Question

我目前正在尝试掌握Surag Nair关于神经网络的工作。

在（https://github.com/suragnair/alpha-zero-general/blob/master/othello/keras/OthelloNNet.py）中有以下代码行：

self.input_boards = keras.Input(shape=(self.board_x, self.board_y))    # s: batch_size x board_x x board_y
x_image = keras.layers.Reshape((self.board.grid_shape, 1))(self.input_boards)  # ??

x_image = (shape)(tensor)如何有效通话？输入返回张量，整形返回形状。据我了解，张量是潜在的高级矩阵和运算，可能会在以后完成。但是，如果我以后要调用它，输入形状必须是张量的参数，而不是相反？

我尝试对其进行测试，结果如下：

class Connect4NN():
    def __init__(self, board, args):
        self.board = board
        self.action_size = board.action_space 
        self.args = args

        #Neural Net
        self.input_boards = keras.Input(shape = (self.board.grid_shape) ) #shape: batch size x X x Y (batch size not needed here)

        #tf.Tensorobject represents a partially defined computation that will eventually produce a value.
        self.input_boards = keras.Input(shape=self.board.grid_shape)    # s: batch_size x board_x x board_y
        x_image = keras.layers.Reshape((self.board.grid_shape, 1))(self.input_boards)  # ?? 

        print("input_boards : {}".format(self.input_boards))
        print("x_image: {}".format(x_image))
        return

在控制台中：

b = Connect4()

args = "bla"

nn = Connect4NN(b,args)
Traceback (most recent call last):

  File "<ipython-input-20-7d7efde846db>", line 1, in <module>
    nn = Connect4NN(b,args)

  File "D:/Programming/code/conn4neuralnet.py", line 22, in __init__
    x_image = keras.layers.Reshape((self.board.grid_shape, 1))(self.input_boards)  # ??

  File "D:\Anaconda\lib\site-packages\keras\engine\base_layer.py", line 457, in __call__
    output = self.call(inputs, **kwargs)

  File "D:\Anaconda\lib\site-packages\keras\layers\core.py", line 401, in call
    return K.reshape(inputs, (K.shape(inputs)[0],) + self.target_shape)

  File "D:\Anaconda\lib\site-packages\keras\backend\tensorflow_backend.py", line 1969, in reshape
    return tf.reshape(x, shape)

  File "D:\Anaconda\lib\site-packages\tensorflow\python\ops\gen_array_ops.py", line 7178, in reshape
    "Reshape", tensor=tensor, shape=shape, name=name)

  File "D:\Anaconda\lib\site-packages\tensorflow\python\framework\op_def_library.py", line 529, in _apply_op_helper
    (input_name, err))

ValueError: Tried to convert 'shape' to a tensor and failed. Error: Shapes must be equal rank, but are 1 and 0
    From merging shape 1 with other shapes. for 'reshape_1/Reshape/packed' (op: 'Pack') with input shapes: [], [2], [].

添加Connect4位板实现的相关部分：

class Connect4():


    def __init__(self):
        self.board = [0, 0]
        self.height = [0,7,14,21,28,35,42]
        self.nn_height = [0,6,12,18,24,30,36]
        self.counter = 0
        self.moves = []
        self.nn_board = np.zeros(shape = 42, dtype = int)
        self.grid_shape = (6,7)
        self.action_space = 7

Answer 1

您有一个误解，Reshape是执行整形操作的图层，它不会“返回形状”。它采用一个符号输入张量，并根据构造函数的形状返回经过重整的张量。

我们的代码中的问题是形状不正确，形状是带有整数的元组，似乎您的元组内部包含另一个元组，并且不支持。这段代码可以很好地作为示例：

x = Input((2,2))
>>> x
<tf.Tensor 'input_2:0' shape=(?, 2, 2) dtype=float32>

x = Reshape((2,2,1))(x)
>>> x
<tf.Tensor 'reshape_1/Reshape:0' shape=(?, 2, 2, 1) dtype=float32>

var =（形状）（张量）？

1 个答案: