Question

这是一段产生错误的代码：

error[E0277]: the trait bound `[f64; 33]: std::clone::Clone` is not satisfied
 --> <std macros>:2:1
  |
2 | $ crate :: vec :: from_elem ( $ elem , $ n ) ) ; ( $ ( $ x : expr ) , * ) => (
  | ^^^^^^^^^^^^^^^^^^^^^^^^^^^
tes3.rs:4:35: 4:58 note: in this expansion of vec! (defined in <std macros>)
  |
  = help: the following implementations were found:
  = help:   <[T; 0] as std::clone::Clone>
  = help:   <[T; 1] as std::clone::Clone>
  = help:   <[T; 2] as std::clone::Clone>
  = help:   <[T; 3] as std::clone::Clone>
  = help: and 29 others
  = note: required by `std::vec::from_elem`

当我尝试提取import pymc import numpy as np a = pymc.Normal('a', 1, 1) b = np.empty(4, dtype=object) for i in range(4): b[i] = 1*a b[i].__name__ = 'b_%i'%i M = pymc.MCMC([a,b]) M.sample(10) M.trace('b_0') # Causes a KeyError:'b_0'以及所有其他b的痕迹时，我不明白为什么会得到KeyError: 'b_0'。痕迹是不是被保存了？如果是这样，有没有办法直接轻弹一些开关来改变它而不必使用@deminministic来制作对象。

Answer 1

我仔细检查了一下，显然这条痕迹并没有得到保存。另外，＆＃34;标志变量＆＃34;为了保持跟踪不是.trace，它是.keep_trace

PyMC：在拉出轨迹时，直接更改对象的名称并不适用

1 个答案: