I have a 2D tensor A with shape [batch_size, D] , and a 1D tensor B with shape [batch_size]. Each element of B is a column index of A, for each row of A, eg. B[i] in [0,D).
What is the best way in tensorflow to get the values A[B]
For example:
A = tf.constant([[0,1,2],
[3,4,5]])
B = tf.constant([2,1])
with desired output:
some_slice_func(A, B) -> [2,4]
There is another constraint. In practice, batch_size is actually None.
Thanks in advance!
I was able to get it working using a linear index:
def vector_slice(A, B):
""" Returns values of rows i of A at column B[i]
where A is a 2D Tensor with shape [None, D]
and B is a 1D Tensor with shape [None]
with type int32 elements in [0,D)
Example:
A =[[1,2], B = [0,1], vector_slice(A,B) -> [1,4]
[3,4]]
"""
linear_index = (tf.shape(A)[1]
* tf.range(0,tf.shape(A)[0]))
linear_A = tf.reshape(A, [-1])
return tf.gather(linear_A, B + linear_index)
This feels slightly hacky though.
If anyone knows a better (as in clearer or faster) please also leave an answer! (I won't accept my own for a while)
Code for what #Eugene Brevdo said:
def vector_slice(A, B):
""" Returns values of rows i of A at column B[i]
where A is a 2D Tensor with shape [None, D]
and B is a 1D Tensor with shape [None]
with type int32 elements in [0,D)
Example:
A =[[1,2], B = [0,1], vector_slice(A,B) -> [1,4]
[3,4]]
"""
B = tf.expand_dims(B, 1)
range = tf.expand_dims(tf.range(tf.shape(B)[0]), 1)
ind = tf.concat([range, B], 1)
return tf.gather_nd(A, ind)
the least hacky way is probably to build a proper 2d index by concatenating range(batch_size) and B, to get a batch_size x 2 matrix. then pass this to tf.gather_nd.
The simplest approach is to do:
def tensor_slice(target_tensor, index_tensor):
indices = tf.stack([tf.range(tf.shape(index_tensor)[0]), index_tensor], 1)
return tf.gather_nd(target_tensor, indices)
Consider to use tf.one_hot, tf.math.multiply and tf.reduce_sum to solve it.
e.g.
def vector_slice (inputs, inds, axis = None):
axis = axis if axis is not None else tf.rank(inds) - 1
inds = tf.one_hot(inds, inputs.shape[axis])
for i in tf.range(tf.rank(inputs) - tf.rank(inds)):
inds = tf.expand_dims(inds, axis = -1)
inds = tf.cast(inds, dtype = inputs.dtype)
x = tf.multiply(inputs, inds)
return tf.reduce_sum(x, axis = axis)
Related
I'm trying to have a layer in keras that takes a flat tensor x (doesn't have zero value in it and shape = (batch_size, units)) multiplied by a mask (of the same shape), and it will sort it in the way that masked values will be placed first in the output (the order of the elements value doesn't matter). For clarity here is an example (batch_size = 1, units = 8):
It seems simple but the problem is that I can't find a good solution. Any code or idea is appreciated.
My current code is as below, If you know a more efficient way please let me know.
class Sort(keras.layers.Layer):
def call(self, inputs):
x = inputs.numpy()
nonx, nony = x.nonzero() # idxs of nonzero elements
zero = [np.where(x == 0)[0][0], np.where(x == 0)[1][0]] # idx of first zero
x_shape = tf.shape(inputs)
result = np.zeros((x_shape[0], x_shape[1], 2), dtype = 'int') # mapping matrix
result[:, :, 0] += zero[0]
result[:, :, 1] += zero[1]
p = np.zeros((x_shape[0]), dtype = 'int')
for i, j in zip(nonx, nony):
result[i, p[i]] = [i, j]
p[i] += 1
y = tf.gather_nd(inputs, result)
return y
I want to concatenate two tensors checkerboard-ly in tensorflow2, like examples showed below:
example 1:
a = [[1,1],[1,1]]
b = [[0,0],[0,0]]
concated_a_and_b = [[1,0,1,0],[0,1,0,1]]
example 2:
a = [[1,1,1],[1,1,1],[1,1,1]]
b = [[0,0,0],[0,0,0],[0,0,0]]
concated_a_and_b = [[1,0,1,0,1,0],[0,1,0,1,0,1],[1,0,1,0,1,0]]
Is there a decent way in tensorflow2 to concatenate them like this?
A bit of background for this:
I first split a tensor c with a checkerboard mask into two halves a and b. A after some transformation I have to concat them back into oringnal shape and order.
