var =（形状）（张量）？答案

【问题标题】：var = (shape) (tensor)?var =（形状）（张量）？
【发布时间】：2019-07-27 10:24:29
【问题描述】：

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)中有这行代码：

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) 如何是一个有效的呼叫？输入返回一个张量，重塑返回一个形状。据我了解，张量是一个潜在的高级矩阵和一个操作，可能会在以后完成。但如果我想稍后调用它，输入形状必须是张量的参数，而不是相反？

我试过了，结果是这样的：

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], [].

添加我的 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

【问题讨论】：

self.board.grid_shape的值是多少？
返回一个 (6,7) 形式的元组
然后产生 (6,7,1) 你应该做 self.board.grid_shape + (1,)
好的，谢谢！我只是直接在板上添加了一个 x 和 y 组件，以使其更清晰，但这也可以。我对python语法还不是很熟悉，所以像添加元组、尾随逗号、星号等所有这些事情真的很难：P

标签： python tensorflow keras

【解决方案1】：

你有一个误解，Reshape 是一个执行整形操作的层，它不会“返回一个形状”。它接受一个符号输入张量并返回重构后的张量，给定构造函数中的形状。

我们的代码中的问题是形状不正确，一个形状是一个带有整数的元组，看起来你的元组里面包含另一个元组，这是不支持的。此代码作为示例运行良好：

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>

【讨论】：