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)
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
Fairly new to numpy/python here, trying to figure out some less c-like, more numpy-like coding styles.
Background
I've got some code done that takes a fixed set of x values and multiple sets of corresponding y value sets and tries to find which set of the y values are the "most linear".
It does this by going through each set of y values in a loop, calculating and storing the residual from a straight line fit of those y's against the x's, then once the loop has finished finding the index of the minimum residual value.
...sorry this might make a bit more sense with the code below.
import numpy as np
import numpy.polynomial.polynomial as poly
# set of x values
xs = [1,22,33,54]
# multiple sets of y values for each of the x values in 'xs'
ys = np.array([[1, 22, 3, 4],
[2, 3, 1, 5],
[3, 2, 1, 1],
[34,23, 5, 4],
[23,24,29,33],
[5,19, 12, 3]])
# array to store the residual from a linear fit of each of the y's against x
residuals = np.empty(ys.shape[0])
# loop through the xs's and calculate the residual of a linear fit for each
for i in range(ys.shape[0]):
_, stats = poly.polyfit(xs, ys[i], 1, full=True)
residuals[i] = stats[0][0]
# the 'most linear' of the ys's is at np.argmin:
print('most linear at', np.argmin(residuals))
Question
I'd like to know if it's possible to "numpy'ize" that into a single expression, something like
residuals = get_residuals(xs, ys)
...I've tried:
I've tried the following, but no luck (it always passes the full arrays in, not row by row):
# ------ ok try to do it without a loop --------
def wrap(x, y):
_, stats = poly.polyfit(x, y, 1, full=True)
return stats[0][0]
res = wrap(xs, ys) # <- fails as passes ys as full 2D array
res = wrap(np.broadcast_to(xs, ys.shape), ys) # <- fails as passes both as 2D arrays
Could anyone give any tips on how to numpy'ize that?
From the numpy.polynomial.polynomial.polyfit docs (not to be confused with numpy.polyfit which is not interchangable)
:
x : array_like, shape (M,)
y : array_like, shape (M,) or (M, K)
Your ys needs to be transposed to have ys.shape[0] equal to xs.shape
def wrap(x, y):
_, stats = poly.polyfit(x, y.T, 1, full=True)
return stats[0]
res = wrap(xs, ys)
res
Out[]: array([284.57337884, 5.54709898, 0.41399317, 91.44641638,
6.34982935, 153.03515358])
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)
In pygame, I have a surface:
im = pygame.image.load('foo.png').convert_alpha()
im = pygame.transform.scale(im, (64, 64))
How can I get a grayscale copy of the image, or convert the image data to grayscale? I have numpy.
Use a Surfarray, and filter it with numpy or Numeric:
def grayscale(self, img):
arr = pygame.surfarray.array3d(img)
#luminosity filter
avgs = [[(r*0.298 + g*0.587 + b*0.114) for (r,g,b) in col] for col in arr]
arr = numpy.array([[[avg,avg,avg] for avg in col] for col in avgs])
return pygame.surfarray.make_surface(arr)
After a lot of research, I came up with this solution, because answers to this question were too slow for what I wanted this feature to:
def greyscale(surface: pygame.Surface):
start = time.time() # delete me!
arr = pygame.surfarray.array3d(surface)
# calulates the avg of the "rgb" values, this reduces the dim by 1
mean_arr = np.mean(arr, axis=2)
# restores the dimension from 2 to 3
mean_arr3d = mean_arr[..., np.newaxis]
# repeat the avg value obtained before over the axis 2
new_arr = np.repeat(mean_arr3d[:, :, :], 3, axis=2)
diff = time.time() - start # delete me!
# return the new surface
return pygame.surfarray.make_surface(new_arr)
I used time.time() to calculate the time cost for this approach, so for a (800, 600, 3) array it takes: 0.026769161224365234 s to run.
As you pointed out, here is a variant preserving the luminiscence:
def greyscale(surface: pygame.Surface):
arr = pygame.surfarray.pixels3d(surface)
mean_arr = np.dot(arr[:,:,:], [0.216, 0.587, 0.144])
mean_arr3d = mean_arr[..., np.newaxis]
new_arr = np.repeat(mean_arr3d[:, :, :], 3, axis=2)
return pygame.surfarray.make_surface(new_arr)
The easiest way is to iterate over all the pixels in your image and call .get_at(...) and .set_at(...).
This will be pretty slow, so in answer to your implicit suggestion about using NumPy, look at http://www.pygame.org/docs/tut/surfarray/SurfarrayIntro.html. The concepts and most of the code are identical.