如何在一个单词中找到加倍？

Question

我的任务是检查双倍的单词列表并将结果输出到字典中。

首先，我试图执行此代码：

for word in wordsList:
    for letter in word:
        if word[letter] == word[letter+1]:

但只要我弄错了，我就改变了一下：

wordsList = ["creativity", "anna", "civic", "apology", "refer", "mistress", "rotor", "mindset"]
dictionary = {}

for word in wordsList:
    for letter in range(len(word)-1):
        if word[letter] == word[letter+1]:
            dictionary[word] = ("This word has a doubling")
        else:
            dictionary[word] = ("This word has no doubling")
print(dictionary)

现在它有效，但不是很好。我真的需要建议！提前致谢

我期待输出{创造力：'这个词没有加倍'}，{anna：'这个病房加倍'}等。

Answer 1

wordsList = ["creativity", "anna", "civic", "apology", "refer", "mistress", "rotor", "mindset"]
dictionary = {}

for word in wordsList:
    for index, char in enumerate(word[:-1]):    # Take one less to prevent out of bounds
        if word[index + 1] == char:             # Check char with next char
            dictionary[word] = True
            break                               # The one you forgot, with the break you wouldnt override the state
    else:
        dictionary[word] = False

print(dictionary)

你忘记了休息声明;循环将继续并覆盖您找到的结论。我自己实现了你的casus。

Answer 2

wordsList = ["creativity", "anna", "civic", "apology", "refer", "mistress", "rotor", "mindset"]
dictionaries = []
for word in wordsList: #for example:creativity
    alphabets=[]
    for x in word:
        alphabets+=x #now alphabets==['c','r','e','a','t','i','v','i','t','y']
    num=0
    while True:
        x=alphabets[0]  #for example 'c'
        alphabets.remove(x)  #now alphabets==['r','e','a','t','i','v','i','t','y']
        if x in alphabets:  # if alphabets contain 'c'
            doubling=True   # it is doubling
            break                
        else:
            doubling=False  #continue checking
            if len(alphabets)==1:  #if there is only one left
                break              #there is no doubling
    if doubling:
        dictionaries.append({word:"This word has a doubling"})
    else:
        dictionaries.append({word:"This word has no doubling"})
print(dictionaries)