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I encountered a problem to reshape an intermediate 4D tensorflow tensor X to a 3D tensor Y, where
X is of shape ( batch_size, nb_rows, nb_cols, nb_filters )
Y is of shape ( batch_size, nb_rows*nb_cols, nb_filters )
batch_size = None
Of course, when nb_rows and nb_cols are known integers, I can reshape X without any problem. However, in my application I need to deal with the case
nb_rows = nb_cols = None
What should I do? I tried Y = tf.reshape( X, (-1, -1, nb_filters)) but it clearly fails to work.
For me, this operation is deterministic because it always squeezes the two middle axes into a single one while keeping the first axis and the last axis unchanged. Can anyone help me?
In this case you can access to the dynamic shape of X through tf.shape(X):
shape = [tf.shape(X)[k] for k in range(4)]
Y = tf.reshape(X, [shape[0], shape[1]*shape[2], shape[3]])
Say I have 2 tensors, one with shape (10,1) and another one with shape (10, 11, 1)... what I want is to multiply those broadcasting along the first axis, and not the last one, as used to
tf.zeros([10,1]) * tf.ones([10,12,1])
however this is not working... is there a way to do it without transposing it using perm?
You cannot change the broadcasting rules, but you can prevent broadcasting by doing it yourself. Broadcasting takes effect if the ranks are different.
So instead of permuting the axes, you can also repeat along a new axis:
import tensorflow as tf
import einops as ops
a = tf.zeros([10, 1])
b = tf.ones([10, 12, 1])
c = ops.repeat(a, 'x z -> x y z', y=b.shape[1]) * b
c.shape
>>> TensorShape([10, 12, 1])
For the above example, you need to do tf.zeros([10,1])[...,None] * tf.ones([10,12,1]) to satisfy broadcasting rules: https://numpy.org/doc/stable/user/basics.broadcasting.html#general-broadcasting-rules
If you want to do this for any random shapes, you can do the multiplication with the transposed shape, so that the last dimensions of both the matrices match, obeying broadcasting rule and then do the transpose again, to get back to the required output,
tf.transpose(a*tf.transpose(b))
Example,
a = tf.ones([10,])
b = tf.ones([10,11,12,13,1])
tf.transpose(b)
#[1, 13, 12, 11, 10]
(a*tf.transpose(b))
#[1, 13, 12, 11, 10]
tf.transpose(a*tf.transpose(b)) #Note a is [10,] not [10,1], otherwise you need to add transpose to a as well.
#[10, 11, 12, 13, 1]
Another approach is to expanding the axis:
a = tf.ones([10])[(...,) + (tf.rank(b)-1) * (tf.newaxis,)]
I'm trying to writting a layer to merge 2 tensors with such a formula
The shapes of x[0] and x[1] are both (?, 1, 500).
M is a 500*500 Matrix.
I want the output to be (?, 500, 500) which is theoretically feasible in my opinion. The layer will output (1,500,500) for every pair of inputs, as (1, 1, 500) and (1, 1, 500). As the batch_size is variable, or dynamic, the output must be (?, 500, 500).
However, I know little about axes and I have tried all the combinations of axes but it doesn't make sense.
I try with numpy.tensordot and keras.backend.batch_dot(TensorFlow). If the batch_size is fixed, taking a =
(100,1,500) for example, batch_dot(a,M,(2,0)), the output can be (100,1,500).
Newbie for Keras, sorry for such a stupid question but I have spent 2 days to figure out and it drove me crazy :(
def call(self,x):
input1 = x[0]
input2 = x[1]
#self.M is defined in build function
output = K.batch_dot(...)
return output
Update:
Sorry for being late. I try Daniel's answer with TensorFlow as Keras's backend and it still raises a ValueError for unequal dimensions.
I try the same code with Theano as backend and now it works.
>>> import numpy as np
>>> import keras.backend as K
Using Theano backend.
