用户输入的逐个频率字排序的程序然后输出是字典

Question

我是python中的新手，现在我还在学习，我对字典有疑问。 我想创建用户将输入字符串的程序，然后它将计算该字符串中的字符。嗯也许我会用图片来解释 this is the preview about program that i want。我真的需要帮助，谢谢你们。顺便说一句，我使用python 3.6

Answer 1

老实说，我认为你的结果是错误的......

这是我的结果：

==> "china chinese"
(1) char-frequency dict:
{'n': 2, 'e': 2, 'a': 1, 's': 1, 'h': 2, 'c': 2, 'i': 2}
(2) frequency-char dict:
{1: ['a', 's'], 2: ['n', 'e', 'h', 'c', 'i']}
(3) result:
OrderedDict([(1, ['a', 's']), (2, ['c', 'e', 'h', 'i', 'n'])])


==> "united states"
(1) char-frequency dict:
{'n': 1, 'e': 2, 'a': 1, 's': 2, 'u': 1, 'd': 1, 'i': 1, 't': 3}
(2) frequency-char dict:
{1: ['n', 'a', 'u', 'd', 'i'], 2: ['e', 's'], 3: ['t']}
(3) result:
OrderedDict([(1, ['a', 'd', 'i', 'n', 'u']), (2, ['e', 's']), (3, ['t'])])

代码：

#!/usr/bin/python3
# 2017.10.30 19:10:16 CST
from collections import OrderedDict

def inverseCharFrequency(s):
    print("\n\n==> \"{}\"".format(s))
    ## assert you are counting the lower letters
    ## get the char-frequency dict
    char_freq = {}
    for c in s:
        if c >='a' and c <='z':
            char_freq[c] = char_freq.get(c,0) + 1

    ## let's print
    print("(1) char-frequency dict:\n{}".format(char_freq))

    ## get the frequency-char dict
    char_freq2 = {}
    for c, freq in char_freq.items():
        if char_freq2.get(freq, None):
            char_freq2[freq].append(c)
        else:
            char_freq2[freq] = [c]

    ## let's print
    print("(2) frequency-char dict:\n{}".format(char_freq2))

    ## why do you want to sort? Ok, just sort!
    res = OrderedDict()
    for k in sorted(char_freq2):
        res[k] = sorted(char_freq2[k])

    ## let's print
    print("(3) result:\n{}".format(res))

    return res


if __name__ == "__main__":
    s = "china chinese"
    res = inverseCharFrequency(s)

    s = "united states"
    res = inverseCharFrequency(s)