从行向量创建块矩阵的最佳方法是什么？ [英] What is the best way to create a block matrix form a row vector?

问题描述

我有以下numpy行矩阵。

I have the following numpy row matrix.

X = np.array([1,2,3])

我要创建一个块矩阵，如下所示：

I want to create a block matrix as follows:

1 0 0
2 1 0
3 2 1
0 3 2
0 0 3

如何使用numpy做到这一点？

How can I do this using numpy?

推荐答案

方法1：使用 np.lib.stride_tricks.as_strided -

from numpy.lib.stride_tricks import as_strided as strided

def zeropad_arr_v1(X):
    n = len(X)
    z = np.zeros(len(X)-1,dtype=X.dtype)
    X_ext = np.concatenate(( z, X, z))

    s = X_ext.strides[0]
    return strided(X_ext[n-1:], (2*n-1,n), (s,-s), writeable=False)

请注意，这将创建只读输出。如果以后需要写，只需在末尾附加 .copy（）即可进行复制。

Note that this would create a read-only output. If you need to write into later on, simply make a copy by appending .copy() at the end.

方法2：使用带零的连接，然后剪切/切片-

Approach #2 : Using concatenation with zeros and then clipping/slicing -

def zeropad_arr_v2(X):
    n = len(X)
    X_ext = np.concatenate((X, np.zeros(n,dtype=X.dtype)))
    return np.tile(X_ext, n)[:-n].reshape(-1,n,order='F')

方法＃1作为基于跨步的方法应该在性能上非常有效。

Approach #1 being a strides-based method should be very efficient on performance.

示例运行-

In [559]: X = np.array([1,2,3])

In [560]: zeropad_arr_v1(X)
Out[560]: 
array([[1, 0, 0],
       [2, 1, 0],
       [3, 2, 1],
       [0, 3, 2],
       [0, 0, 3]])

In [561]: zeropad_arr_v2(X)
Out[561]: 
array([[1, 0, 0],
       [2, 1, 0],
       [3, 2, 1],
       [0, 3, 2],
       [0, 0, 3]])

运行时间测试

In [611]: X = np.random.randint(0,9,(1000))

# Approach #1 (read-only)
In [612]: %timeit zeropad_arr_v1(X)
100000 loops, best of 3: 8.74 µs per loop

# Approach #1 (writable)
In [613]: %timeit zeropad_arr_v1(X).copy()
1000 loops, best of 3: 1.05 ms per loop

# Approach #2
In [614]: %timeit zeropad_arr_v2(X)
1000 loops, best of 3: 705 µs per loop

# @user8153's solution
In [615]: %timeit hstack_app(X)
100 loops, best of 3: 2.26 ms per loop

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