>>> from keras.layers import Input
>>> x1 = Input(shape=[1,500,])
>>> M = K.variable(np.ones([1,500,500]))
>>> firstMul = K.batch_dot(x1, M, axes=[1,2])
I don't know how to print tensors' shape in theano. It's definitely harder than tensorflow for me... However it works.
For that I scan 2 versions of codes for Tensorflow and Theano. Following are differences.
In this case, x = (?, 1, 500), y = (1, 500, 500), axes = [1, 2]
In tensorflow_backend:
return tf.matmul(x, y, adjoint_a=True, adjoint_b=True)
In theano_backend:
return T.batched_tensordot(x, y, axes=axes)
(If following changes of out._keras_shape don't make influence on out's value.)
Your multiplications should select which axes it uses in the batch dot function.
Axis 0 - the batch dimension, it's your ?
Axis 1 - the dimension you say has length 1
Axis 2 - the last dimension, of size 500
You won't change the batch dimension, so you will use batch_dot always with axes=[1,2]
But for that to work, you must ajust M to be (?, 500, 500).
For that define M not as (500,500), but as (1,500,500) instead, and repeat it in the first axis for the batch size:
import keras.backend as K
#Being M with shape (1,500,500), we repeat it.
BatchM = K.repeat_elements(x=M,rep=batch_size,axis=0)
#Not sure if repeating is really necessary, leaving M as (1,500,500) gives the same output shape at the end, but I haven't checked actual numbers for correctness, I believe it's totally ok.
#Now we can use batch dot properly:
firstMul = K.batch_dot(x[0], BatchM, axes=[1,2]) #will result in (?,500,500)
#we also need to transpose x[1]:
x1T = K.permute_dimensions(x[1],(0,2,1))
#and the second multiplication:
result = K.batch_dot(firstMul, x1T, axes=[1,2])
I prefer using TensorFlow so I tried to figure it out with TensorFlow in past few days.
The first one is much similar to Daniel's solution.
x = tf.placeholder('float32',shape=(None,1,3))
M = tf.placeholder('float32',shape=(None,3,3))
tf.matmul(x, M)
# return: <tf.Tensor 'MatMul_22:0' shape=(?, 1, 3) dtype=float32>
It needs to feed values to M with fit shapes.
sess = tf.Session()
sess.run(tf.matmul(x,M), feed_dict = {x: [[[1,2,3]]], M: [[[1,2,3],[0,1,0],[0,0,1]]]})
# return : array([[[ 1., 4., 6.]]], dtype=float32)
Another way is simple with tf.einsum.
x = tf.placeholder('float32',shape=(None,1,3))
M = tf.placeholder('float32',shape=(3,3))
tf.einsum('ijk,lm->ikl', x, M)
# return: <tf.Tensor 'MatMul_22:0' shape=(?, 1, 3) dtype=float32>
Let's feed some values.
sess.run(tf.einsum('ijk,kl->ijl', x, M), feed_dict = {x: [[[1,2,3]]], M: [[1,2,3],[0,1,0],[0,0,1]]})
# return: array([[[ 1., 4., 6.]]], dtype=float32)
Now M is a 2D tensor and no need to feed batch_size to M.
What's more, now it seems such a question can be solved in TensorFlow with tf.einsum. Does it mean it's a duty for Keras to invoke tf.einsum in some situations? At least I find no where Keras calls tf.einsum. And in my opinion, when batch_dot 3D tensor and 2D tensor Keras behaves weirdly. In Daniel's answer, he pads M to (1,500,500) but in K.batch_dot() M will be adjusted to (500,500,1) automatically. I find tf will adjust it with Broadcasting rules and I'm not sure Keras does the same.
I want to extract elements from 3d-tensor to make 2d-tensor.
I have 3d-tensor (2000 X 2 X 3) and indices 1d-tensor for 3-dim (2000).
The indices tensor surely contain indices 0~2 of 2000 elements.
In fact, I want same result with A[:,:,inds].
How to do this with tf.gather_nd, please, help me.