What I mean by checkerboard-ly:
Step 1: Generate a matrix with alternated values
You can do this by first concatenating into [1, 0] pairs, and then by applying a final reshape.
Step 2: Reverse some rows
I split the matrix into two parts, reverse the second part and then rebuild the full matrix by picking alternatively from the first and second part
Code sample:
import math
import numpy as np
import tensorflow as tf
a = tf.ones(shape=(3, 4))
b = tf.zeros(shape=(3, 4))
x = tf.expand_dims(a, axis=-1)
y = tf.expand_dims(b, axis=-1)
paired_ones_zeros = tf.concat([x, y], axis=-1)
alternated_values = tf.reshape(paired_ones_zeros, [-1, a.shape[1] + b.shape[1]])
num_samples = alternated_values.shape[0]
middle = math.ceil(num_samples / 2)
is_num_samples_odd = middle * 2 != num_samples
# Gather first part of the matrix, don't do anything to it
first_elements = tf.gather_nd(alternated_values, [[index] for index in range(middle)])
# Gather second part of the matrix and reverse its elements
second_elements = tf.reverse(tf.gather_nd(alternated_values, [[index] for index in range(middle, num_samples)]), axis=[1])
# Pick alternatively between first and second part of the matrix
indices = np.concatenate([[[index], [index + middle]] for index in range(middle)], axis=0)
if is_num_samples_odd:
indices = indices[:-1]
output = tf.gather_nd(
tf.concat([first_elements, second_elements], axis=0),
indices
)
print(output)
I know this is not a decent way as it will affect time and space complexity. But it solves the above problem
def concat(tf1, tf2):
result = []
for (index, (tf_item1, tf_item2)) in enumerate(zip(tf1, tf2)):
item = []
for (subitem1, subitem2) in zip(tf_item1, tf_item2):
if index % 2 == 0:
item.append(subitem1)
item.append(subitem2)
else:
item.append(subitem2)
item.append(subitem1)
concated_a_and_b.append(item)
return concated_a_and_b
In a problem I want to solve using Tensorflow, I want to build a n-dimensional rank tensor that is 'diagonal' by blocks. That is, I want to generate a tensor object from a concatenation of low order tensors.
I have tried to define the whole tf.Variable tensor and then to impose the value 0 to some variables but Tensorflow does not allow assignments when working with variable tensors.
Moreover, I would want to create 'diagonal' tensors with the same independent variables, as, for example, using a stacked 2D representation, being A a 2 dimensional tensor:
T = [A, 0;0 , A]
My current source code:
shape1 = [3,3,10,10]
shape2 = [3,3]
i1 = tf.truncated_normal(shape1, stddev=1.0, dtype = tf.float32)
i2 = tf.truncated_normal(shape2, stddev=1.0, dtype = tf.float32)
A = tf.Variable(i1)
V = tf.Variable(i2)
for i in range(10):
for j in range(10):
if i != j:
A[:,:,i,j] = tf.zeros((3,3))
else:
A[:,:,i,j] = V
Of course, this code returns the error Variable object does not support item assignment.
What I want, at the end of the day, is to define a variable tensor such as:
T[:,:,i,j] = tf.zeros([D0,D1]), if i != j
and
T[:,:,i,j] = A, if i = j
with A = tf.variable([D0,D1])
Thank you very much in advance!
One way would be to use tf.stack, which converts a list of tensors of dimension n to a tensor of dimension n+1.
l = []
for i in range(10):
li = [V * 0.0 if i != j else V for j in range(10)]
Ai = tf.stack(li)
l.append(Ai)
A = tf.stack(l)
In general when we multiply a vector v of dimension 1*n with a tensor T of dimension m*n*k, we expect to get a matrix/tensor of dimension m*k/m*1*k. This means that our tensor has m slices of matrices with dimension n*k, and v is multiplied to each matrix and the resulting vectors are stacked together. In order to do this multiplication in tensorflow, I came up with the following formulation. I am just wondering if there is any built-in function that does this standard multiplication straightforward?
