I try to implement roulette wheel selection in Tensorflow. So I started with this:
x = tf.random_uniform([tf.shape(probabilities)[0]])
cumsum = tf.cumsum(probabilities, axis=1) # cumulative sum
b = tf.greater_equal(x, cumsum) # Boolean values now
...
indices = tf.where(b) # this given indices for all the True values, I need only the first one per row
indices = indices[:,1] # we only need column index
Any suggestions for this? Or a better procedure to do the roulette wheel selection?
So a small example to make it more clear
probabilities = [[0.2 0.3 0.5],
[0.1 0.6 0.3],
[0.5 0.4 0.1]]
x = [0.27, 0.86, 0.73] # drawn randomly
Then I want as output [1, 2, 1]
As far as I understand, you want to draw the samples from multinomial distribution. To do that, it is easiest to simply use tf.multinomial:
samples = tf.multinomial(tf.log(probabilities), 1)
Possibly followed by reshaping:
samples_vector = tf.reshape(samples, [-1])
Related
I have n networks, each with the same input / output. I want to randomly select one of the outputs according to a categorical distribution. Tfp.Categorical outputs only integers and I tried to do something like
act_dist = tfp.distributions.Categorical(logits=act_logits) # act_logits are all the same, so the distribution is uniform
rand_out = act_dist.sample()
x = nn_out1 * tf.cast(rand_out == 0., dtype=tf.float32) + ... # for all my n networks
But rand_out == 0. is always false, as well as the other conditions.
Any idea for achieving what I need?
You might also look at MixtureSameFamily, which does a gather under the covers for you.
nn_out1 = tf.expand_dims(nn_out1, axis=2)
...
outs = tf.concat([nn_out1, nn_nout2, ...], axis=2)
probs = tf.tile(tf.reduce_mean(tf.ones_like(nn_out1), axis=1, keepdims=True) / n, [1, n]) # trick to have ones of shape [None,1]
dist = tfp.distributions.MixtureSameFamily(
mixture_distribution=tfp.distributions.Categorical(probs=probs),
components_distribution=tfp.distributions.Deterministic(loc=outs))
x = dist.sample()
I think you need to use tf.equal, because Tensor == 0 is always False.
Separately though, you might want to use OneHotCategorical. For training, you might also try using RelaxedOneHotCategorical.
I want to split a long vector into smaller unequal pieces, do a summation on each piece and gather the results into a new vector.
I need to do this in pytorch but I am also interested to see how this is done with numpy.
This can easily be accomplish by splitting the vector.
sizes = [3, 7, 5, 9]
X = torch.ones(sum(sizes))
Y = torch.tensor([s.sum() for s in torch.split(X, sizes)])
or with np.ones and np.split.
Is there a more efficient way to do this?
Edit:
Inspired by the first comment:
indices = np.cumsum([0]+sizes)[:-1]
Y = np.add.reduceat(X, indices.tolist())
solves it for numpy. I am still looking for a solution with pytorch.
index_add_ is your friend!
# inputs
sizes = torch.tensor([3, 7, 5, 9], dtype=torch.long)
x = torch.ones(sizes.sum())
# prepare an index vector for summation (what elements of x are summed to each element of y)
ind = torch.zeros(sizes.sum(), dtype=torch.long)
ind[torch.cumsum(sizes, dim=0)[:-1]] = 1
ind = torch.cumsum(ind, dim=0)
# prepare the output
y = torch.zeros(len(sizes))
# do the actual summation
y.index_add_(0, ind, x)
How can I make x from y where
x = tf.constant([[1,5,3], [100,20,3]])
y = ([[0,5,0], [100,0,0]])
So it basically preserves only the max values and makes other elements zero. Using tf.argmax we can get the max indices but don't really know how to make y from it.
Could you please help?
And would such y has its proper gradient (i.e., at the max element gradient 1 and at others gradient 0).
Not sure if this is the optimized way but you can do it with tf.gather_nd and tf.scatter_nd. 1) use tf.argmax to construct the indices corresponding to the maximum values; 2) extract the maximum values using tf.gather_nd and indices; 3) make a new tensor with the indices and updates using tf.scatter_nd.
x = tf.constant([[1,5,3], [100,20,3]])
with tf.Session() as sess:
indices = tf.stack([tf.range(x.shape[0], dtype=tf.int64), tf.argmax(x, axis=1)], axis=1)
updates = tf.gather_nd(x, indices)
output = tf.scatter_nd(indices, updates, x.shape)
print(sess.run(output))
#[[ 0 5 0]
# [100 0 0]]
I am wondering if it possible to apply a mask before performing theano.tensor.nnet.softmax?
This is the behavior I am looking for:
>>>a = np.array([[1,2,3,4]])
>>>m = np.array([[1,0,1,0]]) # ignore index 1 and 3
>>>theano.tensor.nnet.softmax(a,m)
array([[ 0.11920292, 0. , 0.88079708, 0. ]])
Note that a and m are matrices, so I would like the softmax with work on an entire matrix and perform row-wise masked softmax.
Also the output should be the same shape as a, so the solution can not do advanced indexing e.g. theano.tensor.softmax(a[0,[0,2]])
def masked_softmax(a, m, axis):
e_a = T.exp(a)
masked_e = e_a * m
sum_masked_e = T.sum(masked_e, axis, keepdims=True)
return masked_e / sum_masked_e
theano.tensor.switch is one way to do this.
In the computational graph you can do the following:
a_mask = theano.tensor.switch(m, a, np.NINF)
sm = theano.tensor.softmax(a_mask)
hope it helps others.
Is there a neat way to compute a color histogram of an image? Maybe by abusing the internal code of tf.histogram_summary? From what I've seen, this code is not very modular and calls directly some C++ code.
Thanks in advance.
I would use tf.unsorted_segment_sum, where the "segment IDs" are computed from the color values and the thing you sum is a tf.ones vector. Note that tf.unsorted_segment_sum is probably better thought of as "bucket sum". It implements dest[segment] += thing_to_sum -- exactly the operation you need for a histogram.
In slightly pseudocode (meaning I haven't run this):
binned_values = tf.reshape(tf.floor(img_r * (NUM_BINS-1)), [-1])
binned_values = tf.cast(binned_values, tf.int32)
ones = tf.ones_like(binned_values, dtype=tf.int32)
counts = tf.unsorted_segment_sum(ones, binned_values, NUM_BINS)
You could accomplish this in one pass instead of separating out the r, g, and b values with a split if you wanted to cleverly construct your "ones" to look like "100100..." for red, "010010" for green, etc., but I suspect it would be slower overall, and harder to read. I'd just do the split that you proposed above.
This is what I'm using right now:
# Assumption: img is a tensor of the size [img_width, img_height, 3], normalized to the range [-1, 1].
with tf.variable_scope('color_hist_producer') as scope:
bin_size = 0.2
hist_entries = []
# Split image into single channels
img_r, img_g, img_b = tf.split(2, 3, img)
for img_chan in [img_r, img_g, img_b]:
for idx, i in enumerate(np.arange(-1, 1, bin_size)):
gt = tf.greater(img_chan, i)
leq = tf.less_equal(img_chan, i + bin_size)
# Put together with logical_and, cast to float and sum up entries -> gives count for current bin.
hist_entries.append(tf.reduce_sum(tf.cast(tf.logical_and(gt, leq), tf.float32)))
# Pack scalars together to a tensor, then normalize histogram.
hist = tf.nn.l2_normalize(tf.pack(hist_entries), 0)
tf.histogram_fixed_width
might be what you are looking for...
Full documentation on
https://www.tensorflow.org/api_docs/python/tf/histogram_fixed_width