Does Tensorflow have an inverse of tf.unravel_index? - tensorflow

tf.unravel_index takes a flat index into an array and a shape, and returns the tuple that represents that index in the array.
with tf.Session() as sess:
t = tf.unravel_index(tf.constant(15), [2, 3, 4])
print(sess.run(t)) # [1 0 3]
Is there an inverse? tf.ravel_index doesn't exist, but maybe it's under a different name?

Here is how you can do the equivalent to np.ravel_multi_index in TensorFlow:
import tensorflow as tf
# multi_index is 2D (num dimensions x num indices), dims is 1D
# Does not check for out of bounds indices
def tf_ravel_multi_index(multi_index, dims):
strides = tf.cumprod(dims, exclusive=True, reverse=True)
return tf.reduce_sum(multi_index * tf.expand_dims(strides, 1), axis=0)
# Test
with tf.Graph().as_default(), tf.Session() as sess:
# Shape tensor
shape = tf.constant([4, 5, 6])
# Some flat indices
idx = tf.constant([23, 56, 4, 17])
# Unravel indices
unravel = tf.unravel_index(idx, shape)
# Ravel indices again
idx2 = tf_ravel_multi_index(unravel, shape)
# Check result
print(*sess.run((unravel, idx2)), sep='\n')
# [[0 1 0 0]
# [3 4 0 2]
# [5 2 4 5]]
# [23 56 4 17]

Related

TensorFlow shuffle sub-tensor in place

I have a tensor X of shape (N,...) and a boolean index mask mask of shape N. I want to shuffle the subarray of X given by mask along the first axis.
How can this be done non-eagerly and, if possible, in place?
Note: I do not need gradients.
You can do that like this:
import tensorflow as tf
def shuffle_mask(x, mask, seed=None):
n = tf.size(mask)
# Get masked indices
idx_masked = tf.cast(tf.where(mask), n.dtype)
# Shuffle masked indices
idx_masked_shuffled = tf.random.shuffle(tf.squeeze(idx_masked, 1), seed=seed)
# Scatter shuffled indices into place
idx_masked_shuffled_scat = tf.scatter_nd(idx_masked, idx_masked_shuffled, [n])
# Combine shuffled and non-shuffled indices
idx_shuffled = tf.where(mask, idx_masked_shuffled_scat, tf.range(n))
# Gather using resulting indices
return tf.gather(x, idx_shuffled)
# Test
with tf.Graph().as_default(), tf.Session() as sess:
tf.random.set_random_seed(0)
x = tf.constant([[0, 1], [2, 3], [4, 5], [6, 7], [8, 9]])
mask = tf.constant([True, False, True, True, False])
y = shuffle_mask(x, mask)
print(sess.run(y))
# [[6 7]
# [2 3]
# [0 1]
# [4 5]
# [8 9]]
You cannot do the operation "in place", as there are no in-place operations at all in TensorFlow. Tensors are constant, so you will always be replacing one tensor with another.

Tensorflow randomly sample from each row

Suppose I have a tensor A of shape (m, n), I would like to randomly sample k elements (without replacement) from each row, resulting in a tensor B of shape (m, k). How to do that in tensorflow?
An example would be:
A: [[1,2,3], [4,5,6], [7,8,9], [10,11,12]]
k: 2
B: [[1,3],[5,6],[9,8],[12,10]]
This is a way to do that:
import tensorflow as tf
with tf.Graph().as_default(), tf.Session() as sess:
tf.random.set_random_seed(0)
a = tf.constant([[1,2,3], [4,5,6], [7,8,9], [10,11,12]], tf.int32)
k = tf.constant(2, tf.int32)
# Tranpose, shuffle, slice, undo transpose
aT = tf.transpose(a)
aT_shuff = tf.random.shuffle(aT)
at_shuff_k = aT_shuff[:k]
result = tf.transpose(at_shuff_k)
print(sess.run(result))
# [[ 3 1]
# [ 6 4]
# [ 9 7]
# [12 10]]

shuffling two tensors in the same order

As above. I tried those to no avail:
tf.random.shuffle( (a,b) )
tf.random.shuffle( zip(a,b) )
I used to concatenate them and do the shuffling, then unconcatenate / unpack. But now I'm in a situation where (a) is 4D rank tensor while (b) is 1D, so, no way to concatenate.
I also tried to give the seed argument to the shuffle method so it reproduces the same shuffling and I use it twice => Failed. Also tried to do the shuffling myself with randomly shuffled range of numbers, but TF is not as flexible as numpy in fancy indexing and stuff ==> failed.
What I'm doing now is, convert everything back to numpy then use shuffle from sklearn then go back to tensors by recasting. It is sheer stupid way. This is supposed to happen inside a graph.
You could just shuffle the indices and then use tf.gather() to extract values corresponding to those shuffled indices:
TF2.x (UPDATE)
import tensorflow as tf
import numpy as np
x = tf.convert_to_tensor(np.arange(5))
y = tf.convert_to_tensor(['a', 'b', 'c', 'd', 'e'])
indices = tf.range(start=0, limit=tf.shape(x)[0], dtype=tf.int32)
shuffled_indices = tf.random.shuffle(indices)
shuffled_x = tf.gather(x, shuffled_indices)
shuffled_y = tf.gather(y, shuffled_indices)
print('before')
print('x', x.numpy())
print('y', y.numpy())
print('after')
print('x', shuffled_x.numpy())
print('y', shuffled_y.numpy())
# before
# x [0 1 2 3 4]
# y [b'a' b'b' b'c' b'd' b'e']
# after
# x [4 0 1 2 3]
# y [b'e' b'a' b'b' b'c' b'd']
TF1.x
import tensorflow as tf
import numpy as np
x = tf.placeholder(tf.float32, (None, 1, 1, 1))
y = tf.placeholder(tf.int32, (None))
indices = tf.range(start=0, limit=tf.shape(x)[0], dtype=tf.int32)
shuffled_indices = tf.random.shuffle(indices)
shuffled_x = tf.gather(x, shuffled_indices)
shuffled_y = tf.gather(y, shuffled_indices)
Make sure that you compute shuffled_x, shuffled_y in the same session run. Otherwise they might get different index orderings.
# Testing
x_data = np.concatenate([np.zeros((1, 1, 1, 1)),
np.ones((1, 1, 1, 1)),
2*np.ones((1, 1, 1, 1))]).astype('float32')
y_data = np.arange(4, 7, 1)
print('Before shuffling:')
print('x:')
print(x_data.squeeze())
print('y:')
print(y_data)
with tf.Session() as sess:
x_res, y_res = sess.run([shuffled_x, shuffled_y],
feed_dict={x: x_data, y: y_data})
print('After shuffling:')
print('x:')
print(x_res.squeeze())
print('y:')
print(y_res)
Before shuffling:
x:
[0. 1. 2.]
y:
[4 5 6]
After shuffling:
x:
[1. 2. 0.]
y:
[5 6 4]

