求一段字符串中每个字符出现的次数，并排序输出（用Python实现）

统计字符串中每个字符的个数 排序输出结果

• 将字符串去重存为列表
• 将列表排序
• 迭代列表 根据列表的值 去统计字符串中的字符
• 将结果存储为字典
• 输出结果

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## 如果有更简单的方法请大佬评论告诉我，谢谢?

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"""
求一个字符串 每个字符的个数。
"""
str1 = """
To be, or not to be—— that is the question;
Whether‘tis nobler in the mind to suffer
The slings and arrows of outrageous fortune,
Or to take arms against a sea of troubles,
And by opposing end them?
To die——to sleep——
No more; and by a sleep to say we end
The heart-ache and the thousand natural shocks
That flesh is heir to, 'tis a consummation
Devoutly to be wish'd.
"""
print("共%d个字符" % len(str1))

# 将字符串去重
set1 = set(str1)

# 将去重后的集合转换为列表并排序
list1 = list(set1)
list1.sort()

dict1 = dict()

# 迭代列表 并统计字符数 存入字典
for i in list1:
    num = str1.count(i)
    dict1.setdefault(i, num)

print(dict1)

输出结果为：

共370个字符
{’\n’: 11, ’ ‘: 60, "’": 2, ‘,’: 4, ‘-’: 1, ‘.’: 1, ‘;’: 2, ‘?’: 1, ‘A’: 1, ‘D’: 1, ‘N’: 1, ‘O’: 1, ‘T’: 5, ‘W’: 1, ‘a’: 22, ‘b’: 7, ‘c’: 3, ‘d’: 10, ‘e’: 33, ‘f’: 6, ‘g’: 4, ‘h’: 17, ‘i’: 14, ‘k’: 2, ‘l’: 8, ‘m’: 6, ‘n’: 19, ‘o’: 29, ‘p’: 4, ‘q’: 1, ‘r’: 15, ‘s’: 24, ‘t’: 29, ‘u’: 10, ‘v’: 1, ‘w’: 3, ‘y’: 4, ‘—’: 6, ‘‘’: 1}
Process finished with exit code 0

exit(⭐)