Question

你好，我有下面的代码，但是它没有按预期工作，我看不到我在哪里出错了，我正在使用python 3 missing_letters方法无法正常工作，因为第二个输出错误，我需要它接受输入并使用字母字符串作为比较来返回缺少的字母

alphabet = "abcdefghijklmnopqrstuvwxyz"   

test_miss = ["zzz","subdermatoglyphic","the quick brown fox jumps over the 

def missing_letters(string):
    compare = histogram(string) 
    for key in sorted(compare): 
        if key in alphabet: 
            result = alphabet.replace(key,'') 
            return result # print result
        else:
            return None

for letters in test_miss: 
    if missing_letters(letters) == None:
        print(letters, 'uses all the letters')   
    else:
        print(letters , 'is missing letters', missing_letters(letters))

# output: zzz is missing letters abcdefghijklmnopqrstuvwxy                                                                               
# subdermatoglyphic is missing letters bcdefghijklmnopqrstuvwxyz                                                                 
# the quick brown fox jumps over the lazy dog uses all the letters

Answer 1

您不需要计算字母的数量，可以直接使用一组字母（输入string中每个字母仅出现一次）。一旦有了该集合，就可以从字母表中减去该集合，最后得到一个包含缺少字母的集合：

from string import ascii_lowercase

alphabet = set(ascii_lowercase)

def missing_letters(string):
    return alphabet - set(string.lower())

print(missing_letters("abcdefghijklmnopqrstuv"))

将输出{'w', 'y', 'x', 'z'}。请注意，ascii_lowercase只是一个包含所有小写字母的字符串。

Answer 2

主要问题是您return 在内部，即在检查了第一个字母之后。另外，似乎您正在迭代单词中的字母并检查它们是否在字母中，而不是反过来，这会产生更多感觉。另外，似乎不必创建字母频率的直方图。

您可以尝试以下操作：

def missing_letters(string):
    missing = []
    for key in alphabet:
        if key not in string:
            missing.append(key)
    return ''.join(missing) or None

或者您也可以将字母和字符串都转换为set，并使用设置差异-：

def missing_letters(string):
    return ''.join(sorted(set(alphabet) - set(string))) or None

Answer 3

使用集进行会员资格测试：首先将alphabet设置为一个集

In [10]: alphabet = "abcdefghijklmnopqrstuvwxyz"

In [11]: alphabet = set(alphabet)

对于字符串“ ab”，“直方图返回”

In [12]: compare = {'a':1,'b':1}

使用设置方法确定差异

In [13]: alphabet.difference(compare.keys())
Out[13]: 
{'c',
 'd',
 'e',
 'f',
 'g',
 'h',
 'i',
 'j',
 'k',
 'l',
 'm',
 'n',
 'o',
 'p',
 'q',
 'r',
 's',
 't',
 'u',
 'v',
 'w',
 'x',
 'y',
 'z'}

Answer 4

您的方法'def missing_letters（string）：'是错误的。我已经更正了，下面的代码/方法将起作用，请尝试

alphabet = "abcdefghijklmnopqrstuvwxyz"   

test_miss = ["zzz","subdermatoglyphic","the quick brown fox jumps over the lazy dog"]     

def missing_letters(string):
    result = alphabet
    compare = set(string)# use histogram to make a dictionary
    compare=sorted(compare)
    for key in compare: # iterate and sort in alphabetical order
        if key in alphabet: # check if the key in alphabet
            result = result.replace(key,'')# remove it from alphabet
    return result

# Test the function          
print(missing_letters("ab")) # output: bcdefghijklmnopqrstuvwxyz

##### test_miss for loop #####

for letters in test_miss: 
    if len( missing_letters(letters) ) == 0:
        print(letters, 'uses all the letters')   
    else:
        print(letters , 'is missing letters', missing_letters(letters))

# output: zzz is missing letters abcdefghijklmnopqrstuvwxy                                                                               
# subdermatoglyphic is missing letters bcdefghijklmnopqrstuvwxyz                                                                 
# the quick brown fox jumps over the lazy dog uses all the letters

python缺少字母方法

4 个答案: