while循环表示returns每个字符出现的次数

Question

查看下面的 class 我需要编写代码来定义“字符频率”，因此它 returns 类似于下面的示例。

class EnhancedString:
    def __init__(self, contents):
        self._contents = str(contents)
    def __str__(self):
        return str(self._contents)
    def __len__(self):
        return len(self._contents)
    def set_contents(self, contents):
        self._contents = str(contents)
    def get_contents(self):
        return self._contents
    **def character_frequency(self):**
 
"""
        Returns a dictionary containing all characters found in _contents as keys, with integer values corresponding to the number of occurrences of each character. Use a while loop to traverse _contents.
        :return: A dictionary with characters as keys and integers as values
        Example use:
            >>> c = EnhancedString("")
            >>> print(c.character_frequency())
            {}
            >>> a = EnhancedString("abcdefg")
            >>> print(a.character_frequency())
            {'a': 1, 'b': 1, 'c': 1, 'd': 1, 'e': 1, 'f': 1, 'g': 1} >>> b = EnhancedString("aabcdefgaa")
            >>> print(b.character_frequency())
            {'a': 4, 'b': 1, 'c': 1, 'd': 1, 'e': 1, 'f': 1, 'g': 1} """

Answer 1

collections.Counter 将计算集合中的可哈希项目，在您的情况下，是字符串中的单个字符。

import collections

class EnhancedString:
    def __init__(self, contents):
        self._contents = str(contents)
    def __str__(self):
        return str(self._contents)
    def __len__(self):
        return len(self._contents)
    def set_contents(self, contents):
        self._contents = str(contents)
    def get_contents(self):
        return self._contents
    def character_frequency(self):
        return collections.Counter(self._contents)

s = EnhancedString("I am the very model of a modern major general")
print(s.character_frequency())

如果您想在没有任何额外模块的情况下执行此操作，您可以创建自己的字典并添加到其中。

    def character_frequency(self):
        counts = {}
        for c in self._contents:
            if c not in counts:
                counts[c] = 1
            else:
                counts[c] += 1
        return counts

或者使用异常块

    def character_frequency(self):
        counts = {}
        for c in self._contents:
            try:
                counts[c] += 1
            except KeyError:
                counts[c] = 1
        return counts

“while”循环更混乱。当您可以使用“for”进行迭代时，无需为集合建立索引。

    def character_frequency(self):
        counts = {}
        i = 0
        end = len(self._contents)
        while i < end:
            c = self._contents[i]
            try:
                counts[c] += 1
            except KeyError:
                counts[c] = 1
            i += 1
        return counts

Answer 2

这个工具不适合这项工作，但如果您需要使用 while 语句来完成它并且您希望您的代码有点 Pythonic，您可以使用迭代器：

content   = "hello world"
counts    = dict.fromkeys(content,0)  # initialize all character counts to zero
iContent  = iter(content)             # iterator on characters
while True:                           # dummy loop, why not at this point ;)
    try: counts[next(iContent)] += 1  # count characters
    except StopIteration: break       # stop when all characters counted

print(counts)
{'h': 1, 'e': 1, 'l': 3, 'o': 2, ' ': 1, 'w': 1, 'r': 1, 'd': 1}

while循环表示returns每个字符出现的次数

While loop that returns the number of occurrences of each character

python

while-loop