T = tf.Variable(tf.random_normal((m,n,k)), name="tensor")
v = tf.Variable(tf.random_normal((1,n)), name="vector")
c = tf.stack([v,v]) # m times, here set m=2
output = tf.matmul(c,T)
You can do it with:
tf.reduce_sum(tf.expand_dims(v,2)*T,1)
Code:
m, n, k = 2, 3, 4
T = tf.Variable(tf.random_normal((m,n,k)), name="tensor")
v = tf.Variable(tf.random_normal((1,n)), name="vector")
c = tf.stack([v,v]) # m times, here set m=2
out1 = tf.matmul(c,T)
out2 = tf.reduce_sum(tf.expand_dims(v,2)*T,1)
with tf.Session() as sess:
sess.run(tf.global_variables_initializer())
n_out1 = sess.run(out1)
n_out2 = sess.run(out2)
#both n_out1 and n_out2 matches
Not sure if there is a better way, but it sounds like you could use tf.map_fn like this:
output = tf.map_fn(lambda x: tf.matmul(v, x), T)
Given a 2-dimensional tensor t, what's the fastest way to compute a tensor h where
h[i, :] = tf.histogram_fixed_width(t[i, :], vals, nbins)
I.e. where tf.histogram_fixed_width is called per row of the input tensor t?
It seems that tf.histogram_fixed_width is missing an axis parameter that works like, e.g., tf.reduce_sum's axis parameter.
tf.histogram_fixed_width works on the entire tensor indeed. You have to loop through the rows explicitly to compute the per-row histograms. Here is a complete working example using TensorFlow's tf.while_loop construct :
import tensorflow as tf
t = tf.random_uniform([2, 2])
i = 0
hist = tf.constant(0, shape=[0, 5], dtype=tf.int32)
def loop_body(i, hist):
h = tf.histogram_fixed_width(t[i, :], [0.0, 1.0], nbins=5)
return i+1, tf.concat_v2([hist, tf.expand_dims(h, 0)], axis=0)
i, hist = tf.while_loop(
lambda i, _: i < 2, loop_body, [i, hist],
shape_invariants=[tf.TensorShape([]), tf.TensorShape([None, 5])])
sess = tf.InteractiveSession()
print(hist.eval())
Inspired by keveman's answer and because the number of rows of t is fixed and rather small, I chose to use a combination of tf.gather to split rows and tf.pack to join rows. It looks simple and works, will see if it is efficient...
t_histo_rows = [
tf.histogram_fixed_width(
tf.gather(t, [row]),
vals, nbins)
for row in range(t_num_rows)]
t_histo = tf.pack(t_histo_rows, axis=0)
I would like to propose another implementation.
This implementation can also handle multi axes and unknown dimensions (batching).
def histogram(tensor, nbins=10, axis=None):
value_range = [tf.reduce_min(tensor), tf.reduce_max(tensor)]
if axis is None:
return tf.histogram_fixed_width(tensor, value_range, nbins=nbins)
else:
if not hasattr(axis, "__len__"):
axis = [axis]
other_axis = [x for x in range(0, len(tensor.shape)) if x not in axis]
swap = tf.transpose(tensor, [*other_axis, *axis])
flat = tf.reshape(swap, [-1, *np.take(tensor.shape.as_list(), axis)])
count = tf.map_fn(lambda x: tf.histogram_fixed_width(x, value_range, nbins=nbins), flat, dtype=(tf.int32))
return tf.reshape(count, [*np.take([-1 if a is None else a for a in tensor.shape.as_list()], other_axis), nbins])
The only slow part here is tf.map_fn but it is still faster than the other solutions mentioned.
If someone knows a even faster implementation please comment since this operation is still very expensive.
answers above is still slow running in GPU. Here i give an another option, which is faster(at least in my running envirment), but it is limited to 0~1 (you can normalize the value first). the train_equal_mask_nbin can be defined once in advance
def histogram_v3_nomask(tensor, nbins, row_num, col_num):
#init mask
equal_mask_list = []
for i in range(nbins):
equal_mask_list.append(tf.ones([row_num, col_num], dtype=tf.int32) * i)
#[nbins, row, col]
#[0, row, col] is tensor of shape [row, col] with all value 0
#[1, row, col] is tensor of shape [row, col] with all value 1
#....
train_equal_mask_nbin = tf.stack(equal_mask_list, axis=0)
#[inst, doc_len] float to int(equaly seg float in bins)
int_input = tf.cast(tensor * (nbins), dtype=tf.int32)
#input [row,col] -> copy N times, [nbins, row_num, col_num]
int_input_nbin_copy = tf.reshape(tf.tile(int_input, [nbins, 1]), [nbins, row_num, col_num])
#calculate histogram
histogram = tf.transpose(tf.count_nonzero(tf.equal(train_equal_mask_nbin, int_input_nbin_copy), axis=2))
return histogram
With the advent of tf.math.bincount, I believe the problem has become much simpler.
Something like this should work:
def hist_fixed_width(x,st,en,nbins):
x=(x-st)/(en-st)
x=tf.cast(x*nbins,dtype=tf.int32)
x=tf.clip_by_value(x,0,nbins-1)
return tf.math.bincount(x,minlength=nbins,axis=-1)