If the tensor is really of size 2000 x 2 x 3, you can use tf.gather:
a = tf.random_normal([2000, 2, 3])
b = tf.gather(a, [1007, 8, 7, 9])
b.get_shape()
TensorShape([Dimension(4), Dimension(2), Dimension(3)])
If it is of shape 2 x 3 x 2000, you can either:
change the shape to 2000 x 2 x 3 using tf.reshape and then do the same, as above
use use regular python mechanism for indexing and then pack the tensor back:
a = tf.random_normal([2, 3, 2000])
indices = [1007, 8, 7, 9]
subtensor = [a[:, :, i] for i in indices]
b = tf.pack(subtensor, axis=2)
b.get_shape()
<tf.Tensor 'pack:0' shape=(2, 3, 4) dtype=float32>
I have a problem with which I've been struggling. It is related to tf.matmul() and its absence of broadcasting.
I am aware of a similar issue on https://github.com/tensorflow/tensorflow/issues/216, but tf.batch_matmul() doesn't look like a solution for my case.
I need to encode my input data as a 4D tensor:
X = tf.placeholder(tf.float32, shape=(None, None, None, 100))
The first dimension is the size of a batch, the second the number of entries in the batch.
You can imagine each entry as a composition of a number of objects (third dimension). Finally, each object is described by a vector of 100 float values.
Note that I used None for the second and third dimensions because the actual sizes may change in each batch. However, for simplicity, let's shape the tensor with actual numbers:
X = tf.placeholder(tf.float32, shape=(5, 10, 4, 100))
These are the steps of my computation:
compute a function of each vector of 100 float values (e.g., linear function)
W = tf.Variable(tf.truncated_normal([100, 50], stddev=0.1))
Y = tf.matmul(X, W)
problem: no broadcasting for tf.matmul() and no success using tf.batch_matmul()
expected shape of Y: (5, 10, 4, 50)
applying average pooling for each entry of the batch (over the objects of each entry):
Y_avg = tf.reduce_mean(Y, 2)
expected shape of Y_avg: (5, 10, 50)
I expected that tf.matmul() would have supported broadcasting. Then I found tf.batch_matmul(), but still it looks like doesn't apply to my case (e.g., W needs to have 3 dimensions at least, not clear why).
BTW, above I used a simple linear function (the weights of which are stored in W). But in my model I have a deep network instead. So, the more general problem I have is automatically computing a function for each slice of a tensor. This is why I expected that tf.matmul() would have had a broadcasting behavior (if so, maybe tf.batch_matmul() wouldn't even be necessary).
Look forward to learning from you!
Alessio
You could achieve that by reshaping X to shape [n, d], where d is the dimensionality of one single "instance" of computation (100 in your example) and n is the number of those instances in your multi-dimensional object (5*10*4=200 in your example). After reshaping, you can use tf.matmul and then reshape back to the desired shape. The fact that the first three dimensions can vary makes that little tricky, but you can use tf.shape to determine the actual shapes during run time. Finally, you can perform the second step of your computation, which should be a simple tf.reduce_mean over the respective dimension. All in all, it would look like this:
X = tf.placeholder(tf.float32, shape=(None, None, None, 100))
W = tf.Variable(tf.truncated_normal([100, 50], stddev=0.1))
X_ = tf.reshape(X, [-1, 100])
Y_ = tf.matmul(X_, W)
X_shape = tf.gather(tf.shape(X), [0,1,2]) # Extract the first three dimensions
target_shape = tf.concat(0, [X_shape, [50]])
Y = tf.reshape(Y_, target_shape)
Y_avg = tf.reduce_mean(Y, 2)
As the renamed title of the GitHub issue you linked suggests, you should use tf.tensordot(). It enables contraction of axes pairs between two tensors, in line with Numpy's tensordot(). For your case:
X = tf.placeholder(tf.float32, shape=(5, 10, 4, 100))
W = tf.Variable(tf.truncated_normal([100, 50], stddev=0.1))
Y = tf.tensordot(X, W, [[3], [0]]) # gives shape=[5, 10, 4, 50]