Python 3.x keras中的Conv2D图层输出形状

我想手动输入keras Conv2D图层

我取MNIST数据集

Conv2D只接受张量，因此我使用keras的

输入

命令将

x\u列

更改为

x\u列

张量

我的输入采用keras说明中给出的格式

(samples,rows, cols,channels) 

输入示例：

(60000,128,128,1) 

我希望输出类似于：

(None, 26, 26, 32) 

我得到：

shape=(?, 59998, 26, 32)

我做错了什么

我的代码：

import keras
from keras.datasets import mnist
from keras.layers import Conv2D
from keras import backend as K
from keras.layers import Input
batch_size = 128
num_classes = 10
epochs = 1
# input image dimensions
img_rows, img_cols = 28, 28
# the data, shuffled and split between train and test sets
(x_train, y_train), (x_test, y_test) = mnist.load_data()
if K.image_data_format() == 'channels_first':
    x_train = x_train.reshape(x_train.shape[0], 1, img_rows, img_cols)
    x_test = x_test.reshape(x_test.shape[0], 1, img_rows, img_cols)
    input_shape = (1, img_rows, img_cols)
else:
    x_train = x_train.reshape(x_train.shape[0], img_rows, img_cols, 1)
    x_test = x_test.reshape(x_test.shape[0], img_rows, img_cols, 1)
    input_shape = (img_rows, img_cols, 1)

x_train = x_train.astype('float32')
x_test = x_test.astype('float32')
x_train /= 255
x_test /= 255
print('x_train shape:', x_train.shape)
print(x_train.shape[0], 'train samples')
print(x_test.shape[0], 'test samples')

x_train_tensor=Input(shape=(60000,28,28), name='x_train')
A=Conv2D(32, kernel_size=(3, 3),
       activation='relu',
       input_shape=input_shape)(x_train_tensor)

---

样本数量不是

输入形状的一部分，在您的情况下，您犯了两个错误。第一个是输入形状错误，第二个是指定两个输入形状，一个在输入构造函数中，另一个在Conv2D实例中：

x_train_tensor=Input(shape=(28, 28, 1), name='x_train')
A=Conv2D(32, kernel_size=(3, 3), activation='relu')(x_train_tensor)

样本数量不是
输入形状的一部分，在您的情况下，您犯了两个错误。第一个是输入形状错误，第二个是指定两个输入形状，一个在输入构造函数中，另一个在Conv2D实例中：

x_train_tensor=Input(shape=(28, 28, 1), name='x_train')
A=Conv2D(32, kernel_size=(3, 3), activation='relu')(x_train_tensor)