如何將列表轉換爲多值字典

Question

我有一個列表，我想要轉換爲4值字典，其中每個列表中的第一個值是關鍵字的列表。因此，例如，名單將是：

[['267-10-7633', '66', '85', '74', 0], ['709-40-8165', '71', '96', '34', 0]] 

，我想這是

{"267-10-7633":[66,85,74,0], "709-40-8165", [71,96,34,0] } 

Answer 1

您可以使用字典解析：

lst = [['267-10-7633', '66', '85', '74', 0], ['709-40-8165', '71', '96', '34', 0]] 

{k: v for k, *v in lst}  
# {'267-10-7633': ['66', '85', '74', 0], '709-40-8165': ['71', '96', '34', 0]} 

---

如果你是在python2 ，似乎你不能使用*v來解壓縮多個元素：

{x[0]: x[1:] for x in lst} 
# {'267-10-7633': ['66', '85', '74', 0], '709-40-8165': ['71', '96', '34', 0]} 

這裏沒有照顧到類型轉換。我想你可以參考其他的答案，如何做到這一點。

Answer 2

一個dict理解編譯字典有list理解的字符串轉換爲int：

> lst = [['267-10-7633', '66', '85', '74', 0], ['709-40-8165', '71', '96', '34', 0]] 
> {l[0]: [int(x) for x in l[1:]] for l in lst} 
{'267-10-7633': [66, 85, 74, 0], '709-40-8165': [71, 96, 34, 0]} 

Answer 3

一個簡單而直接的解決方案。

lst = [['267-10-7633', '66', '85', '74', 0], ['709-40-8165', '71', '96', '34', 0]] 

# create an empty dict 
new_dict = {} 

# iterate through the list 
for item in lst: 

    # key is first element in the inner list 
    # value is second element in the inner list 
    key = item[0] 
    value = item[1:] 
    new_dict[key] = value 

print new_dict 

Answer 4

列表解析適合在這種情況下

{element[0]: [int(x) for x in element[1:]] for element in\ 
    [['267-10-7633', '66', '85', '74', 0], ['709-40-8165', '71', '96', '34', 0]]} 

Answer 5

一種簡單的方法：

your_list = [['267-10-7633', '66', '85', '74', 0], ['709-40-8165', '71', '96', '34', 0]] 

dictionary = {} 
for item in your_list: 
    dictionary[item[0]] = [int(i) for i in item[1:]] 

print(dictionary) 

---

list隨着和dict理解：

dictionary = {item[0]: [int(i) for i in item[1:]] for item in your_list} 
print(dictionary) 

---

在這兩種情況下，輸出：

{'267-10-7633': [66, 85, 74, 0], '709-40-8165': [71, 96, 34, 0]} 

Answer 6

ll = [['267-10-7633', '66', '85', '74', 0], ['709-40-8165', '71', '96', '34', 0]] 

mydict = {} 
for item in ll: 
    key,*values = item 
    mydict[key] = values 

print(mydict)