TensorFlow邏輯運算（（A == B）&&（C == D））導致“形狀不兼容：[2]與[3]”

Question

我正在嘗試構建以下邏輯表達式-

tf.logical_and(tf.equal(tf.argmax(y_conv,0), tf.argmax(y_,0)), 
                tf.equal(tf.argmax(y_conv,1), tf.argmax(y_,1)), name=None)

但這會導致以下錯誤-

不兼容的形狀：[2]與[3]

tf.equal(tf.argmax(y_conv,0), tf.argmax(y_,0))和tf.equal(tf.argmax(y_conv,1), tf.argmax(y_,1))工作正常，僅tf.logical_and會發生錯誤。 tf.logical_and期望布爾張量，而tf.equal返回布爾張量，因此所有參數都順序正確，因此不確定為什么失敗。

給予一定的情況下，原來的代碼如下，我只是想更新correct_prediction包括兩個0和1 tf.argmax

UPDATE1開始（這將添加所有變量聲明）

sess = tf.InteractiveSession()
def weight_variable(shape):
    initial = tf.truncated_normal(shape, stddev=0.1)
    return tf.Variable(initial)
def bias_variable(shape):
    initial = tf.constant(0.1, shape=shape)
    return tf.Variable(initial)
def conv2d(x, W): return tf.nn.conv2d(x, W, strides=[1, 1, 1, 1], padding='SAME')
def max_pool_2x2(x): return tf.nn.max_pool(x, ksize=[1, 2, 2, 1], strides=[1, 2, 2, 1], padding='SAME')
W_conv1 = weight_variable([3, 3, 1, 32])
b_conv1 = bias_variable([32])
x = tf.placeholder(tf.float32, shape=[None, 9])
x_image = tf.reshape(x, [-1,3,3,1])
h_conv1 = tf.nn.relu(conv2d(x_image, W_conv1) + b_conv1)
h_pool1 = max_pool_2x2(h_conv1)
h_pool2_flat = tf.reshape(h_conv1, [-1, 3 * 3 * 32])
h_fc1 = tf.nn.relu(tf.matmul(h_pool2_flat, W_fc1) + b_fc1)
W_fc2 = weight_variable([64, 2])
b_fc2 = bias_variable([2])
h_fc1_drop = tf.nn.dropout(h_fc1, keep_prob)

UPDATE1 End（這將添加所有變量聲明）

這是問題所在的位置-

y_ = tf.placeholder(tf.float32, shape=[None, 2])
y_conv = tf.matmul(h_fc1_drop, W_fc2) + b_fc2
#This works - correct_prediction = tf.equal(tf.argmax(y_conv,0), tf.argmax(y_,0)) . Changed it to -
correct_prediction = tf.logical_and(tf.equal(tf.argmax(y_conv,0), tf.argmax(y_,0)), 
                tf.equal(tf.argmax(y_conv,1), tf.argmax(y_,1)), 
accuracy = tf.reduce_mean(tf.cast(correct_prediction, tf.float32))
sess.run(tf.global_variables_initializer())
train_step.run(feed_dict={x: xtrain, y_: ytrain, keep_prob: 0.5})
#In debugging mode, code breaks at the below line
print("test accuracy %g"%accuracy.eval(feed_dict={x: xtest, y_: ytest, keep_prob: 1.0}))

如何調試此錯誤？

Answer 1

出現問題是因為tf.equal()是元素操作，並且它返回張量的形狀與其參數相同。 修復你表達的最簡單方法是使用tf.reduce_all()聚合的結果tf.equal()計算之前下降到標tf.logical_and()如下所示：

tf.logical_and(
    tf.reduce_all(tf.equal(tf.argmax(y_conv, 0), tf.argmax(y_, 0))), 
    tf.reduce_all(tf.equal(tf.argmax(y_conv,1), tf.argmax(y_,1))))