Question

我如何获得这样的字典：

{'SEA': {'SFO': 6.020797289396148, 'DEN': 12.041594578792296, 'OAK': 6.029925372672534}, 'SFO': {'SEA': 6.020797289396148, 'DEN': 9.7082439194738}}

是这样的：

[['SEA', ['SFO', 6.020797289396148, 'DEN', 12.041594578792296, 'OAK', 6.029925372672534]], ['SFO', ['SEA', 6.020797289396148, 'DEN', 9.7082439194738]]]

Answer 1

将嵌套的dictionary comprehension与zip和dict配合使用：

l = [['SEA', ['SFO', 6.020797289396148, 'DEN', 12.041594578792296, 'OAK', 6.029925372672534]], ['SFO', ['SEA', 6.020797289396148, 'DEN', 9.7082439194738]]]
d = {v[0]:{k:v1 for k,v1 in zip(v[1][::2], v[1][1::2])} for v in l}

或使用如下所示的for循环：

d = {}
for v in l:
    temp_d = {}
    for k,v1 in zip(v[1][::2],v[1][1::2]):
        temp_d[k]=v1
    d[v[0]]=temp_d

或者：

d = {v[0]:dict(zip(v[1][::2],v[1][1::2])) for v in l}

print(d)
{'SEA': {'SFO': 6.020797289396148,
  'DEN': 12.041594578792296,
  'OAK': 6.029925372672534},
 'SFO': {'SEA': 6.020797289396148, 'DEN': 9.7082439194738}}

Answer 2

您可以使用dict理解，

In [95]: {key:dict(zip(values[::2], values[1::2])) for key,values in a}
Out[95]: 
{'SEA': {'DEN': 12.041594578792296,
  'OAK': 6.029925372672534,
  'SFO': 6.020797289396148},
 'SFO': {'DEN': 9.7082439194738, 'SEA': 6.020797289396148}}

zip(values[::2], values[1::2])将创建这样的元组列表，

[('SFO', 6.020797289396148),
  ('DEN', 12.041594578792296),
  ('OAK', 6.029925372672534)]

dict(zip(values[::2], values[1::2]))将生成一个与此类似的字典，

{'DEN': 12.041594578792296,
  'OAK': 6.029925372672534,
  'SFO': 6.020797289396148}

Answer 3

简单的方法

d = {'SEA': {'SFO': 6.020797289396148, 'DEN': 12.041594578792296, 'OAK': 6.029925372672534}, 'SFO': {'SEA': 6.020797289396148, 'DEN': 9.7082439194738}}

l = []
for a in d.keys():
    l2 = []
    d2 = d[a]
    for b in d2.keys():
        l2.extend([b,d2[b]])
    l.extend([a,l2])
print(l)

你得到 ['SEA'，['SFO'，6.020797289396148，'DEN'，12.041594578792296，'OAK'，6.029925372672534]，'SFO'，['SEA'，6.020797289396148，'DEN'，9.7082439194738]]

Answer 4

[['SEA', ['SFO', 6.020797289396148, 'DEN', 12.041594578792296, 'OAK', 6.029925372672534]], ['SFO', ['SEA', 6.020797289396148, 'DEN', 9.7082439194738]]]

在此显示Array of Array的意思是[ '1',['a',['a1','a2','a3']] ] 主Arrya第0个索引['1']是正常值，另一个索引是数组['a',['a1','a2','a3']]

但是在第二种情况下，它是JSON数据

{'SEA': {'SFO': 6.020797289396148, 'DEN': 12.041594578792296, 'OAK': 6.029925372672534}, 'SFO': {'SEA': 6.020797289396148, 'DEN': 9.7082439194738}}

示例：

{ "SEA" :{'SFO':6.020797289396148}}

SEA-> {}-SFO-> 6.020797289396148 基本上第二种方法包含一个对象作为该对象中的对象的对象值包含值6.020797289396148

SEA.SFO使用此即可获得此6.020797289396148的值

将列表列表变成单个字典

4 个答案: