有没有更简单快捷的方法来获取索引dict，其中包含列表或numpy数组中相同元素的索引

Question

Description:

I have a large array with simple integers(positive and not large) like 1, 2, ..., etc. For example: [1, 1, 2, 2, 1, 2]. I want to get a dict in which use a single value from the list as the dict's key, and use the indexes list of this value as the dict's value.

Question:

Is there a simpler and faster way to get the expected results in python? (array can be a list or a numpy array)

Code:

a = [1, 1, 2, 2, 1, 2]
results = indexes_of_same_elements(a)
print(results)

Expected results:

{1:[0, 1, 4], 2:[2, 3, 5]}

Answer 1

You can avoid iteration here using vectorized methods, in particular np.unique + np.argsort :

idx = np.argsort(a)
el, c = np.unique(a, return_counts=True)

out = dict(zip(el, np.split(idx, c.cumsum()[:-1])))

{1: array([0, 1, 4], dtype=int64), 2: array([2, 3, 5], dtype=int64)} 

---

Performance

a = np.random.randint(1, 100, 10000)

In [183]: %%timeit
     ...: idx = np.argsort(a)
     ...: el, c = np.unique(a, return_counts=True)
     ...: dict(zip(el, np.split(idx, c.cumsum()[:-1])))
     ...:
897 µs ± 41.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [184]: %%timeit
     ...: results = {}
     ...: for i, k in enumerate(a):
     ...:     results.setdefault(k, []).append(i)
     ...:
2.61 ms ± 18.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 2

It is pretty trivial to construct the dict:

In []:
results = {}
for i, k in enumerate(a):
    results.setdefault(k, []).append(i)   # str(k) if you really need the key to be a str
print(results)

Out[]:
{1: [0, 1, 4], 2: [2, 3, 5]}

You could also use results = collections.defaultdict(list) and then results[k].append(i) instead of results.setdefault(k, []).append(i)

Answer 3

We can exploit the fact that the elements are "simple" (ie nonnegative and not too large?) integers.

The trick is to construct a sparse matrix with just one element per row and then to transform it to a column wise representation. This is typically faster than argsort because this transform is O(M + N + nnz), if the sparse matrix is MxN with nnz nonzeros.

from scipy import sparse

def use_sprsm():
    x = sparse.csr_matrix((a, a, np.arange(a.size+1))).tocsc()
    idx, = np.where(x.indptr[:-1] != x.indptr[1:])
    return {i: a for i, a in zip(idx, np.split(x.indices, x.indptr[idx[1:]]))}

# for comparison

def use_asort():
    idx = np.argsort(a)
    el, c = np.unique(a, return_counts=True)
    return dict(zip(el, np.split(idx, c.cumsum()[:-1])))

Sample run:

>>> a = np.random.randint(0, 100, (10_000,))
>>> 
# sanity check, note that `use_sprsm` returns sorted indices
>>> for k, v in use_asort().items():
...     assert np.array_equal(np.sort(v), use_sprsm()[k])
... 
>>> timeit(use_asort, number=1000)
0.8930604780325666
>>> timeit(use_sprsm, number=1000)
0.38419671391602606