如果我有一个大小为300个点的
换句话说:我想先选取前10个点,然后跳过20个点,再获取10个点,然后跳过10个...直到数组的末尾。
numpy.ndarray
(目前是1 x 300),我想选择每30个点中的10个,我该怎么做?换句话说:我想先选取前10个点,然后跳过20个点,再获取10个点,然后跳过10个...直到数组的末尾。
a.reshape(-1,30)[:,:10]
优点是输出将成为对输入的视图,因此几乎免费且不需要额外的内存开销。让我们进行一个示例运行以展示并证明这些内容 -
In [43]: np.random.seed(0)
In [44]: a = np.random.randint(0,9,(1,300))
In [48]: np.shares_memory(a,a.reshape(10,30)[0,:,:10])
Out[48]: True
.ravel()
-a.reshape(-1,30)[:,:10].ravel()
时间 -
In [38]: a = np.random.randint(0,9,(300))
# @sacul's soln
In [39]: %%timeit
...: msk = [True] * 10 + [False] * 20
...: out = a[np.tile(msk, len(a)//len(msk))]
100000 loops, best of 3: 7.6 µs per loop
# From this post
In [40]: %timeit a.reshape(-1,30)[:,:10].ravel()
1000000 loops, best of 3: 1.07 µs per loop
In [41]: a = np.random.randint(0,9,(3000000))
# @sacul's soln
In [42]: %%timeit
...: msk = [True] * 10 + [False] * 20
...: out = a[np.tile(msk, len(a)//len(msk))]
100 loops, best of 3: 3.66 ms per loop
# From this post
In [43]: %timeit a.reshape(-1,30)[:,:10].ravel()
100 loops, best of 3: 2.32 ms per loop
# If you are okay with `2D` output, it is virtually free
In [44]: %timeit a.reshape(-1,30)[:,:10]
1000000 loops, best of 3: 519 ns per loop
1D
数组的通用情况A. 元素数量是块长度的倍数
对于一个元素数量为n
的1D
数组a
,如果要从每个包含n
个元素的块中选择m
个元素,并获得一个1D
数组输出,则需要:
a.reshape(-1,n)[:,:m].ravel()
ravel()
函数会进行复制来实现扁平化。因此,如果可能的话,请保留未扁平化的 2D
版本以提高内存效率。In [59]: m,n = 2,5
In [60]: N = 25
In [61]: a = np.random.randint(0,9,(N))
In [62]: a
Out[62]:
array([5, 0, 3, 3, 7, 3, 5, 2, 4, 7, 6, 8, 8, 1, 6, 7, 7, 8, 1, 5, 8, 4,
3, 0, 3])
# Select 2 elements off each block of 5 elements
In [63]: a.reshape(-1,n)[:,:m].ravel()
Out[63]: array([5, 0, 3, 5, 6, 8, 7, 7, 8, 4])
B. 元素的通用编号
我们将利用np.lib.stride_tricks.as_strided
,受到这篇帖子
的启发,从每个包含n
个元素的块中选择m
个元素-
def skipped_view(a, m, n):
s = a.strides[0]
strided = np.lib.stride_tricks.as_strided
shp = ((a.size+n-1)//n,n)
return strided(a,shape=shp,strides=(n*s,s), writeable=False)[:,:m]
def slice_m_everyn(a, m, n):
a_slice2D = skipped_view(a,m,n)
extra = min(m,len(a)-n*(len(a)//n))
L = m*(len(a)//n) + extra
return a_slice2D.ravel()[:L]
skipped_view
让我们可以查看输入数组,可能还可以查看未分配给输入数组的内存区域,但之后我们会对其进行展平和切片以将其限制在所需的输出范围内,这会生成一个副本。In [170]: np.random.seed(0)
...: a = np.random.randint(0,9,(16))
In [171]: a
Out[171]: array([5, 0, 3, 3, 7, 3, 5, 2, 4, 7, 6, 8, 8, 1, 6, 7])
# Select 2 elements off each block of 5 elements
In [172]: slice_m_everyn(a, m=2, n=5)
Out[172]: array([5, 0, 3, 5, 6, 8, 7])
In [173]: np.random.seed(0)
...: a = np.random.randint(0,9,(19))
In [174]: a
Out[174]: array([5, 0, 3, 3, 7, 3, 5, 2, 4, 7, 6, 8, 8, 1, 6, 7, 7, 8, 1])
# Select 2 elements off each block of 5 elements
In [175]: slice_m_everyn(a, m=2, n=5)
Out[175]: array([5, 0, 3, 5, 6, 8, 7, 7])
msk = [True] * 10 + [False] * 20
arr[np.tile(msk, len(arr)//len(msk))]
最简单的例子:
从一个包含30个元素的数组中选择1个元素,然后跳过2个元素:
>>> arr
array([6, 7, 2, 7, 1, 9, 1, 4, 4, 8, 6, 5, 2, 6, 3, 6, 8, 5, 6, 7, 2, 1, 9,
6, 7, 2, 1, 8, 2, 2])
msk = [True] * 1 + [False] * 2
>>> arr[np.tile(msk, len(arr)//len(msk))]
array([6, 7, 1, 8, 2, 6, 6, 1, 7, 8])
解释:
msk
是一个布尔掩码。
>>> msk
[True, False, False]
您可以使用np.tile
重复该掩码,直到它与您的原始数组长度相同(即您的数组长度除以掩码长度):
>>> np.tile(msk, len(arr)//len(msk))
array([ True, False, False, True, False, False, True, False, False,
True, False, False, True, False, False, True, False, False,
True, False, False, True, False, False, True, False, False,
True, False, False], dtype=bool)
然后,只需按布尔值进行索引,这是 numpy
擅长的简单操作。
IIUC
get = 10
skip = 20
k = [item for z in [np.arange(get) + idx for idx in np.arange(0, x.size, skip+get)] for item in z]
然后只需切片
x[k]
例子:
x = np.arange(100)
x[k]
array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 30, 31, 32, 33, 34, 35, 36,
37, 38, 39, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 90, 91, 92, 93,
94, 95, 96, 97, 98, 99])