Mixing
tensorflow placeholder shape [None] [None,1] difference
I am new to tensorflow. I experimented a DQN algorithm with a section involving
a = tf.placeholder(tf.int32, shape = [None],name='A')
q = tf.reduce_sum(critic_q * tf.one_hot(a,n_outputs),axis=1,keepdims=True,name='Q')#Q value for chosen action
y = tf.placeholder(tf.float32, shape = [None],name='Y')
learning_rate = 1e-4
cost = tf.reduce_mean(tf.square(y-q))#mean squared error
global_step = tf.Variable(0,trainable=False,name='global_step')
optimizer = tf.train.AdamOptimizer(learning_rate)
training_op = optimizer.minimize(cost,global_step=global_step)
and initialized the input y with y_batch=np.zeros(nbatch). The network hardly trained at all.
Then, I switched to defining y as
y = tf.placeholder(tf.float32, shape = [None,1],name='Y')
and initialized the input with y_batch=np.zeros(nbatch).reshape(-1,1), which worked nicely.
What was happening in the first implementation?
python tensorflow
asked Jan 1 at 6:57
user15988user15988
32
add a comment |
I am new to tensorflow. I experimented a DQN algorithm with a section involving
a = tf.placeholder(tf.int32, shape = [None],name='A')
q = tf.reduce_sum(critic_q * tf.one_hot(a,n_outputs),axis=1,keepdims=True,name='Q')#Q value for chosen action
y = tf.placeholder(tf.float32, shape = [None],name='Y')
learning_rate = 1e-4
cost = tf.reduce_mean(tf.square(y-q))#mean squared error
global_step = tf.Variable(0,trainable=False,name='global_step')
optimizer = tf.train.AdamOptimizer(learning_rate)
training_op = optimizer.minimize(cost,global_step=global_step)
and initialized the input y with y_batch=np.zeros(nbatch). The network hardly trained at all.
Then, I switched to defining y as
y = tf.placeholder(tf.float32, shape = [None,1],name='Y')
and initialized the input with y_batch=np.zeros(nbatch).reshape(-1,1), which worked nicely.
What was happening in the first implementation?
python tensorflow
asked Jan 1 at 6:57
user15988user15988
32
add a comment |
I am new to tensorflow. I experimented a DQN algorithm with a section involving
a = tf.placeholder(tf.int32, shape = [None],name='A')
q = tf.reduce_sum(critic_q * tf.one_hot(a,n_outputs),axis=1,keepdims=True,name='Q')#Q value for chosen action
y = tf.placeholder(tf.float32, shape = [None],name='Y')
learning_rate = 1e-4
cost = tf.reduce_mean(tf.square(y-q))#mean squared error
global_step = tf.Variable(0,trainable=False,name='global_step')
optimizer = tf.train.AdamOptimizer(learning_rate)
training_op = optimizer.minimize(cost,global_step=global_step)
and initialized the input y with y_batch=np.zeros(nbatch). The network hardly trained at all.
Then, I switched to defining y as
y = tf.placeholder(tf.float32, shape = [None,1],name='Y')
and initialized the input with y_batch=np.zeros(nbatch).reshape(-1,1), which worked nicely.
What was happening in the first implementation?
python tensorflow
asked Jan 1 at 6:57
user15988user15988
32
I am new to tensorflow. I experimented a DQN algorithm with a section involving
a = tf.placeholder(tf.int32, shape = [None],name='A')
q = tf.reduce_sum(critic_q * tf.one_hot(a,n_outputs),axis=1,keepdims=True,name='Q')#Q value for chosen action
y = tf.placeholder(tf.float32, shape = [None],name='Y')
learning_rate = 1e-4
cost = tf.reduce_mean(tf.square(y-q))#mean squared error
global_step = tf.Variable(0,trainable=False,name='global_step')
optimizer = tf.train.AdamOptimizer(learning_rate)
training_op = optimizer.minimize(cost,global_step=global_step)
and initialized the input y with y_batch=np.zeros(nbatch). The network hardly trained at all.
Then, I switched to defining y as
y = tf.placeholder(tf.float32, shape = [None,1],name='Y')
and initialized the input with y_batch=np.zeros(nbatch).reshape(-1,1), which worked nicely.
What was happening in the first implementation?
python tensorflow
python tensorflow
asked Jan 1 at 6:57
user15988user15988
32
asked Jan 1 at 6:57
user15988user15988
32
asked Jan 1 at 6:57
user15988user15988
32
asked Jan 1 at 6:57
user15988user15988
32
asked Jan 1 at 6:57
user15988user15988
32
32
add a comment |
add a comment |
1 Answer
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Every tensor has a rank (number of dimensions) and a set of dimensions.
A placeholder with shape [1] is a placeholder with rank 1 and the dimension in position 0 of 1.
A placeholder with shape [None, 1] is a placeholder with rank 2, hence it has 2 dimensions. The first dimension (index 0) has unknown size (it will be resolved at runtime) while the second dimension (index 1) has the known size of 1.
In order to be compatible, tensors must have the same rank a dimensions.
You can read a more complete assessment about the tensors shape here: https://pgaleone.eu/tensorflow/2018/07/28/understanding-tensorflow-tensors-shape-static-dynamic/#tensors-the-basic
answered Jan 1 at 13:01
nessunonessuno
14.7k34348
I suppose you are implying that q has a dimension [None,1] as well. What happens when you make an operation of y-q, when y has dimension [None]?
– user15988
Jan 1 at 18:40
Tensorflow operations use the broadcasting semantics, that's well explained here: docs.scipy.org/doc/numpy-1.13.0/user/basics.broadcasting.html (it uses the same semantics of numpy). Hence, if there's a1dimension, the data of one operand is duplicated and the operation is repeated
– nessuno
Jan 1 at 19:51
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Every tensor has a rank (number of dimensions) and a set of dimensions.
A placeholder with shape [1] is a placeholder with rank 1 and the dimension in position 0 of 1.
A placeholder with shape [None, 1] is a placeholder with rank 2, hence it has 2 dimensions. The first dimension (index 0) has unknown size (it will be resolved at runtime) while the second dimension (index 1) has the known size of 1.
In order to be compatible, tensors must have the same rank a dimensions.
You can read a more complete assessment about the tensors shape here: https://pgaleone.eu/tensorflow/2018/07/28/understanding-tensorflow-tensors-shape-static-dynamic/#tensors-the-basic
answered Jan 1 at 13:01
nessunonessuno
14.7k34348
I suppose you are implying that q has a dimension [None,1] as well. What happens when you make an operation of y-q, when y has dimension [None]?
– user15988
Jan 1 at 18:40
Tensorflow operations use the broadcasting semantics, that's well explained here: docs.scipy.org/doc/numpy-1.13.0/user/basics.broadcasting.html (it uses the same semantics of numpy). Hence, if there's a1dimension, the data of one operand is duplicated and the operation is repeated
– nessuno
Jan 1 at 19:51
add a comment |
Every tensor has a rank (number of dimensions) and a set of dimensions.
A placeholder with shape [1] is a placeholder with rank 1 and the dimension in position 0 of 1.
A placeholder with shape [None, 1] is a placeholder with rank 2, hence it has 2 dimensions. The first dimension (index 0) has unknown size (it will be resolved at runtime) while the second dimension (index 1) has the known size of 1.
In order to be compatible, tensors must have the same rank a dimensions.
You can read a more complete assessment about the tensors shape here: https://pgaleone.eu/tensorflow/2018/07/28/understanding-tensorflow-tensors-shape-static-dynamic/#tensors-the-basic
answered Jan 1 at 13:01
nessunonessuno
14.7k34348
I suppose you are implying that q has a dimension [None,1] as well. What happens when you make an operation of y-q, when y has dimension [None]?
