信不信由你，所有字符都已经隐式分配了一个数字:它们的 ASCII 字符代码.您可以使用 ord() 函数访问它们，或者直接比较它们:

>>>一个">乙"错误的>>>b">一种"真的

但请注意，大写字母编码为 65 - 90，而小写字母编码为 97 - 122，因此:

>>>C">乙"错误的

您必须确保比较所有大写字母或所有小写字母.

这是一个可能的函数，它使用上述信息来检查给定的字符串是否按字母顺序排列，只是为了让您开始:

def isAlphabetical(word):对于 xrange(len(word) - 1) 中的 i:如果单词[i] >字[i+1]:返回错误返回真

I have a text file full of words. I need to go through the text file and count the number of words that are spelled in alphabetical order. I am struggling to think of way to figure out if a word is in alphabetical order.

I have already looked around and have seen that it can be done with sorted. However, we haven't yet learned about sort in class so I can't use it. Help/tips much appreciated.

Do I need to do something like assigning each letter of the alphabet a number?

Believe it or not, all characters are already implicitly assigned a number: their ASCII character codes. You can access them by using the ord() function, or compare them directly:

>>> "a" > "b"
False

>>> "b" > "a"
True

Beware though, capital letters are coded 65 - 90, while lowercase letters are coded 97 - 122, so:

>>> "C" > "b"
False

You have to ensure that you are comparing all uppercase or all lowercase letters.

Here's one possible function that uses the above information to check if a given string is in alphabetical order, just to get you started:

def isAlphabetical(word):
    for i in xrange(len(word) - 1):
        if word[i] > word[i+1]:
            return False
    return True