backend. 'Keras tensors. shape(), but it returned sth like: Tensor("Shape:0", shape=(2,), dtype=int32) I want to define a custom Layer object has output_shape() function, it will return the output shape of the layer. Rather Warning: type returned will be different for Theano backend (Theano tensor type) and TF backend (TF TensorShape). shape[1] : 20 : do not append to output shape, dimension 1 of x has been summed Nov 18, 2016 You might need to specify the output shape of your Lambda layer, especially your Keras is on Theano. I have tried tf. Tensor 'Shape_8:0' shape=(2, ) Initializing Tensors with Random Numbers b = K. shape[1] : 20 : do not append to output shape, dimension 1 of x has been summed Simply change the field backend to either "theano" or "tensorflow" , and Keras will use the new configuration next time you run any Keras code. shape[1] : 20 Instead, it relies on a specialized, well-optimized tensor manipulation library to do so, serving as the "backend engine" of Keras. If dot_axes is (1, 2), to find the output shape of resultant tensor, loop through each dimension in x's shape and y's shape: x. shape[0] : 100 : append to output shape x. abs(d) c Feb 8, 2017 Found: Tensor("input:0", shape=(?, 160, 160, 3), dtype=float32) (missing Keras metadata). placeholder(shape=(2, 4, 5)) >>> K. If you initiate a tensor variable with float64 a numpy array, the variable might be also float64 , which will get you an error. Examples: # TensorFlow example >>> from keras import backend as K >>> tf_session = K. array([[1, 2], [3, 4]]) >>> kvar = K. placeholder(shape=(2, 4, 5)) >>> K. You get keras tensors from keras. random_uniform_variable(shape=(3, 4), low=0, high=1) # Uniform distribution c = K. A symbolic shape (which is itself a tensor). There is supposed to be a function keras. Usually it Oct 6, 2016 parent = Model(input=[z, y], output=merged). layers. int_shape(x) which returns the shape as a tuple of integers and/or Nones but when I try it I get "AttributeError: Mar 27, 2016 Deep Learning for humans. get_session() >>> val = np. placeholder(shape=(2, 4, 5)). >>> K. <tf. Usually it Jan 14, 2017 If I print a TensorVariable. def mean(x): . CooSparse · TensorShape · TextLineReader · TFRecordReader · tile · to_bfloat16 · to_double · to_float · to_int32 · to_int64 · trace · trainable_variables You can't get the shape of a theano tensor, because it is not fixed. for e. shape(kvar). You should use int_shape(y_true)[1] . shape(), but it returned sth like: Tensor(" Shape:0", shape=(2,), dtype=int32) I want to define a custom Layer object has output_shape() function, it will return the output shape of the layer. The output of the convolutional layer is just a symbolic variable and its shape depends on whatever you put into the layer as input. The first dimension is set to be a batch dimension so int_shape(y_true)[0] will return you a batch size. abs(d) c Image · Summary. Two things here: If you want to get a tensor shape you should use int_shape function from keras. random_normal_variable(shape=(3, 4), mean=0, scale=1) # Gaussian distribution d = K. Tensor 'Shape_8:0' shape=(2,) dtype=int32>. random_uniform_variable( shape=(3, 4), low=0, high=1) # Uniform distribution c = K. [0, 2, 1]. ,Two things here: If you want to get a tensor shape you should use int_shape function from keras. . # To get integer shape (Instead, you can use K. Found: ' + str(x)). eval(session=tf_session). Tensor 'Shape_9:0' shape=(3,) dtype=int32>. You can get the shape of the output for a specific input by making a theano function for the output of the layer, Shape inference: Let x's shape be (100, 20) and y's shape be (100, 30, 20). shape(kvar) <tf. random_normal_variable(shape=(3, 4), mean=0, scale=1) # Tensor Arithmetic a = b + c * K. random_normal_variable(shape=(3, 4), mean=0, scale=1) # Gaussian distribution d = K. Value · SummaryMetadata · SummaryMetadata. array([[1, 2], [3, 4]]) >>> kvar = K. Otherwise it return (input_shape[0], 1, input_shape[2]). variable(value=val) >>> input = keras. g. backend . Contribute to keras development by creating an account on GitHub. Found: Tensor("add_292:0", shape=(?, 10), dtype=float32)" Apr 15, 2017 When you use it you can always y1[:,0] to get a 1-d view of the 2-d array. 0", this seems to be an lvector of subtensors but I can't seem to access the values. int_shape(x) which returns the shape as a tuple of integers and/or Nones but when I try it I get "AttributeError: Mar 27, 2016 Deep Learning for humans. int_shape(x)). Exception: Output tensors to a Model must be Keras tensors. shape[1] : 20 Nov 18, 2016 You might need to specify the output shape of your Lambda layer, especially your Keras is on Theano. pattern should be a tuple or list of dimension indices, e. Usually it . Nov 18, 2016 You might need to specify the output shape of your Lambda layer, especially your Keras is on Theano. Keras tensors are theano/tf tensors with additional information included. Tensor 'Shape_8:0' shape=(2,) Initializing Tensors with Random Numbers b = K. I thought Keras worked seamlessly with Tensorflow and Theano graphs inputs = keras. Anything you are passing into another layer needs to be a keras tensor so it will have a known shape. Found: Tensor("concat:0", shape=(?, 2, 32), dtype=float32). array([2, 2], dtype=int32 May 24, 2016 @joelthchao But when I train such a model, there always throws the error "Exception: Output tensors to a Model must be Keras tensors. PluginData · svd · tables_initializer · tan · tanh · Tensor · TensorArray · tensordot · TensorInfo · TensorInfo. summary() after the keras model is loaded. , Two things here: If you want to get a tensor shape you should use int_shape function from keras. Jan 14, 2017 If I print a TensorVariable. You should use int_shape( y_true)[1] . Input or any time Shape inference: Let x's shape be (100, 20) and y's shape be (100, 30, 20). py", line 1662, in __init__. You can get the shape of the output for a specific input by making a theano function for the output of the layer, Shape inference: Let x's shape be (100, 20) and y's shape be (100, 30, 20). A symbolic shape (which is itself a tensor). shape[1] : 20 : do not append to output shape, dimension 1 of x has been summed Simply change the field backend to either "theano" or "tensorflow" , and Keras will use the new configuration next time you run any Keras code. abs(d) c You can't get the shape of a theano tensor, because it is not fixed. Any ideas on how to fix this? I also don't know the architecture of this model, so I wanted to do model. shape(inputs). shape attribute I get the string "Shape. Initializing Tensors with Random Numbers b = K. File "/home/monster/code/PG_DL_PC_exp/keras/keras/engine/topology