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[英]How to unfold a python dictionary with lists as values into a list of key-value tuples with two elements?
[英]How to flatten the value (list of lists of tuples) of a key in Python dictionary?
我在 python 中有一个字典,看起来像这样:
{(-1, 1): (0, 1),
(0, 0): [([([(1, 0), (0, 1)], (0, 1))], (1, 0))],
(0, 1): [([([((-1, 1), (0, 2))], (1, 1))], (0, 0))],
(0, 2): (0, 1)}
我不希望它有所有这些额外的括号和圆括号。 这是我用来创建这本字典的代码:
if condition1==True:
if condition2==True:
if (x,y) in adjList_dict: ##if the (x,y) tuple key is already in the dict
##add tuple neighbours[i] to existing list of tuples
adjList_dict[(x,y)]=[(adjList_dict[(x,y)],neighbours[i])]
else:
adjList_dict.update( {(x,y) : neighbours[i]} )
我只是想创建一个字典,其中键是元组,每个键的值是一个元组列表。
例如我想要这个结果: (0, 0): [(1, 0), (0, 1), (0, 1), (1, 0)]
我可以展平输出还是应该在创建字典时更改某些内容?
您可以使用递归,然后测试实例是否是一个包含 int 值的简单元组,例如:
sample = {(-1, 1): (0, 1),
(0, 0): [([([(1, 0), (0, 1)], (0, 1))], (1, 0))],
(0, 1): [([([((-1, 1), (0, 2))], (1, 1))], (0, 0))],
(0, 2): (0, 1)}
def flatten(data, output):
if isinstance(data, tuple) and isinstance(data[0], int):
output.append(data)
else:
for e in data:
flatten(e, output)
output = {}
for key, values in sample.items():
flatten_values = []
flatten(values, flatten_values)
output[key] = flatten_values
print(output)
>>> {(-1, 1): [(0, 1)], (0, 0): [(1, 0), (0, 1), (0, 1), (1, 0)], (0, 1): [(-1, 1), (0, 2), (1, 1), (0, 0)], (0, 2): [(0, 1)]}
您可以使用带有字典理解的递归方法:
d = {(-1, 1): (0, 1),
(0, 0): [([([(1, 0), (0, 1)], (0, 1))], (1, 0))],
(0, 1): [([([((-1, 1), (0, 2))], (1, 1))], (0, 0))],
(0, 2): (0, 1)}
def flatten(e):
if isinstance(e[0], int):
yield e
else:
for i in e:
yield from flatten(i)
{k: list(flatten(v)) for k, v in d.items()}
输出:
{(-1, 1): [(0, 1)],
(0, 0): [(1, 0), (0, 1), (0, 1), (1, 0)],
(0, 1): [(-1, 1), (0, 2), (1, 1), (0, 0)],
(0, 2): [(0, 1)]}
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