我正在嘗試編寫一個(gè)鑒別器，用于評(píng)估圖像的色塊。因此，我從輸入生成32x32不重疊的色塊，然后將它們串聯(lián)在新軸上。我使用時(shí)間分布層的原因是最后，鑒別器應(yīng)該將整個(gè)圖像評(píng)估為真還是假。因此，我試圖分別對(duì)每個(gè)補(bǔ)丁執(zhí)行前向傳遞，然后通過lambda層平均各個(gè)補(bǔ)丁的鑒別器輸出：def my_average(x): x = K.mean(x, axis=1) return xdef my_average_shape(input_shape): shape = list(input_shape) del shape[1] return tuple(shape)def defineD(input_shape): a = Input(shape=(256, 256, 1)) cropping_list = [] n_patches = 256/32 for x in range(256/32): for y in range(256/32): cropping_list += [ K.expand_dims( Cropping2D((( x * 32, 256 - (x+1) * 32), ( y * 32, 256 - (y+1) * 32)))(a) , axis=1) ] x = Concatenate(1)(cropping_list) x = TimeDistributed(Conv2D(4 * 8, 3, padding='same'))(x) # x = TimeDistributed(MaxPooling2D())(x) x = TimeDistributed(LeakyReLU())(x) # 16 x = TimeDistributed(Conv2D(4 * 16, 3, padding='same'))(x) x = TimeDistributed(MaxPooling2D())(x) x = TimeDistributed(LeakyReLU())(x) # 8 x = TimeDistributed(Conv2D(4 * 32, 3, padding='same'))(x) x = TimeDistributed(MaxPooling2D())(x) x = TimeDistributed(LeakyReLU())(x) # 4 x = TimeDistributed(Flatten())(x) x = TimeDistributed(Dense(2, activation='sigmoid'))(x) x = Lambda(my_average, my_average_shape)(x) return keras.models.Model(inputs=a, outputs=x)由于某種原因，我得到以下錯(cuò)誤：File "testing.py", line 41, in <module> defineD((256,256,1) ) File "testing.py", line 38, in defineD return keras.models.Model(inputs=a, outputs=x) File "/usr/local/lib/python2.7/dist-packages/keras/legacy/interfaces.py", line 91, in wrapper return func(*args, **kwargs) File "/usr/local/lib/python2.7/dist-packages/keras/engine/network.py", line 93, in __init__ self._init_graph_network(*args, **kwargs) File "/usr/local/lib/python2.7/dist-packages/keras/engine/network.py", line 237, in _init_graph_network self.inputs, self.outputs)
2 回答
LEATH
TA貢獻(xiàn)1936條經(jīng)驗(yàn) 獲得超7個(gè)贊
您需要將裁剪操作放在一個(gè)函數(shù)中，然后在Lambda圖層中使用該函數(shù)：
def my_cropping(a):
cropping_list = []
n_patches = 256/32
for x in range(256//32):
for y in range(256//32):
cropping_list += [
K.expand_dims(
Cropping2D((( x * 32, 256 - (x+1) * 32), ( y * 32, 256 - (y+1) * 32)))(a)
, axis=1)
]
return cropping_list
要使用它：
cropping_list = Lambda(my_cropping)(a)
眼眸繁星
TA貢獻(xiàn)1873條經(jīng)驗(yàn) 獲得超9個(gè)贊
我遇到了同樣的問題，它確實(shí)通過在 @today 提議的張量周圍包裹一個(gè) Lambda 層來解決。
感謝您的提示，它為我指明了正確的方向。我想將向量變成對(duì)角矩陣
我想將矢量與正方形圖像連接起來，并通過在diag矩陣中旋轉(zhuǎn)矢量來實(shí)現(xiàn)。它適用于以下代碼段：
def diagonalize(vector):
diagonalized = tf.matrix_diag(vector) # make diagonal matrix from vector
out_singlechan = tf.expand_dims(diagonalized, -1) # append 1 channel to get compatible to the multichannel image dim
return out_singlechan
lstm_out = Lambda(diagonalize, output_shape=(self.img_shape[0],self.img_shape[1],1))(lstm_out)
添加回答
舉報(bào)
0/150
提交
取消