Keras - pad tensor with values on the borders

I have image with size that's not even, so when convolution scales it down by a factor of 2, and then I do Conv2DTranspose, I don't get consistent sizes, which is a problem.
So I thought I'd pad the intermediate tensor with an extra row and column, with values same as what I see on the edges, for minimal disruption. How do I do this in Keras, is it even possible? What are my alternatives?
With Tensorflow for background, you could use tf.concat() to add to your tensor a duplicate of the row/column.
Supposing you want to duplicate the last row/column:
import tensorflow as tf
from keras.layers import Lambda, Input
from keras.models import Model
import numpy as np
def duplicate_last_row(tensor):
return tf.concat((tensor, tf.expand_dims(tensor[:, -1, ...], 1)), axis=1)
def duplicate_last_col(tensor):
return tf.concat((tensor, tf.expand_dims(tensor[:, :, -1, ...], 2)), axis=2)
# --------------
# Demonstrating with TF:
x = tf.convert_to_tensor([[[1, 2, 3], [4, 5, 6]],
[[10, 20, 30], [40, 50, 60]]])
x = duplicate_last_row(duplicate_last_col(x))
with tf.Session() as sess:
print(sess.run(x))
# [[[ 1 2 3 3]
# [ 4 5 6 6]
# [ 4 5 6 6]]
#
# [[10 20 30 30]
# [40 50 60 60]
# [40 50 60 60]]]
# --------------
# Using as a Keras Layer:
inputs = Input(shape=(5, 5, 3))
padded = Lambda(lambda t: duplicate_last_row(duplicate_last_col(t)))(inputs)
model = Model(inputs=inputs, outputs=padded)
model.compile(optimizer="adam", loss='mse', metrics=['mse'])
batch = np.random.rand(2, 5, 5, 3)
x = model.predict(batch, batch_size=2)
print(x.shape)
# (2, 6, 6, 3)

How do I select certain columns of a 2D tensor in TensorFlow?

As generalized slicing is being worked on in this issue, what would be the best way to achieve an op gathering columns of a 2D tensor (matrix)? For example, for tensor t:
1 2 3 4
5 6 7 8
and indices [1,3], I would like to get:
2 4
6 8
which is equivalent to numpy t[:, [1,3]].
Meanwhile the gather method has an axis parameter.
import tensorflow as tf
params = tf.constant([[1,2,3],[4,5,6]])
indices = [0,2]
op = tf.gather(params, indices, axis=1)
produces the output
[[1 3]
[4 6]]
There is a function named tf.nn.embedding_lookup(params, ind) which retrieves the rows of the params tensor.
To achieve what you want, we can first transpose the tensor t from which you want to select certain columns from. Then look up the rows of tf.transpose(t) (columns of t). After the selection, we transpose the result back.
import tensorflow as tf
t = tf.constant([[1, 2, 3],
[4, 5, 6]])
ind = tf.constant([0, 2])
result = tf.transpose(tf.nn.embedding_lookup(tf.transpose(t), ind))
with tf.Session() as sess:
print(sess.run(result))
So far, I created a workaround by flattening the input and using gather:
def gather_cols(params, indices, name=None):
"""Gather columns of a 2D tensor.
Args:
params: A 2D tensor.
indices: A 1D tensor. Must be one of the following types: ``int32``, ``int64``.
name: A name for the operation (optional).
Returns:
A 2D Tensor. Has the same type as ``params``.
"""
with tf.op_scope([params, indices], name, "gather_cols") as scope:
# Check input
params = tf.convert_to_tensor(params, name="params")
indices = tf.convert_to_tensor(indices, name="indices")
try:
params.get_shape().assert_has_rank(2)
except ValueError:
raise ValueError('\'params\' must be 2D.')
try:
indices.get_shape().assert_has_rank(1)
except ValueError:
raise ValueError('\'indices\' must be 1D.')
# Define op
p_shape = tf.shape(params)
p_flat = tf.reshape(params, [-1])
i_flat = tf.reshape(tf.reshape(tf.range(0, p_shape[0]) * p_shape[1],
[-1, 1]) + indices, [-1])
return tf.reshape(tf.gather(p_flat, i_flat),
[p_shape[0], -1])
Which for:
params = tf.constant([[1, 2, 3],
[4, 5, 6]])
indices = [0, 2]
op = gather_cols(params, indices)
produces the expected output:
[[1 3]
[4 6]]