Vectorized way to construct a block Hankel matrix in numpy (or scipy)
Question:
I want to contrsuct the following matrix :
[v0 v1 v2 v3 .... v(M-d+1)
v1 .
v2 . .
. .
.
vd . . v(M) ]
where each v(k) is a (ndarray) vector, say from a matrix
X = np.random.randn(100, 8)
M = 7
d = 3
v0 = X[:, 0]
v1 = X[:, 1]
...
Using a for loop, I can do something like this for example:
v1 = np.array([1, 2, 3]).reshape((-1, 1))
v2 = np.array([10, 20, 30]).reshape((-1, 1))
v3 = np.array([100, 200, 300]).reshape((-1, 1))
v4 = np.array([100.1, 200.1, 300.1]).reshape((-1, 1))
v5 = np.array([1.1, 2.2, 3.3]).reshape((-1, 1))
X = np.hstack((v1, v2, v3, v4, v5))
d = 2
X_ = np.zeros((d * X.shape[0], X.shape[1]+1-d))
for i in range (d):
X_[i*X.shape[0]:(i+1) * X.shape[0], :] = X[:X.shape[0], i:i+(X.shape[1]+1-d)]
And I get :
X_ = array([[ 1. , 10. , 100. , 100.1],
[ 2. , 20. , 200. , 200.1],
[ 3. , 30. , 300. , 300.1],
[ 10. , 100. , 100.1, 1.1],
[ 20. , 200. , 200.1, 2.2],
[ 30. , 300. , 300.1, 3.3]]) #Which is the wanted matrix
Is there any way to construct this matrix in a vectorized way (which I imagine would be faster than for loops when it comes to large matrices ?).
Thank you.
Answers:
This looks about optimal; you did a good job vectorizing it already. The only improvement I can make is to replace np.zeros
with np.empty
, which skips initializing the array. I tried using np.vstack
and np.lib.stride_tricks.sliding_window_view
(after https://stackoverflow.com/a/60581287) and got the same performance as the for loop with np.empty
.
# sliding window:
X_ = np.lib.stride_tricks.sliding_window_view(X, (X.shape[0], X.shape[1]+1-d)).reshape(d*X.shape[0], -1)
# np.vstack:
X_ = np.vstack([X[:, i:i+(X.shape[1]+1-d)] for i in range(d)])
I didn’t understand the second part of the question exactly, but I was looking to create a Hankel-like matrix (non-square) just like the first part of the question really fast and I couldn’t find any answer for it. Here is how I did it:
N = 7
K = 4
L = N-K+1
B = np.vstack([range(K)]*L)
print(B.T+range(K))
Output:
[[0 1 2 3]
[1 2 3 4]
[2 3 4 5]
[3 4 5 6]]
You can then use the above matrix to index your 1-dimensional array.
I want to contrsuct the following matrix :
[v0 v1 v2 v3 .... v(M-d+1)
v1 .
v2 . .
. .
.
vd . . v(M) ]
where each v(k) is a (ndarray) vector, say from a matrix
X = np.random.randn(100, 8)
M = 7
d = 3
v0 = X[:, 0]
v1 = X[:, 1]
...
Using a for loop, I can do something like this for example:
v1 = np.array([1, 2, 3]).reshape((-1, 1))
v2 = np.array([10, 20, 30]).reshape((-1, 1))
v3 = np.array([100, 200, 300]).reshape((-1, 1))
v4 = np.array([100.1, 200.1, 300.1]).reshape((-1, 1))
v5 = np.array([1.1, 2.2, 3.3]).reshape((-1, 1))
X = np.hstack((v1, v2, v3, v4, v5))
d = 2
X_ = np.zeros((d * X.shape[0], X.shape[1]+1-d))
for i in range (d):
X_[i*X.shape[0]:(i+1) * X.shape[0], :] = X[:X.shape[0], i:i+(X.shape[1]+1-d)]
And I get :
X_ = array([[ 1. , 10. , 100. , 100.1],
[ 2. , 20. , 200. , 200.1],
[ 3. , 30. , 300. , 300.1],
[ 10. , 100. , 100.1, 1.1],
[ 20. , 200. , 200.1, 2.2],
[ 30. , 300. , 300.1, 3.3]]) #Which is the wanted matrix
Is there any way to construct this matrix in a vectorized way (which I imagine would be faster than for loops when it comes to large matrices ?).
Thank you.
This looks about optimal; you did a good job vectorizing it already. The only improvement I can make is to replace np.zeros
with np.empty
, which skips initializing the array. I tried using np.vstack
and np.lib.stride_tricks.sliding_window_view
(after https://stackoverflow.com/a/60581287) and got the same performance as the for loop with np.empty
.
# sliding window:
X_ = np.lib.stride_tricks.sliding_window_view(X, (X.shape[0], X.shape[1]+1-d)).reshape(d*X.shape[0], -1)
# np.vstack:
X_ = np.vstack([X[:, i:i+(X.shape[1]+1-d)] for i in range(d)])
I didn’t understand the second part of the question exactly, but I was looking to create a Hankel-like matrix (non-square) just like the first part of the question really fast and I couldn’t find any answer for it. Here is how I did it:
N = 7
K = 4
L = N-K+1
B = np.vstack([range(K)]*L)
print(B.T+range(K))
Output:
[[0 1 2 3]
[1 2 3 4]
[2 3 4 5]
[3 4 5 6]]
You can then use the above matrix to index your 1-dimensional array.