仅使用部分数据预突变

Question

import itertools

alphabet = {
    "A": 1,
    "B": 2,
    "C": 3,
    "D": 4,
    "E": 5,
    "F": 6,
    "G": 7,
    "H": 8,
    "I": 9,
    "J": 10,
    "K": 11,
    "L": 12,
    "M": 13,
    "N": 14,
    "O": 15,
    "P": 16,
    "Q": 17,
    "R": 18,
    "S": 19,
    "T": 20,
    "U": 21,
    "V": 22,
    "W": 23,
    "X": 24,
    "Y": 25,
    "Z": 26
}


def gen_combination_values(n):
    comb_values = {}
    for i in range(n + 1):
        for product in itertools.product(alphabet.keys(), repeat=i):
            prod = list(product)
            prod.sort()
            value = 0
            comb = ""
            for char in prod:
                value += alphabet[char]
                comb += char
            try:
                comb_values[value].add(comb)
            except KeyError:
                comb_values[value] = set()
                comb_values[value].add(comb)
    print(comb_values)
    return comb_values

我有这个脚本我只想使用排列的一部分怎么办？

我希望脚本的下一部分仅使用最多 10 个结果

{0: {''}, 1: {'A'}, 2: {'B', 'AA'}, 3: {'C', 'AB', 'BA'}, 4: {'BB' , 'AC', 'D', 'CA'}, 5: {'AD', 'BC', 'E', 'CB', 'DA'}, 6: {'F', 'CC', ' AE ',' DB ',' BD ',' EA '}, 7: {' CD ',' AF ',' G ',' EB ',' FA ',' BE ',' DC '}, 8: {'AG', 'CE', 'EC', 'H', 'BF', 'FB', 'DD', 'GA'}, 9: {'AH', 'DE', 'HA', ' BG ',' CF ',' ED ',' I ',' GB ',' FC '}, 10: {' EE ',' HB ',' DF ',' FD ',' IA ',' AI ' , 'J', 'BH', 'CG', 'GC'},

其余不用

11: {'FE', 'BI', 'CH', 'AJ', 'JA', 'DG', 'GD', 'IB', 'HC', 'EF'}, 12: {'EG' , 'CI', 'IC', 'BJ', 'FF', 'DH', 'HD', 'JB', 'GE'}, 13: {'EH', 'HE', 'DI', ' CJ ',' FG ',' JC ',' ID ',' GF '}, 14: {' DJ ',' EI ',' IE ',' HF ',' JD ',' GG ',' FH ' }, 15: {'EJ', 'IF', 'GH', 'JE', 'FI', 'HG'}, 16: {'IG', 'JF', 'FJ', 'GI', ' HH '}, 17: {' JG ',' GJ ',' HI ',' IH '}, 18: {' JH ',' II ',' HJ '}, 19: {' IJ ',' JI ' }, 20: {'JJ'}}

如何实现？基本上这是整个代码我不知道如何处理所有失败的尝试

我只查找要从代码中使用的前十个预突变

Answer 1

您的 main 代码中有一个小错误，但我认为这应该可以满足您的要求：

import itertools

alphabet = {
    "A": 1,
    "B": 2,
    "C": 3,
    "D": 4,
    "E": 5,
    "F": 6,
    "G": 7,
    "H": 8,
    "I": 9,
    "J": 10
}


def gen_combination_values(n):
    comb_values = {}
    for i in range(n + 1):
        for product in itertools.product(alphabet.keys(), repeat=i):
            value = 0
            comb = ""
            for char in product:
                value += alphabet[char]
                comb += char
            try:
                comb_values[value].add(comb)
            except KeyError:
                comb_values[value] = set()
                comb_values[value].add(comb)
    print(comb_values)
    return comb_values


def gematria(word):
    result = 0
    for char in word:
        result += alphabet[char]
    return result


if __name__ == '__main__':
    values = gen_combination_values(2)
    for i,(k,v) in enumerate(values.items()):
        if k <= 10:
            print(i)

Answer 2

我建议使用 gen_combination_values() 函数的 re-write。构建字典的方法比较简单（我在这段代码中展示）。然后，您需要根据您的限制（默认为 10）构造一个 return 的新字典。

import itertools

ALPHABET = {k:i for i, k in enumerate('ABCDEFGHIJ', 1)}

def gen_combination_values(n, limit=10):
    comb_values = {}
    for i in range(n + 1):
        for product in itertools.product(ALPHABET, repeat=i):
            value = sum(ALPHABET[c] for c in product)
            comb = ''.join(c for c in product)
            comb_values.setdefault(value, set()).add(comb)
    return {k:v for k, v in comb_values.items() if k <= limit}

print(gen_combination_values(2))

输出：

{0: {''}, 1: {'A'}, 2: {'B', 'AA'}, 3: {'C', 'BA', 'AB'}, 4: {'D', 'AC', 'BB', 'CA'}, 5: {'DA', 'E', 'AD', 'BC', 'CB'}, 6: {'CC', 'AE', 'EA', 'BD', 'DB', 'F'}, 7: {'AF', 'FA', 'CD', 'BE', 'G', 'EB', 'DC'}, 8: {'AG', 'GA', 'CE', 'EC', 'FB', 'DD', 'H', 'BF'}, 9: {'CF', 'FC', 'AH', 'BG', 'ED', 'DE', 'GB', 'HA', 'I'}, 10: {'DF', 'EE', 'BH', 'IA', 'CG', 'AI', 'HB', 'GC', 'J', 'FD'}}

仅使用部分数据预突变

use only part of the data premutation

python

itertools

more-itertools