Question

我取一列的第3个值（第三个）并将这些值放在3个新列的行中。并将新旧列合并为一个新的矩阵A

在col nr 1和2中的col nr3值中输入时间序列

[x x 1]
[x x 2]
[x x 3]

输出：矩阵A

[x x 1 0 0 0]
[x x 2 0 0 0]
[x x 3 1 2 3]
[x x 4 2 3 4]

为简洁起见，首先代码生成矩阵6行/ 3列。我想用最后一列填充3个额外的列并将其合并到一个新的矩阵A中。这个矩阵A预先填充了2行以抵消起始位置。

我在下面的代码中实现了这个想法，处理大型数据集需要很长时间。如何提高转换速度

import  numpy as np

matrix = np.arange(18).reshape((6, 3))

nr=3 
A = np.zeros((nr-1,nr))

for x in range( matrix.shape[0]-nr+1):
    newrow =  (np.transpose( matrix[x:x+nr,2:3] ))
    A = np.vstack([A , newrow])

total= np.column_stack((matrix,A))
print (total)

Answer 1

这是一种方法，使用broadcasting获取那些滑动窗口元素，然后只是一些堆叠来获取A -

col2 = matrix[:,2]
nrows = col2.size-nr+1
out = np.zeros((nr-1+nrows,nr))
col2_2D = np.take(col2,np.arange(nrows)[:,None] + np.arange(nr))
out[nr-1:] = col2_2D

这是使用NumPy strides获取col2_2D的有效替代方法 -

n = col2.strides[0]
col2_2D = np.lib.stride_tricks.as_strided(col2, shape=(nrows,nr), strides=(n,n))

最好将大小为total的零输出数组初始化，然后使用col2_2D为其分配值，最后使用输入数组matrix。

运行时测试

作为功能的方法 -

def org_app1(matrix,nr):    
    A = np.zeros((nr-1,nr))
    for x in range( matrix.shape[0]-nr+1):
        newrow =  (np.transpose( matrix[x:x+nr,2:3] ))
        A = np.vstack([A , newrow])
    return A

def vect_app1(matrix,nr):    
    col2 = matrix[:,2]
    nrows = col2.size-nr+1
    out = np.zeros((nr-1+nrows,nr))
    col2_2D = np.take(col2,np.arange(nrows)[:,None] + np.arange(nr))
    out[nr-1:] = col2_2D
    return out

def vect_app2(matrix,nr):    
    col2 = matrix[:,2]
    nrows = col2.size-nr+1
    out = np.zeros((nr-1+nrows,nr))
    n = col2.strides[0]
    col2_2D = np.lib.stride_tricks.as_strided(col2, \
                        shape=(nrows,nr), strides=(n,n))
    out[nr-1:] = col2_2D
    return out

计时和验证 -

In [18]: # Setup input array and params
    ...: matrix = np.arange(1800).reshape((60, 30))
    ...: nr=3
    ...: 

In [19]: np.allclose(org_app1(matrix,nr),vect_app1(matrix,nr))
Out[19]: True

In [20]: np.allclose(org_app1(matrix,nr),vect_app2(matrix,nr))
Out[20]: True

In [21]: %timeit org_app1(matrix,nr)
1000 loops, best of 3: 646 µs per loop

In [22]: %timeit vect_app1(matrix,nr)
10000 loops, best of 3: 20.6 µs per loop

In [23]: %timeit vect_app2(matrix,nr)
10000 loops, best of 3: 21.5 µs per loop

In [28]: # Setup input array and params
    ...: matrix = np.arange(7200).reshape((120, 60))
    ...: nr=30
    ...: 

In [29]: %timeit org_app1(matrix,nr)
1000 loops, best of 3: 1.19 ms per loop

In [30]: %timeit vect_app1(matrix,nr)
10000 loops, best of 3: 45 µs per loop

In [31]: %timeit vect_app2(matrix,nr)
10000 loops, best of 3: 27.2 µs per loop

将列值Numpy转换为行值

1 个答案: