使用直方图函数检查列表中的重复字母

Question

我正在尝试完成我的作业，但在整合所需的直方图函数时遇到了困难。

这是我必须使用的代码：

alphabet = "abcdefghijklmnopqrstuvwxyz"
test_dups = ["zzz","dog","bookkeeper","subdermatoglyphic","subdermatoglyphics"]
test_miss = ["zzz","subdermatoglyphic","the quick brown fox jumps over the lazy dog"]

def histogram(s):
     d = dict()
     for c in s:
          if c not in d:
               d[c] = 1
          else:
               d[c] += 1
     return d

我需要编写一个名为has_duplicates() 的函数，它接受一个字符串参数，如果字符串有任何重复字符，则返回True。否则，它应该返回False。

通过使用上面的histogram() 函数创建直方图来实现has_duplicates()。不要使用教科书中给出的has_duplicates() 的任何实现。相反，您的实现应该使用直方图中的计数来确定是否有任何重复。

在提供的test_dups 列表中的字符串上编写一个循环。根据该字符串的 has_duplicates() 的返回值打印列表中的每个字符串以及它是否有任何重复项。例如，aaa 和 abc 的输出将如下所示。

aaa has duplicates
abc has no duplicates

为 test_dups 中的每个字符串打印一条类似上述的行。

编写一个名为missing_letters 的函数，它接受一个字符串参数并返回一个新字符串，其中包含不在参数字符串中的所有字母。返回字符串中的字母应按字母顺序排列。

我的实现应该使用来自histogram() 函数的直方图。它还应该使用全局变量alphabet。它应该直接使用这个全局变量，而不是通过参数或本地副本。它应该遍历alphabet 中的字母以确定输入参数中缺少哪些字母。

函数missing_letters 应该将丢失的字母列表组合成一个字符串并返回该字符串。

在列表 test_miss 中的字符串上编写一个循环，并为每个字符串调用 missing_letters。为列出缺失字母的每个字符串打印一行。例如，对于字符串“aaa”，输出应如下所示。

aaa 缺少字母 bcdefghijklmnopqrstuvwxyz

如果字符串包含alphabet 中的所有字母，则输出应显示它使用了所有字母。例如，字符串 alphabet 本身的输出将如下所示。

"abcdefghijklmnopqrstuvwxyz uses all the letters"

为 test_miss 中的每个字符串打印一条类似上述的行。

据我所知……

def has_duplicates(t):
    if histogram(t) > 1:
        return True
    else:
        return False

结果：

'>' not supported between instances of 'str' and 'int'

Answer 1

以下内容应提供所需的结果：

alphabet = "abcdefghijklmnopqrstuvwxyz"
test_dups = ["zzz","dog","bookkeeper","subdermatoglyphic","subdermatoglyphics"]
test_miss = ["zzz","subdermatoglyphic","the quick brown fox jumps over the lazy dog"]

def histogram(s):
     d = dict()
     for c in s:
          if c not in d:
               d[c] = 1
          else:
               d[c] += 1
     return d

def has_duplicates(s):
    # Return False if each letter in s is not distinct
    return len(histogram(s)) != len(s)

def missing_letters(s):
    h = histogram(s)
    rv = ''
    # Loop over letters in alphabet, if the letter is not in the histogram then
    # append to the return string.
    for c in alphabet:
        if c not in h:
            rv = rv + c
    return rv

# Loop over test strings as required.
for s in test_miss:
    miss = missing_letters(s)
    if miss:
        print(f"{s} is missing letters {miss}.")
    else:
        print(f"{s} uses all the letters.")

输出：

zzz is missing letters abcdefghijklmnopqrstuvwxy.
subdermatoglyphic is missing letters fjknqvwxz.
the quick brown fox jumps over the lazy dog uses all the letters.

Answer 2

alphabet = "abcdefghijklmnopqrstuvwxyz"
test_dups = ["zzz", "dog", "bookkeeper", "subdermatoglyphic", "subdermatoglyphics"]
test_miss = ["zzz", "subdermatoglyphic", "the quick brown fox jumps over the lazy dog"]


def histogram(string):
    d = dict()
    for char in string:
        if char not in d:
            d[char] = 1
        else:
            d[char] += 1
    return d


# Part 1
def has_duplicate(string):
    h = histogram(string)
    for k, v in h.items():
        if v > 1:
            return True
    return False


for string in test_dups:
    if has_duplicate(string):
        print(string, "has duplicates")
    else:
        print(string, "has no duplicates")

# Part 2


def missing_letters(string):
    h = histogram(string)
    new_list = []
    for char in alphabet:
        if char not in h:
            new_list.append(char)
    return "".join(new_list)


for string in test_miss:
    new_list = missing_letters(string)
    if len(new_list):
        print(string, "is missing letters", new_list)
    else:
        print(string, "uses all letters")