1

这里跟进,我有如下代码:

@jit(float_[:,:,:](float_[:,:], int_[:], int_)) 
def train_function(X, y, H):
    # do lots of stuff, including setting the arrays g and g_per_round like this:
    g = np.zeros((no_features, no_classes))
    g_per_round = np.zeros((H, no_features, no_classes))

    # do more stuff, then:    
        g_h = None
        j = 0
        print "Calculating regression coefficients per class. .."
        # building the parameters per j class
        for y1_w in zip(z.T, weights.T):
            y1, w = y1_w 
            temp_g = sm.WLS(y1, X, w).fit()  # Step 2(a)(ii)
            if g_h is None: # sometimes g *is* None, and that's fine
                   g_h = temp_g.params # this is an array of floats
            else:
                    g_h = np.c_[g_h, temp_g.params]
            j = j + 1

        if np.allclose(g,0) or g is None:
            g = g_h
        else:            
            g = g + g_h 

    # do lots more stuff, then finally:
    return g_per_round

class GentleBoostC(object):
    # init functions and stuff
    def train(self, X, y, H):
        self.g_per_round = train_function(X, y, H)    

现在我收到以下错误:

 @jit(float_[:,:,:](float_[:,:], int_[:], int_))
 more lines, etc etc etc, last few lines:
    unresolved_types, var_name)
  File "C:\Users\app\Anaconda\lib\site-packages\numba\typesystem\ssatypes.py", line 767, in promote_arrays
    assert_equal(non_array_types[0])
  File "C:\Users\app\Anaconda\lib\site-packages\numba\typesystem\ssatypes.py", line 764, in assert_equal
    var_name, result_type, other_type))
TypeError: Arrays must have consistent types in assignment for variable 'g': 'float64[:, :]' and 'none'

@jit在尝试添加以加速我的代码之前,我实际上对此没有任何问题。

4

2 回答 2

2

问题是 numba 无法知道g_h最终None分配给它的时间,g因为它的类型g_h取决于运行时流控制。换句话说,如果g_h 永远不可能是float64,那么它必须假设有时不是。

这是numba 的记录限制和一般类型推理系统的限制:

但是,有一些限制,即变量必须在控制流合并点具有可统一的类型。例如,以下代码将无法编译:

@jit def incompatible_types(arg):
    if arg > 10:
        x = "hello"
    else:
        x = 1

    return x        # ERROR! Inconsistent type for x!

解决方案是初始化g_h为兼容类型而不是= None.

Numba 的类型推断实际上非常智能,因此您可以在许多情况下在特定局部变量上混合类型而不会出现问题,只要在返回之前可以统一类型即可。阅读有关类型的 Numba 文档以获取更多信息。

于 2014-09-05T15:05:05.927 回答
2

问题是 Numba 推断g_hNoneType; 将其初始化为向量,它将正确编译:

g_h = np.zeroes((H, no_features, no_classes))
于 2014-09-05T15:02:04.297 回答