Summarize the problem
I am working with high dimensional tensors in pytorch and I need to index one tensor with the argmax values from another tensor. So I need to index tensor y of dim [3,4] with the results from the argmax of tensor xwith dim [3,4]. If tensors are:
import torch as T
# Tensor to get argmax from
# expected argmax: [2, 0, 1]
x = T.tensor([[1, 2, 8, 3],
[6, 3, 3, 5],
[2, 8, 1, 7]])
# Tensor to index with argmax from preivous
# expected tensor to retrieve [2, 4, 9]
y = T.tensor([[0, 1, 2, 3],
[4, 5, 6, 7],
[8, 9, 10, 11]])
# argmax
x_max, x_argmax = T.max(x, dim=1)
I would like an operation that given the argmax indexes of x, or x_argmax, retrieves the values in tensor y in the same indexes x_argmax indexes.
Describe what you’ve tried
This is what I have tried:
# What I have tried
print(y[x_argmax])
print(y[:, x_argmax])
print(y[..., x_argmax])
print(y[x_argmax.unsqueeze(1)])
I have been reading a lot about numpy indexing, basic indexing, advanced indexing and combined indexing. I have been trying to use combined indexing (since I want a slice in first dimension of the tensor and the indexes values on the second one). But I have not been able to come up with a solution for this use case.
You are looking for torch.gather:
idx = torch.argmax(x, dim=1, keepdim=true) # get argmax directly, w/o max
out = torch.gather(y, 1, idx)
Resulting with
tensor([[2],
[4],
[9]])
How about y[T.arange(3), x_argmax]?
That does the job for me...
Explanation: You take dimensional information away when you invoke T.max(x, dim=1), so this information needs to be restored explicitly.
Related
I have extracted the embeddings for a particular entity X from every sentence in my dataset. Where X is mentioned more than once within the same sentence, this yields an embedding for each mention: I'd like to put these through an average pooling layer to arrive at a single embedding for X in each sentence.
Simplified working example:
import tensorflow as tf
embeddings = tf.constant([[1, 1, 1],
[2, 2, 2],
[4, 4, 4],
[5, 5, 5]])
# Let's imagine rows [1, 1, 1] & [4, 4, 4]
# correspond to embeddings for X from the same sentence
# We can indicate sentence belonging through an sent_idxs variable:
sent_idxs = tf.constant([0, 1, 0, 2])
With the help of related stackoverflow questions (Torch - How to calculate average of tensors with the same indexes, Summing over specific indices PyTorch (similar to scatter_add)), I could average embeddings corresponding to the same sentence like this:
unique_idxs, _, counts = tf.unique_with_counts(sent_idxs) # counts = ([2, 1, 1])
result_holder = tf.zeros([unique_idxs.shape[0], embeddings.shape[1]], dtype= embeddings.dtype)
embeddings = tf.tensor_scatter_nd_add(result_holder, tf.expand_dims(sent_idxs, axis=1), embeddings)
embeddings /= counts[:, None]
However, I would prefer to re-shape my original embeddings to instead perform the averaging with AveragePooling2D or AveragePooling1D, and I'm really struggling with imagining the appropriate shape to enable this.
Both tf.scatter_add and tf.scatter_nd allow indices to be a matrix. It is clear from the documentation of tf.scatter_nd that the last dimension of indices contains values that are used to index a tensor of shape shape. The other dimensions of indices define the number of elements/slices to be scattered. Suppose updates has a rank N. First k dimensions of indices (except the last dimension) should match with first k dimensions of updates. The last (N-k) dimensions of updates should match with the last (N-k) dimensions of shape.
This implies that tf.scatter_nd can be used to perform an N-dimensional scatter. However, tf.scatter_add also takes matrices as indices. But, its not clear which dimensions of indices correspond to the number of scatters to be performed and how do these dimensions align with updates. Can someone provide a clear explanation possibly with examples?
#shaunshd , I finally fully understand the 3 tensors relationship in tf.scatter_nd_*() arguments, especially when the indices have multi-demensions. e.g:
indices = tf.constant([[0, 0, 0], [1, 1, 1], [2, 2, 2], [3, 3, 3], [3,3,2]], dtype=tf.int32)
Please don't expect tf.rank(indices)>2, tf.rank(indices)==2 is permanently true;
The following is my test codes to show more complex test case than the examples provided in tensroflow's official website:
def testScatterNDUpdate(self):
ref = tf.Variable(np.zeros(shape=[4, 4, 4], dtype=np.float32))
indices = tf.constant([[0, 0, 0], [1, 1, 1], [2, 2, 2], [3, 3, 3], [3,3,2]], dtype=tf.int32)
updates = tf.constant([1,2,3,4,5], dtype=tf.float32)
#shape = (4,4,4)
print(tf.tensor_scatter_nd_update(ref, indices, updates))
print(ref.scatter_nd_update(indices, updates))
#print(updates.shape[-1]==shape[-1], updates.shape[0]<=shape[0])
#conditions are:
# updates.shape[0]==indices[0]
# indices[1]<=len(shape)
# tf.rank(indices)==2
You also could understand the indices with the following psudo codes:
def scatter_nd_update(ref, indices, updates):
for i in range(tf.shape(indices)[0]):
ref[indices[i]]=updates[i]
return ref
Comapring with numpy's fancy indexing feature, tensorflow's indexing features are still very difficult to use and have different using style, not unified as same as numpy yet. Hope the situation could be better in tf3.x
Let's assume, I have a 3 dimensional tensor of shape [a, b, c].
