如何获得字典键组合的唯一值集？

Question

我正在使用 Python 3.7。我有一个字典列表，例如

my_dict = [{"a": 1, "b": 5, "c": 6}, {"a": 1, "b": 5, "c": 2}, {"a": 2, "b": 1, "c": 6}]

如果我想获取单个键的唯一值集，例如"a"，我可以

set(d['a'] for d in my_dict)

但是我如何获得键组合的唯一值集，比如 "a" 和 "b"？在上面的例子中，答案是

[[1, 2], [1, 5]]

Answer 1

同理；您也只需要遍历键即可。

[set(d[k] for d in my_dict) for k in ["a", "b"]]

Answer 2

我不完全确定这些要求（有什么独特之处？），但根据我的解释，以下作品：

> {tuple(d[i] for i in ["a", "b"]) for d in my_dict}
{(1, 5), (2, 1)}

它还执行以下操作：

> my_dict2 = [{"a": 1, "b": 5, "c": 6}, {"a": 5, "b": 1, "c": 2}, {"a": 2, "b": 1, "c": 6}]

> {tuple(d[i] for i in ["a", "b"]) for d in my_dict2}
{(1, 5), (2, 1), (5, 1)}

相对于

> [set(d[k] for d in my_dict2) for k in ["a", "b"]]
[{1, 2, 5}, {1, 5}]

Answer 3

OP 要求提供一些未完全指定的内容，但通过示例进行了说明。到目前为止，已经发布了两个答案。 None 其中与请求的示例匹配，因此我提供了一个替代方案，它确实与 OP 中的示例匹配。

对于使用的两种情况

my_dict = [{"a": 1, "b": 5, "c": 6}, {"a": 1, "b": 5, "c": 2}, {"a": 2, "b": 1, "c": 6}]   # In the OP
my_dict2 = [{"a": 1, "b": 5, "c": 6}, {"a": 5, "b": 1, "c": 2}, {"a": 2, "b": 1, "c": 6}]   # In one of the answers

之前提供的2个选项，加上当前的一个，给出

print([set(d[k] for d in my_dict) for k in ["a", "b"]])         # Answer #1
print({tuple(d[i] for i in ["a", "b"]) for d in my_dict})       # Answer #2
print([list(set(d[k] for d in my_dict)) for k in ["a", "b"]])   # Current Answer

print([set(d[k] for d in my_dict2) for k in ["a", "b"]])        # Answer #1
print({tuple(d[i] for i in ["a", "b"]) for d in my_dict2})      # Answer #2
print([list(set(d[k] for d in my_dict2)) for k in ["a", "b"]])  # Current Answer

[{1, 2}, {1, 5}]
{(1, 5), (2, 1)}
[[1, 2], [1, 5]]           <--- As requested
[{1, 2, 5}, {1, 5}]
{(1, 5), (5, 1), (2, 1)}
[[1, 2, 5], [1, 5]]

Answer 4

您可以使用运算符 itemgetter 从字典中获取多个值：

from operator import itemgetter

zipped = zip(*map(itemgetter('a', 'b'), my_dict))
list(map(set, zipped))
# [{1, 2}, {1, 5}]

如何获得字典键组合的唯一值集？

How do I get the unique set of values for a combination of dictionary keys?

python

dictionary

unique

python-3.x

python-3.7