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Efficiently compute ordering permutations in a numpy array

I have a numpy array. What is the fastest way to calculate all permutations of orderings.

I mean, given the first element in my array, I need a list of all the elements that follow it sequentially. Then, given the second element, a list of all subsequent elements.

So, given my list: b, c and d, follow a. c and d follow b, and d follows c.

x = np.array(["a", "b", "c", "d"])

Thus, the potential conclusion is as follows:

[
    ["a","b"],
    ["a","c"],
    ["a","d"],

    ["b","c"],
    ["b","d"],

    ["c","d"],
]

I will need to do this several million times, so I am looking for an effective solution.

I tried something like:

im = np.vstack([x]*len(x))
a = np.vstack(([im], [im.T])).T
results = a[np.triu_indices(len(x),1)]

but actually it is slower than a cycle ...

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itertools, chain.from_iterable combinations np.fromiter . Python, NumPy:

>>> from itertools import combinations, chain
>>> arr = np.fromiter(chain.from_iterable(combinations(x, 2)), dtype=x.dtype)
>>> arr.reshape(arr.size/2, 2)
array([['a', 'b'],
       ['a', 'c'],
       ['a', 'd'],
       ..., 
       ['b', 'c'],
       ['b', 'd'],
       ['c', 'd']], 
      dtype='|S1')

:

>>> x = np.array(["a", "b", "c", "d"]*100)
>>> %%timeit
    im = np.vstack([x]*len(x))
    a = np.vstack(([im], [im.T])).T
    results = a[np.triu_indices(len(x),1)]
... 
10 loops, best of 3: 29.2 ms per loop
>>> %%timeit
    arr = np.fromiter(chain.from_iterable(combinations(x, 2)), dtype=x.dtype)
    arr.reshape(arr.size/2, 2)
... 
100 loops, best of 3: 6.63 ms per loop
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, , tri . - Python, , .

, , .

, ; , , triu_indices.

:

def triangalize_1(x):
    xs, ys = numpy.triu_indices(len(x), 1)
    return numpy.array([x[xs], x[ys]]).T

x .

x , :

triu_cache = {}
def triangalize_1(x):
    if len(x) in triu_cache:
        xs, ys = triu_cache[len(x)]

    else:
        xs, ys = numpy.triu_indices(len(x), 1)
        triu_cache[len(x)] = xs, ys

    return numpy.array([x[xs], x[ys]]).T

x - .

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