numpy strides 只能在子数组内跨步吗?
Can numpy strides stride only within subarrays?
我有一个非常大的 numpy 数组(145000 行 * 550 列)。我想在子数组中创建滚动切片。我试图用一个函数来实现它。函数 lagged_vals
的行为符合预期,但 np.lib.stride_tricks
的行为与我希望的不同 -
def lagged_vals(series,l):
# Garbage implementation but still right
return np.concatenate([[x[i:i+l] for i in range(x.shape[0]) if i+l <= x.shape[0]] for x in series]
,axis = 0)
# Sample 2D numpy array
something = np.array([[1,2,2,3],[2,2,3,3]])
lagged_vals(something,2) # Works as expected
# array([[1, 2],
# [2, 2],
# [2, 3],
# [2, 2],
# [2, 3],
# [3, 3]])
np.lib.stride_tricks.as_strided(something,
(something.shape[0]*something.shape[1],2),
(8,8))
# array([[1, 2],
# [2, 2],
# [2, 3],
# [3, 2], <--- across subarray stride, which I do not want
# [2, 2],
# [2, 3],
# [3, 3])
如何删除 np.lib.stride_tricks
实施中的特定行?我如何为一个大的 numpy 数组缩放这个跨数组步幅删除?
当然,np.lib.stride_tricks.as_strided
可以做到这一点。这是一种方式 -
from numpy.lib.stride_tricks import as_strided
L = 2 # window length
shp = a.shape
strd = a.strides
out_shp = shp[0],shp[1]-L+1,L
out_strd = strd + (strd[1],)
out = as_strided(a, out_shp, out_strd).reshape(-1,L)
样本输入、输出-
In [177]: a
Out[177]:
array([[0, 1, 2, 3],
[4, 5, 6, 7]])
In [178]: out
Out[178]:
array([[0, 1],
[1, 2],
[2, 3],
[4, 5],
[5, 6],
[6, 7]])
请注意,整形的最后一步强制它在那里制作副本。但如果我们需要最终输出为 2D
,那是无法避免的。如果我们对 3D
输出没有问题,请跳过该重塑,从而实现 view
,如示例案例 -
所示
In [181]: np.shares_memory(a, out)
Out[181]: False
In [182]: as_strided(a, out_shp, out_strd)
Out[182]:
array([[[0, 1],
[1, 2],
[2, 3]],
[[4, 5],
[5, 6],
[6, 7]]])
In [183]: np.shares_memory(a, as_strided(a, out_shp, out_strd) )
Out[183]: True
我有一个非常大的 numpy 数组(145000 行 * 550 列)。我想在子数组中创建滚动切片。我试图用一个函数来实现它。函数 lagged_vals
的行为符合预期,但 np.lib.stride_tricks
的行为与我希望的不同 -
def lagged_vals(series,l):
# Garbage implementation but still right
return np.concatenate([[x[i:i+l] for i in range(x.shape[0]) if i+l <= x.shape[0]] for x in series]
,axis = 0)
# Sample 2D numpy array
something = np.array([[1,2,2,3],[2,2,3,3]])
lagged_vals(something,2) # Works as expected
# array([[1, 2],
# [2, 2],
# [2, 3],
# [2, 2],
# [2, 3],
# [3, 3]])
np.lib.stride_tricks.as_strided(something,
(something.shape[0]*something.shape[1],2),
(8,8))
# array([[1, 2],
# [2, 2],
# [2, 3],
# [3, 2], <--- across subarray stride, which I do not want
# [2, 2],
# [2, 3],
# [3, 3])
如何删除 np.lib.stride_tricks
实施中的特定行?我如何为一个大的 numpy 数组缩放这个跨数组步幅删除?
当然,np.lib.stride_tricks.as_strided
可以做到这一点。这是一种方式 -
from numpy.lib.stride_tricks import as_strided
L = 2 # window length
shp = a.shape
strd = a.strides
out_shp = shp[0],shp[1]-L+1,L
out_strd = strd + (strd[1],)
out = as_strided(a, out_shp, out_strd).reshape(-1,L)
样本输入、输出-
In [177]: a
Out[177]:
array([[0, 1, 2, 3],
[4, 5, 6, 7]])
In [178]: out
Out[178]:
array([[0, 1],
[1, 2],
[2, 3],
[4, 5],
[5, 6],
[6, 7]])
请注意,整形的最后一步强制它在那里制作副本。但如果我们需要最终输出为 2D
,那是无法避免的。如果我们对 3D
输出没有问题,请跳过该重塑,从而实现 view
,如示例案例 -
In [181]: np.shares_memory(a, out)
Out[181]: False
In [182]: as_strided(a, out_shp, out_strd)
Out[182]:
array([[[0, 1],
[1, 2],
[2, 3]],
[[4, 5],
[5, 6],
[6, 7]]])
In [183]: np.shares_memory(a, as_strided(a, out_shp, out_strd) )
Out[183]: True