将此元组列表转换为字典

问题描述

我有一个元组列表。

list = [("a",1),("b",2),("a",3),("c",1)]

我想将此元组列表转换成字典。

输出：

{'a': [1,3,2],'b': [2,1],'c': [1]}

我的代码：

list_a = [("a",1)]
new_dict = {}
value = []
for i in range(len(list_a)):
    key = list_a[i][0]
    
    value.append(list_a[i][1])
    
    new_dict[key] = value
print(new_dict)

但是，我的输出如下：

{'a': [1,2,1,'b': [1,'c': [1,1]}

解决方法

list_a = [("a",1),("b",2),("a",3),("c",1)]
new_dict = {}

for item in list_a:
    # checking the item[0] which gives the first element
    # of a tuple which is there in the key element of 
    # of the new_dict
    if item[0] in new_dict:
        new_dict[item[0]].append(item[1])
    else:
        # add a new data for the new key 
        # which we get by item[1] from the tuple
        new_dict[item[0]] = [item[1]]
print(new_dict)

输出

{'a': [1,3,2],'b': [2,1],'c': [1]}

,

下面是一个选项。美观且可读

阅读下面代码中的注释

list = [("a",1)]

#create dictionary
dic = {}

#iterate each item in the list
for itm in list:
    #check if first item in the tuple is in the keys of the dictionary
    if itm[0] in dic.keys():
        #if so append the list
        dic[itm[0]].append(itm[1])
    else:
        #if not create new key,value pair (value is a list as enclosed with [])
        dic[itm[0]] = [itm[1]]        
        

,

还有其他的builin库可以帮助您解决问题，但与此同时，我们可以使用本机逻辑来解决它。

list_ = [('a',('b',('a',('c',1)]
result = {}
for i in list_:
    result[i[0]] = (result.get(i[0]) or []) + [i[1]]

print(result)  # {'a': [1,'c': [1]}