I want to extract one dimension at model run time, e.g. [4, 2, c], so I would end up with with a 1 dimensional tensor [c].
The parameters of a and b are stored in different tensors of shape [a, 1] and [b, 1], so using tf.slice is not an option, as tf.slice only accepts a 1 dimensional scalar tensor.
Any ideas?
Thanks!
You can use the tf.reshape function, although you need to pass the number of entries in the new tensor as an argument. For instance:
import tensorflow as tf
# Define a 3D tensor
tensor3d = tf.constant([[[1, 2, 3], [4, 5, 6]], [[7, 8, 9], [10, 11, 12]]])
# Convert the 3D tensor to 1D
tensor1d = tf.reshape(tensor3d, [12])
Probably you need to split the tensor with tf.split.
https://www.tensorflow.org/api_docs/python/tf/split
tensorflow.reduce_sum(..) computes the sum of elements across dimensions of a tensor. it is Ok.
But one thing is not clear to me , what is the purpose of saying reduce in the function name ?
Is it related to map_reduce of parallel computation?
Let's say, it distributes the required computation to
different cores , and collect the result from the cores , finally delivers the sum of the collected results ?
Because you can compute the sum along a given dimension (and therefore reduce it). And no it has nothing to do with map-reduce.
Quoting the documentation string of the method:
Reduces input_tensor along the dimensions given in axis. Unless keepdims is true, the rank of the tensor is reduced by 1 for each entry in axis. If keepdims is true, the reduced dimensions are retained with length 1.
Example from the API:
x = tf.constant([[1, 1, 1], [1, 1, 1]])
tf.reduce_sum(x) # 6
tf.reduce_sum(x, 0) # [2, 2, 2]
tf.reduce_sum(x, 1) # [3, 3]
tf.reduce_sum(x, 1, keepdims=True) # [[3], [3]]
tf.reduce_sum(x, [0, 1]) # 6
I have two following tensors (note that they are both Tensorflow tensors which means they are still virtually symbolic at the time I construct the following slicing op before I launch a tf.Session()):
params: has shape (64,784, 256)
indices: has shape (64, 784)
and I want to construct an op that returns the following tensor:
output: has shape (64,784) where
output[i,j] = params_tensor[i,j, indices[i,j] ]
What is the most efficient way in Tensorflow to do so?
ps: I tried with tf.gather but couldn't make use of it to perform the operation I described above.
Many thanks.
-Bests
You can get exactly what you want using tf.gather_nd. The final expression is:
tf.gather_nd(params, tf.stack([tf.tile(tf.expand_dims(tf.range(tf.shape(indices)[0]), 1), [1, tf.shape(indices)[1]]), tf.transpose(tf.tile(tf.expand_dims(tf.range(tf.shape(indices)[1]), 1), [1, tf.shape(indices)[0]])), indices], 2))
This expression has the following explanation:
tf.gather_nd does what you expected and uses the indices to gather the output from the params
tf.stack combines three separate tensors, the last of which is the indices. The first two tensors specify the ordering of the first two dimensions (axis 0 and axis 1 of params/indices)
For the example provided, this ordering is simply 0, 1, 2, ..., 63 for axis 0, and 0, 1, 2, ... 783 for axis 1. These sequences are obtained with tf.range(tf.shape(indices)[0]) and tf.range(tf.shape(indices)[1]), respectively.
For the example provided, indices has shape (64, 784). The other two tensors from the last point above need to have this same shape in order to be combined with tf.stack
First, an additional dimension/axis is added to each of the two sequences using tf.expand_dims.
The use of tf.tile and tf.transpose can be shown by example: Assume the first two axes of params and index have shape (5,3). We want the first tensor to be:
[[0, 0, 0], [1, 1, 1], [2, 2, 2], [3, 3, 3], [4, 4, 4]]
We want the second tensor to be:
[[0, 1, 2], [0, 1, 2], [0, 1, 2], [0, 1, 2], [0, 1, 2]]
These two tensors almost function like specifying the coordinates in a grid for the associated indices.
The final part of tf.stack combines the three tensors on a new third axis, so that the result has the same 3 axes as params.
Keep in mind if you have more or less axes than in the question, you need to modify the number of coordinate-specifying tensors in tf.stack accordingly.
What you want is like a custom reduction function. If you are keeping something like index of maximum value at indices then I would suggest using tf.reduce_max:
max_params = tf.reduce_max(params_tensor, reduction_indices=[2])
Otherwise, here is one way to get what you want (Tensor objects are not assignable so we create a 2d list of tensors and pack it using tf.pack):
import tensorflow as tf
import numpy as np
with tf.Graph().as_default():
params_tensor = tf.pack(np.random.randint(1,256, [5,5,10]).astype(np.int32))
indices = tf.pack(np.random.randint(1,10,[5,5]).astype(np.int32))
output = [ [None for j in range(params_tensor.get_shape()[1])] for i in range(params_tensor.get_shape()[0])]
for i in range(params_tensor.get_shape()[0]):
for j in range(params_tensor.get_shape()[1]):
output[i][j] = params_tensor[i,j,indices[i,j]]
output = tf.pack(output)
with tf.Session() as sess:
params_tensor,indices,output = sess.run([params_tensor,indices,output])
print params_tensor
print indices
print output
I know I'm late, but I recently had to do something similar, and was able to to do it using Ragged Tensors:
output = tf.gather(params, tf.RaggedTensor.from_tensor(indices), batch_dims=-1, axis=-1)
Hope it helps