所以我必須替換名稱列表中的每個元音以匹配加密代碼，所以我需要的是，例如，對于“Andre”，我需要得到：0ndr0、0ndr1、0ndr2、0ndr3 ... 1ndr1， 1ndr2 ... 9ndr5 ... 9ndr9這是我所做的簡化版本：def testVoyellePrenom(): voyelle = ["A", "E", "I", "O", "U", "Y"] myNumbers = [a + 1 for a in range(-1,9)] myNumbers.reverse() pre = "ANDRE" testing = "" for x in pre: if(x in voyelle): for nb in myNumbers: pre = pre.replace(x,str(nb)) x = str(nb) testing = pre print(testing)Output :9NDRE8NDRE7NDRE6NDRE5NDRE4NDRE3NDRE2NDRE1NDRE0NDRE0NDR90NDR80NDR70NDR60NDR50NDR40NDR30NDR20NDR10NDR0Expected output :...0NDR01NDR91NDR81NDR71NDR61NDR51NDR41NDR31NDR21NDR11NDR0....
2 回答
浮云間
TA貢獻1829條經驗 獲得超4個贊
這是一個解決方案。基本邏輯是：
識別元音 ( vowel_inx) 的所有位置。
為這些位置創建所有數字組合的叉積。
循環這個叉積，并通過將數字分配到它們在原始術語中的相應位置來創建新詞。
vowels = ["A", "E", "I", "O", "U", "Y"]
pre = "ANDRE"
import itertools
vowel_inx = [i for i in range(len(pre)) if pre[i] in vowels ]
ranges = [range(10) for _ in vowel_inx]
for comb in itertools.product(*ranges):
pw = list(pre)
for i in range(len(vowel_inx)):
pw[vowel_inx[i]] = str(comb[i])
print ("".join(pw))
輸出是：
0NDR0
0NDR1
0NDR2
0NDR3
0NDR4
0NDR5
0NDR6
0NDR7
0NDR8
0NDR9
1NDR0
1NDR1
1NDR2
...
慕少森
TA貢獻2019條經驗 獲得超9個贊
這是一個正確的方法：
voyelle = ["A", "E", "I", "O", "U", "Y"]
def permute_single_index(word, index):
permutations = []
word_left = word[0:index]
word_right = word[index+1:]
for i in reversed(range(10)):
permutations.append(f"{word_left}{i}{word_right}")
return permutations
def testVoyellePrenom(pre):
upper_pre = pre.upper()
replace_indexes = [i for i, x in enumerate(upper_pre) if x in voyelle]
permutations = [upper_pre]
for ri in replace_indexes:
new_permutations = []
for word in permutations:
new_permutations += permute_single_index(word, ri)
permutations += new_permutations
print(permutations)
print(len(permutations))
def main():
testVoyellePrenom("Andre")
if __name__ == "__main__":
main()
你最終得到 121 個排列。
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