also has 0 vowels. 
It has 3 capital letters. 

我很困惑，因为它知道算元音的数量，包括它在答案中，却没有统计单词中的元音。

我需要做一个小调整还是一个大的改变？

谢谢。

P.S.我如何将问题标记为已回答？

Answer 1

您正在比较该信与列表。相反，请检查该信是否为in该列表。

sum(1 for v in word if v in ["a", more vowels, "U"]) 

此外，您可以通过使用一个字符串，而不是一个列表，并通过套管低信为先，并没有重复自己尽可能多的让你的代码稍短。

num_vowels = sum(1 for v in word if v.lower() in "aeiou") 
print "It also has ", num_vowels, (" vowel." if num_vowels == 1 else " vowels.") 

---

为了找到最常见的字母，你应该使用的字典。只要迭代单词中的字母并增加它们在字典中的数量，然后选择数量最高的字母。

counts = {} 
for x in word: 
    counts[x] = counts.get(x, 0) + 1 
print "most common:", max(word, key=lambda x: counts[x]) 

或者只是使用collections.Counter(word).most_common(1)