从元组列表创建嵌套字典

Question

我有一个元组列表，如下所示。我想创建一个嵌套字典，其中元组中的第一个元素是键，具有相同键的值应该分组到与字典相同的键中。

这是我尝试过的方法，但没有提供我想要的嵌套字典。

data = [(37043, 2862826), (37043,2850223), (37107,2847978), (37107,2848001), (37107,2844725)]

data_dict = defaultdict(list)
for key, value in data:
    value = {value:1}
    data_dict[key].append(value)

print (data_dict)
>>> defaultdict(<class 'list'>, {37043: [{2862826: 1}, {2850223: 1}], 37107: [{2847978: 1}, {2848001: 1}, {2844725: 1}]})

data_dict = defaultdict(dict)    
for key, value in data:
    value = {value:1}
    data_dict[key] = value

print (data_dict)
>>> defaultdict(<class 'dict'>, {37043: {2850223: 1}, 37107: {2844725: 1}})

想要的结果：

 {37043: {2862826: 1, 2850223: 1}, 37107: {2847978:1, 2848001: 1, 2844725: 1}}

Answer 1

如何使用defaultdict(dict):

>>> data_dict = defaultdict(dict)
>>> for key, value in data:
...     data_dict[key][value] = 1
... 
>>> data_dict
defaultdict(<type 'dict'>, {37043: {2862826: 1, 2850223: 1}, 37107: {2848001: 1, 2847978: 1, 2844725: 1}})

Answer 2

下面的代码适合您：

data = [(37043, 2862826), (37043,2850223), (37107,2847978), (37107,2848001), (37107,2844725)]

data_map = {}
for key, value in data:
    data_map.setdefault(key, {})
    data_map[key][value] = 1

print data_map
# {37043: {2862826: 1, 2850223: 1}, 37107: {2848001: 1, 2847978: 1, 2844725: 1}}

Link 到 setdefault 方法文档。

Answer 3

以下是简单的 key-value 赋值：

data = [(37043, 2862826), (37043,2850223), (37107,2847978), (37107,2848001), (37107,2844725)]
output = {}
for key,value in data:
  if key in output:
    if value in output[key]: output[key][value] += 1
    else: output[key][value] = 1
  else: output[key] = {value:1}

输出：

{37043: {2862826: 1, 2850223: 1}, 37107: {2848001: 1, 2847978: 1, 2844725: 1}}

从元组列表创建嵌套字典

Creating a nested dictionary from a list of tuples

python

dictionary

defaultdict