I've seen this syntax to index a tensor in PyTorch, not sure what it means:
v = torch.div(t, n[:, None])
where v, t, and n are tensors.
What is the role of "None" here? I can't seem to find it in the documentation.
Similar to NumPy you can insert a singleton dimension ("unsqueeze" a dimension) by indexing this dimension with None. In turn n[:, None] will have the effect of inserting a new dimension on dim=1. This is equivalent to n.unsqueeze(dim=1):
>>> n = torch.rand(3, 100, 100)
>>> n[:, None].shape
(3, 1, 100, 100)
>>> n.unsqueeze(1).shape
(3, 1, 100, 100)
Here are some other types of None indexings.
In the example above : is was used as a placeholder to designate the first dimension dim=0. If you want to insert a dimension on dim=2, you can add a second : as n[:, :, None].
You can also place None with respect to the last dimension instead. To do so you can use the ellipsis syntax ...:
n[..., None] will insert a dimension last, i.e. n.unsqueeze(dim=-1).
n[..., None, :] on the before last dimension, i.e. n.unsqueeze(dim=-2).
Related
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,)]
Suppose I have a pytorch tensor x of shape [N, N_g, 2]. It can be viewed as N * N_g 2d vectors. Specifically, x[i, j, :] is the 2d vector of the jth group in the ith batch.
Now I am trying to get the coordinates of vectors of top 5 length in each group. So I tried the following:
(i) First I used x_len = (x**2).sum(dim=2).sqrt() to compute their lengths, resulting in x_len.shape==[N, N_g].
(ii) Then I used tk = x_len.topk(5) to get the top 5 lengths in each group.
(iii) The desired output would be a tensor x_top5 of shape [N, 5, 2]. Naturally I thought of using tk.indices to index x so as to obtain x_top5. But I failed as it seems such indexing is not supported.
How can I do this?
A minimal example:
x = torch.randn(10,10,2) # N=10 is the batchsize, N_g=10 is the group size
x_len = (x**2).sum(dim=2).sqrt()
tk = x_len.topk(5)
x_top5 = x[tk.indices]
print(x_top5.shape)
# torch.Size([10, 5, 10, 2])
However, this gives x_top5 as a tensor of shape [10, 5, 10, 2], instead of [10, 5, 2] as desired.
I have a matrix L of shape (2,5,2). The values along the last axis form a probability distribution. I want to sample another matrix S of shape (2, 5) where each entry is one of the following integers: 0, 1.
For example,
L = [[[0.1, 0.9],[0.2, 0.8],[0.3, 0.7],[0.5, 0.5],[0.6, 0.4]],
[[0.5, 0.5],[0.9, 0.1],[0.7, 0.3],[0.9, 0.1],[0.1, 0.9]]]
One of the samples could be,
S = [[1, 1, 1, 0, 1],
[1, 1, 1, 0, 1]]
The distributions are binomial in the above example. However, in general, the last dimension of L can be any positive integer, so the distributions can be multinomial.
The samples need to be generated efficiently within Tensorflow computation graph. I know how to do this using numpy using the functions apply_along_axis and numpy.random.multinomial.
You can use tf.multinomial() here.
You will first need to reshape your input tensor to shape [-1, N] (where N is the last dimension of L):
# L has shape [2, 5, 2]
L = tf.constant([[[0.1, 0.9],[0.2, 0.8],[0.3, 0.7],[0.5, 0.5],[0.6, 0.4]],
[[0.5, 0.5],[0.9, 0.1],[0.7, 0.3],[0.9, 0.1],[0.1, 0.9]]])
dims = L.get_shape().as_list()
N = dims[-1] # here N = 2
logits = tf.reshape(L, [-1, N]) # shape [10, 2]
Now we can apply the function tf.multinomial() to logits:
samples = tf.multinomial(logits, 1)
# We reshape to match the initial shape minus the last dimension
res = tf.reshape(samples, dims[:-1])
Be cautious when using tf.multinomial(). The inputs to the function should be logits and not probability distributions.
However, in your example, the last axis is a probability distribution.
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]
How does the implicit broadcasting in tensorflow using + and * work?
If i Have two tensors, such that
a.get_shape() = [64, 10, 1, 100]
b.get_shape() = [64, 100]
(a+b).get_shape = [64, 10, 64, 100]
(a*b).get_shape = [64, 10, 64, 100]
How does that become [64, 10, 64, 100]??
According to the documentation, operations like add are broadcasting operation.
Quoting the glossary:
Broadcasting operation
An operation that uses numpy-style broadcasting to make the shapes of its tensor arguments compatible.
The numpy-style broadcasting is well documented in the documentation:
In brief:
[...] the smaller array is “broadcast” across the larger array so that they have compatible shapes.
Broadcasting provides a means of vectorizing array operations so that looping occurs in C instead of Python.
I think that the broadcasting isn't doing what you intended. It's actually broadcasting both directions. Let me show you what I mean by modifying your example
a = tf.ones([64, 10, 1, 100])
b = tf.ones([128, 100])
print((a+b).shape) # prints "(64, 10, 128, 100)"
From this we see that it broadcasts by matching the last dimensions first. It's implicitly tiling a across it's third dimension to match the size of b's first dimension, then implicitly adding singletons and tiling b across a's first two dimensions.
What I think you expected to do was to implicitly tile b across a's second dimension. To do that, you need b to be a different shape:
a = tf.ones([64, 10, 1, 100])
b = tf.ones([64, 1, 1, 100])
print((a+b).shape) # prints "(64, 10, 1, 100)"
You can use tf.expand_dims() twice on your b to add the two singleton dimensions to match this shape.
numpy style broadcasting is well documented, but to give a short explanation: the 2 tensors' shapes will be compared start from the last shape backward, then any shape lacked in either tensor will be replicated to be matched.
For example, with
a.get_shape() = [64, 10, 1, 100]
b.get_shape() = [64, 100]
(a*b).get_shape = [64, 10, 64, 100]
a and b have the same last shape==100, then the next to last shape of a is replicated to match b shape==64, b lacks the first two shapes of a and they will be created.
Note that any lacking shape must be 1 or absent, because the whole of lower-level shapes are replicated.