– user15988
Jan 1 at 18:40
Tensorflow operations use the broadcasting semantics, that's well explained here: docs.scipy.org/doc/numpy-1.13.0/user/basics.broadcasting.html (it uses the same semantics of numpy). Hence, if there's a1dimension, the data of one operand is duplicated and the operation is repeated
– nessuno
Jan 1 at 19:51
add a comment |
Every tensor has a rank (number of dimensions) and a set of dimensions.
A placeholder with shape [1] is a placeholder with rank 1 and the dimension in position 0 of 1.
A placeholder with shape [None, 1] is a placeholder with rank 2, hence it has 2 dimensions. The first dimension (index 0) has unknown size (it will be resolved at runtime) while the second dimension (index 1) has the known size of 1.
In order to be compatible, tensors must have the same rank a dimensions.
You can read a more complete assessment about the tensors shape here: https://pgaleone.eu/tensorflow/2018/07/28/understanding-tensorflow-tensors-shape-static-dynamic/#tensors-the-basic
answered Jan 1 at 13:01
nessunonessuno
14.7k34348
Every tensor has a rank (number of dimensions) and a set of dimensions.
A placeholder with shape [1] is a placeholder with rank 1 and the dimension in position 0 of 1.
A placeholder with shape [None, 1] is a placeholder with rank 2, hence it has 2 dimensions. The first dimension (index 0) has unknown size (it will be resolved at runtime) while the second dimension (index 1) has the known size of 1.
In order to be compatible, tensors must have the same rank a dimensions.
You can read a more complete assessment about the tensors shape here: https://pgaleone.eu/tensorflow/2018/07/28/understanding-tensorflow-tensors-shape-static-dynamic/#tensors-the-basic
answered Jan 1 at 13:01
nessunonessuno
14.7k34348
answered Jan 1 at 13:01
nessunonessuno
14.7k34348
answered Jan 1 at 13:01
nessunonessuno
14.7k34348
answered Jan 1 at 13:01
nessunonessuno
14.7k34348
14.7k34348
I suppose you are implying that q has a dimension [None,1] as well. What happens when you make an operation of y-q, when y has dimension [None]?
– user15988
Jan 1 at 18:40
Tensorflow operations use the broadcasting semantics, that's well explained here: docs.scipy.org/doc/numpy-1.13.0/user/basics.broadcasting.html (it uses the same semantics of numpy). Hence, if there's a1dimension, the data of one operand is duplicated and the operation is repeated
– nessuno
Jan 1 at 19:51
add a comment |
I suppose you are implying that q has a dimension [None,1] as well. What happens when you make an operation of y-q, when y has dimension [None]?
– user15988
Jan 1 at 18:40
Tensorflow operations use the broadcasting semantics, that's well explained here: docs.scipy.org/doc/numpy-1.13.0/user/basics.broadcasting.html (it uses the same semantics of numpy). Hence, if there's a1dimension, the data of one operand is duplicated and the operation is repeated
– nessuno
Jan 1 at 19:51
I suppose you are implying that q has a dimension [None,1] as well. What happens when you make an operation of y-q, when y has dimension [None]?
– user15988
Jan 1 at 18:40
I suppose you are implying that q has a dimension [None,1] as well. What happens when you make an operation of y-q, when y has dimension [None]?
– user15988
Jan 1 at 18:40
Tensorflow operations use the broadcasting semantics, that's well explained here: docs.scipy.org/doc/numpy-1.13.0/user/basics.broadcasting.html (it uses the same semantics of numpy). Hence, if there's a
1 dimension, the data of one operand is duplicated and the operation is repeated– nessuno
Jan 1 at 19:51
Tensorflow operations use the broadcasting semantics, that's well explained here: docs.scipy.org/doc/numpy-1.13.0/user/basics.broadcasting.html (it uses the same semantics of numpy). Hence, if there's a
1 dimension, the data of one operand is duplicated and the operation is repeated– nessuno
Jan 1 at 19:51
add a comment |
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