如何在python中從位置找到列表中的元素（我使用index（）找到）

Question

alpha = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

那是清單

for letters in sentence:
    pos2 = alpha.index(letters) + 1
    #to find the positions of each letter
    new_sentence.append(pos2)
    #to add each position to the empty list
print (new_sentence)

這就是我用來查找字母列表中輸入消息中每個字母的位置的方法

現在我想將其轉換回職位的字母。

Answer 1

您可以將其編入索引：

[alpha[p-1] for p in new_sentence]

---

sentence = "helloworld"
# ... run your code to get the new_sentence
''.join(alpha[p-1] for p in new_sentence)

# 'helloworld'

---

如果您打算在原始字母之后找到該字母，則可以使用索引@RushyPanchal的其余部分作為索引：

sentence = "hello world"

# ... run your code to get the new_sentence
''.join(alpha[p % len(alpha)] for p in new_sentence)

# 'ifmmpxpsme'

Answer 2

因為您有索引，所以可以在該索引處獲取值。

print my_list[pos2]

python還有一個內置方法enumerate(enumerable_obj)返回index, value對

for index, value in enumerate(my_list):
    print index, value

Answer 3

alpha_text = ""
for i in new_sentence:
    alpha_text = alpha_text + alpha[i - 1]
print(alpha_text)

因此，您的整個代碼如下所示：

sentence = "lololololololololol" #your text
alpha = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
new_sentence = []
for letters in sentence:
    pos2 = alpha.index(letters) + 1
    #to find the positions of each letter
    new_sentence.append(pos2)

print (new_sentence)
alpha_text =""
for i in new_sentence:
    alpha_text = alpha_text + alpha[i - 1]
print(alpha_text)

輸出：

 [12, 15, 12, 15, 12, 15, 12, 15, 12, 15, 12, 15, 12, 15, 12, 15, 12, 15, 12]

 lololololololololol

Answer 4

@Psidom的答案是從字符串中獲取字符列表的正確方法。

但是，如果只想移動字符，可以使用chr和ord函數：

sentence = "some string"
shifted_sentence = ''.join(chr(ord(c)+1) for c in sentence)

Answer 5

在性能方面，制作字母及其值的字典應該更容易，反之亦然。 這樣，您每次查詢只使用固定時間。 這項更改使您的代碼更具可伸縮性，並且速度更快。

Answer 6

使用enumerate()和filter() ：

>>> sentence = 'hello'

對於此示例是：

>>> new_sentence
[8, 5, 12, 12, 15]

結果如下：

>>> letters_enum = [(j,c) for j,c in enumerate(alpha, 1) if j in new_sentence]
>>> result = []
>>> for i in new_sentence:
...     letter = list(filter(lambda item: item[0] == i, letters_enum))
...     result.extend(letter[0][1]*len(letter))
...
>>>
>>> result
['h', 'e', 'l', 'l', 'o']

Answer 7

您當前的方法效率低下，因為您必須為輸入句子中的每個字母搜索多達26個不同的列表項。 字母“ z”也將失敗，因為列表中“ z”之后沒有任何內容。

既然您已經闡明要嘗試使用關鍵字cipher ，那么更有效的方法是使用str.translate ：

import string

keyword = 'stack'
source = string.ascii_lowercase
target = keyword + ''.join(filter(lambda x: x not in keyword, source))

sentence = 'encode me'
encoded = sentence.translate(str.maketrans(source, target))

print(encoded)

輸出：

klamck jk