Question

从here跟进，我得到的代码如下：

@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'

在尝试添加@jit以加快我的代码之前，我确实没有遇到任何问题。

Answer 1

问题在于Numba推断g_h为NoneType;将它初始化为一个向量，它将正确编译它：

g_h = np.zeroes((H, no_features, no_classes))

Answer 2

问题是当最终将g_h分配给None时，numba无法知道g不会g_h，因为g_h的类型取决于运行时流控制。换句话说，如果@jit def incompatible_types(arg): if arg > 10: x = "hello" else: x = 1 return x # ERROR! Inconsistent type for x! 不能一个float64，那么它必须假定有时不是。

这是一个documented limitation of numba和一般类型推理系统的限制：

但是，有一些限制，即变量必须有控制流合并点的一个统一类型。例如，以下代码将无法编译：
g_h

解决方案是将= None初始化为兼容类型，而不是{{1}}。

Numba的类型推断实际上非常聪明，所以你可以在特定的局部变量上混合类型，在很多情况下只要在返回之前统一类型就可以。请阅读Numba documentation on types了解详情。

TypeError：数组在赋值时必须具有一致的类型

2 个答案: