多亏了philipcloud的伟大的answer to a previous question,我在scikit中挖掘了pairwise_distances
的源代码。在
相关部分是:
def pairwise_distances(X, Y=None, metric="euclidean", n_jobs=1, **kwds):
if metric == "precomputed":
return X
elif metric in PAIRWISE_DISTANCE_FUNCTIONS:
func = PAIRWISE_DISTANCE_FUNCTIONS[metric]
if n_jobs == 1:
return func(X, Y, **kwds)
else:
return _parallel_pairwise(X, Y, func, n_jobs, **kwds)
elif callable(metric):
# Check matrices first (this is usually done by the metric).
X, Y = check_pairwise_arrays(X, Y)
n_x, n_y = X.shape[0], Y.shape[0]
# Calculate distance for each element in X and Y.
# FIXME: can use n_jobs here too
D = np.zeros((n_x, n_y), dtype='float')
for i in range(n_x):
start = 0
if X is Y:
start = i
for j in range(start, n_y):
# distance assumed to be symmetric.
D[i][j] = metric(X[i], Y[j], **kwds)
if X is Y:
D[j][i] = D[i][j]
return D
如果我要计算成对距离矩阵,如:
matrix = pairwise_distances(foo, metric=lambda u,v: haversine(u,v), n_jobs= -1)
其中haversine(u,v)
是一个计算两点间Haversine距离的函数,并且该函数在PAIRWISE_DISTANCE_FUNCTIONS
中是而不是,那么即使n_jobs= -1
,该计算也不会被并行化吗?在
我意识到#FIXME
注释似乎暗示了这一点,但我想确保我不是疯了,因为似乎有点奇怪的是,当你用一个不在PAIRWISE_DISTANCE_FUNCTIONS
中的可调用函数传递n_jobs= -1
时,不会抛出信息性警报,说明计算实际上不会被并行化。在
已确认。如果可调用不在}一起传递将不会导致并行化。在
PAIRWISE_DISTANCE_FUNCTIONS
中,那么将可调用作为metric
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