[英]Multiple coefficient sets for least squares fitting in numpy/scipy
Is there a way to perform multiple simultaneous (but unrelated) least-squares fits with different coefficient matrices in either numpy.linalg.lstsq
or scipy.linalg.lstsq
? 有没有一种方法可以在
numpy.linalg.lstsq
或scipy.linalg.lstsq
执行具有不同系数矩阵的多个同时(但不相关)最小二乘拟合? For example, here is a trivial linear fit that I would like to be able to do with different x-values but the same y-values. 例如,这是一个简单的线性拟合,我希望能够对不同的x值但相同的y值进行拟合。 Currently, I have to write a loop:
当前,我必须编写一个循环:
x = np.arange(12.0).reshape(4, 3)
y = np.arange(12.0, step=3.0)
m = np.stack((x, np.broadcast_to(1, x.shape)), axis=0)
fit = np.stack(tuple(np.linalg.lstsq(w, y, rcond=-1)[0] for w in m), axis=-1)
This results in a set of fits with the same slope and different intercepts, such that fit[n]
corresponds to coefficients m[n]
. 这导致具有相同斜率和不同截距的一组拟合,使得
fit[n]
对应于系数m[n]
。
Linear least squares is not a great example since it is invertible, and both functions have an option for multiple y-values. 线性最小二乘不是一个很好的例子,因为它是可逆的,并且两个函数都可以选择多个y值。 However, it serves to illustrate my point.
但是,它可以说明我的观点。
Ideally, I would like to extend this to any "broadcastable" combination of a
and b
, where a.shape[-2] == b.shape[0]
exactly, and the last dimensions have to either match or be one (or missing). 理想情况下,我想将此扩展到
a
和b
任何“可广播”组合,其中a.shape[-2] == b.shape[0]
恰好一样,并且最后一个尺寸必须匹配或为一个(或失踪)。 I am not really hung up on which dimension of a
is the one representing the different matrices: it was just convenient to make it the first one to shorten the loop. 我并不是很想知道
a
哪个维度代表不同的矩阵:将它作为第一个缩短循环的维度很方便。
Is there a built in method in numpy or scipy to avoid the Python loop? 是否有numpy或scipy中的内置方法来避免Python循环? I am very much interested in using
lstsq
rather than manually transposing, multiplying and inverting the matrices. 我对使用
lstsq
感兴趣,而不是手动转置,乘法和求逆矩阵。
You could use scipy.sparse.linalg.lsqr
together with scipy.sparse.block_diag
. 您可以将
scipy.sparse.linalg.lsqr
与scipy.sparse.block_diag
一起使用。 I'm just not sure it will be any faster. 我只是不确定是否会更快。
Example: 例:
>>> import numpy as np
>>> from scipy.sparse import block_diag
>>> from scipy.sparse import linalg as sprsla
>>>
>>> x = np.random.random((3,5,4))
>>> y = np.random.random((3,5))
>>>
>>> for A, b in zip(x, y):
... print(np.linalg.lstsq(A, b))
...
(array([-0.11536962, 0.22575441, 0.03597646, 0.52014899]), array([0.22232195]), 4, array([2.27188101, 0.69355384, 0.63567141, 0.21700743]))
(array([-2.36307163, 2.27693405, -1.85653264, 3.63307554]), array([0.04810252]), 4, array([2.61853881, 0.74251282, 0.38701194, 0.06751288]))
(array([-0.6817038 , -0.02537582, 0.75882223, 0.03190649]), array([0.09892803]), 4, array([2.5094637 , 0.55673403, 0.39252624, 0.18598489]))
>>>
>>> sprsla.lsqr(block_diag(x), y.ravel())
(array([-0.11536962, 0.22575441, 0.03597646, 0.52014899, -2.36307163,
2.27693405, -1.85653264, 3.63307554, -0.6817038 , -0.02537582,
0.75882223, 0.03190649]), 2, 15, 0.6077437777160813, 0.6077437777160813, 6.226368324510392, 106.63227777368986, 1.3277892240815807e-14, 5.36589277249043, array([0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.]))
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