使用Counter:
from collections import Counter
t_str = "peter piper picked a peck of pickled peppers; a peck of pickled peppers peter picked; if peter piper picked a peck of pickled peppers, where's the peck of pickled peppers peter picked?"
print(Counter(t_str))
输出:
Counter({'e': 33, 'p': 32, ' ': 31, 'c': 12, 'k': 12, 'r': 11, 'i': 11, 'd': 8, 't': 5, 'f': 5, 's': 5, 'o': 4, 'l': 4, 'a': 3, ';': 2, 'h': 2, ',': 1, 'w': 1, "'": 1, '?': 1})
或
print(dict((letter,t_str.count(letter)) for letter in set(t_str)))
输出:
{'?': 1, ',': 1, 'r': 11, 'p': 32, ' ': 31, 'k': 12, 'a': 3, 'l': 4, 'd': 8, 'h': 2, "'": 1, 'i': 11, 'w': 1, 'c': 12, ';': 2, 't': 5, 'o': 4, 's': 5, 'f': 5, 'e': 33}
EDIT(计算来自a-z的字母的出现次数,无论它们在字符串中是否存在,即在这种情况下它等于0):
import string
letter_set = string.ascii_lowercase
print(dict((letter,t_str.count(letter)) for letter in letter_set))
输出:
{'a': 3, 'b': 0, 'c': 12, 'd': 8, 'e': 33, 'f': 5, 'g': 0, 'h': 2, 'i': 11, 'j': 0, 'k': 12, 'l': 4, 'm': 0, 'n': 0, 'o': 4, 'p': 32, 'q': 0, 'r': 11, 's': 5, 't': 5, 'u': 0, 'v': 0, 'w': 1, 'x': 0, 'y': 0, 'z': 0}