你可以這樣做:
def sum_unique(label, weight):
order = np.lexsort(label.T)
label = label[order]
weight = weight[order]
unique = np.ones(len(label), 'bool')
unique[:-1] = (label[1:] != label[:-1]).any(-1)
totals = weight.cumsum()
totals = totals[unique]
totals[1:] = totals[1:] - totals[:-1]
return label[unique], totals
而且使用這樣的:
In [110]: coord = np.random.randint(0, 3, (10, 2))
In [111]: coord
Out[111]:
array([[0, 2],
[0, 2],
[2, 1],
[1, 2],
[1, 0],
[0, 2],
[0, 0],
[2, 1],
[1, 2],
[1, 2]])
In [112]: weights = np.ones(10)
In [113]: uniq_coord, sums = sum_unique(coord, weights)
In [114]: uniq_coord
Out[114]:
array([[0, 0],
[1, 0],
[2, 1],
[0, 2],
[1, 2]])
In [115]: sums
Out[115]: array([ 1., 1., 2., 3., 3.])
In [116]: a = np.zeros((3,3))
In [117]: x, y = uniq_coord.T
In [118]: a[x, y] = sums
In [119]: a
Out[119]:
array([[ 1., 0., 3.],
[ 1., 0., 3.],
[ 0., 2., 0.]])
我只是想到了這一點,它可能是更容易:
In [120]: flat_coord = np.ravel_multi_index(coord.T, (3,3))
In [121]: sums = np.bincount(flat_coord, weights)
In [122]: a = np.zeros((3,3))
In [123]: a.flat[:len(sums)] = sums
In [124]: a
Out[124]:
array([[ 1., 0., 3.],
[ 1., 0., 3.],
[ 0., 2., 0.]])
謝謝,這個作品很棒! – brorfred 2012-02-14 23:07:45