Python 的 itertools.combinations() 函数生成的结果是数字的组合。例如:
a = [7, 5, 5, 4]
b = list(itertools.combinations(a, 2))
# b = [(7, 5), (7, 5), (7, 4), (5, 5), (5, 4), (5, 4)]
但我也想获得如下组合的索引:
index = [(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]
我该怎么做?
Python 的 itertools.combinations() 函数生成的结果是数字的组合。例如:
a = [7, 5, 5, 4]
b = list(itertools.combinations(a, 2))
# b = [(7, 5), (7, 5), (7, 4), (5, 5), (5, 4), (5, 4)]
但我也想获得如下组合的索引:
index = [(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]
我该怎么做?
你可以使用enumerate:
>>> a = [7, 5, 5, 4]
>>> list(itertools.combinations(enumerate(a), 2))
[((0, 7), (1, 5)), ((0, 7), (2, 5)), ((0, 7), (3, 4)), ((1, 5), (2, 5)), ((1, 5), (3, 4)), ((2, 5), (3, 4))]
>>> b = list((i,j) for ((i,_),(j,_)) in itertools.combinations(enumerate(a), 2))
>>> b
[(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]
>>> list(itertools.combinations(range(3), 2))
[(0, 1), (0, 2), (1, 2)]
因此,您可以使用len(a)
:
>>> list(itertools.combinations(range(len(a)), 2))
[(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]
a = [7, 5, 5, 4]
n_combinations = 2
np.array(list(itertools.combinations(enumerate(a), n_combinations)))[...,0]
n_combinations
不需要更改代码。它基本上是使用numpy切片表达的解决方案。
itertools.combinations(range(len(a)), 2)
是什么意思? - jonrsharpe