使用 PYMC3/Theano 广播数学运算

Question

我认为这个问题归结为我对Theano作品缺乏了解。我处于一种情况，我想创建一个变量，该变量是分布和 numpy 数组之间相减的结果。当我将形状参数指定为 1

时，这很好用

import pymc3 as pm
import numpy as np
import theano.tensor as T

X = np.random.randint(low = -10, high = 10, size = 100)

with pm.Model() as model:
    nl = pm.Normal('nl', shape = 1)
    det = pm.Deterministic('det', nl - x)

nl.dshape
(1,)

但是，当我指定形状 > 1 时，这会中断

with pm.Model() as model:
    nl = pm.Normal('nl', shape = 2)
    det = pm.Deterministic('det', nl - X)

ValueError: Input dimension mis-match. (input[0].shape[0] = 2, input[1].shape[0] = 100)

nl.dshape
(2,)

X.shape
(100,)

我尝试转置 X 使其可广播

X2 = X.reshape(-1, 1).transpose()

X2.shape
(1, 100)

但现在它在 .shape[1] 而不是 .shape[0]

处声明不匹配

with pm.Model() as model:
    nl = pm.Normal('nl', shape = 2)
    det = pm.Deterministic('det', nl - X2)

ValueError: Input dimension mis-match. (input[0].shape[1] = 2, input[1].shape[1] = 100)

如果我遍历分布的元素，我可以完成这项工作

distShape = 2
with pm.Model() as model:
    nl = pm.Normal('nl', shape = distShape)

    det = {}
    for i in range(distShape):
        det[i] = pm.Deterministic('det' + str(i), nl[i] - X)

det
{0: det0, 1: det1}

然而，这让人感觉不够优雅，并迫使我对模型的其余部分使用循环。我想知道是否有一种方法可以指定此操作，以便它可以像分发版一样工作。

distShape = 2
with pm.Model() as model:
    nl0 = pm.Normal('nl1', shape = distShape)
    nl1 = pm.Normal('nl2', shape = 1)

    det = pm.Deterministic('det', nl0 - nl1)

Answer 1

你可以做到

X = np.random.randint(low = -10, high = 10, size = 100)
X = x[:,None] # or x.reshape(-1, 1)

然后

with pm.Model() as model:
    nl = pm.Normal('nl', shape = 2)
    det = pm.Deterministic('det', nl - X)

在这种情况下，nl 和 X 的形状将分别为 ((2, 1), (100,))，然后可广播。

注意我们得到了两个 NumPy 数组的相同行为（不仅是一个 Theano 张量和一个 NumPy 数组）

a0 = np.array([1,2])
b0 = np.array([1,2,3,5])
a0 = a0[:,None]  # comment/uncomment this line
print(a0.shape, b0.shape)
b0-a0

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