Slow computation on google colab while solving partial differential equation - tensorflow

I 'm using google colab to solve the homogeneous heat equation. I had made a program earlier with scipy using sparse matrices which worked upto N = 10(hyperparameter) but I need to run it for like N = 4... 1000 and thus it won't work on my pc. I therefore converted the code to tensorflow and here I 'm unable to use sparse matrices like I could in sympy but even the GPU/TPU computation is also slow and slower than my pc. Problems that I'm facing in the code and require solution for
1) tf.contrib is removed and thus I 've to use an older version of tensorflow for odeint function. Where is it in 2.0?
2)If the computation can be computed with sparse matrices it could be good since matrices are tridiagonal.I know about sparse_dense_mul() function but that returns dense tensor and it wouldn't do the job. The "func" function applies time independent boundary conditions and then requires matrix multiplication of (nxn) with (nX1) which gives (nX1) with multiple matrices.
Also the program was running faster without I created the class.
Also it's giving this
WARNING: Logging before flag parsing goes to stderr.
W0829 09:12:24.415445 139855355791232 lazy_loader.py:50]
The TensorFlow contrib module will not be included in TensorFlow 2.0.
For more information, please see:
* https://github.com/tensorflow/community/blob/master/rfcs/20180907-contrib-sunset.md
* https://github.com/tensorflow/addons
* https://github.com/tensorflow/io (for I/O related ops)
If you depend on functionality not listed there, please file an issue.
W0829 09:12:24.645356 139855355791232 deprecation.py:323] From /usr/local/lib/python3.6/dist-packages/tensorflow/contrib/integrate/python/ops/odes.py:233: div (from tensorflow.python.ops.math_ops) is deprecated and will be removed in a future version.
Instructions for updating:
Deprecated in favor of operator or tf.math.divide.
when I run code for loop in range(2, 10) and tqdm does not display and cell keeps running forever but it works fine for in (2, 5) and tqdm bar does appears.
#find a way to use sparse matrices
class Heat:
def __init__(self, N):
self.N = N
self.H = 1/N
self.A = ts.to_dense(ts.SparseTensor(indices=[[0, 0], [0, 1]] + \
[[i, i+j] for i in range(1, N) for j in [-1, 0, 1]] +[[N, N-1], [N, N]],
values=self.H*np.array([1/3, 1/6] + [1/6, 2/3, 1/6]*(N-1) + [1/6, 1/3], dtype=np.float32),
dense_shape=(N+1, N+1 )))
self.D = ts.to_dense(ts.SparseTensor(indices=[[0, 0], [0, 1]] + [[i, i+j] \
for i in range(1, N) for j in [-1, 0, 1]] +[[N, N-1], [N, N]],
values=N*np.array([1-(1), -1 -(-1)] + [-1, 2, -1]*(N-1) + [-1-(-1), 1-(1)], dtype=np.float32),
dense_shape=(N+1, N+1)))
self.domain = tf.linspace(0.0, 1.0, N+1)
def f(k):
if k == 0:
return (1 + math.pi**2)*(math.pi*self.H - math.sin(math.pi*self.H))/(math.pi**2*self.H)
elif k == N:
return -(1 + math.pi**2)*(-math.pi*self.H + math.sin(math.pi*self.H))/(math.pi**2*self.H)
else:
return -2*(1 + math.pi**2)*(math.cos(math.pi*self.H) - 1)*math.sin(math.pi*self.H*k)/(math.pi**2*self.H)
self.F = tf.constant([f(k) for k in range(N+1)], shape=(N+1,), dtype=tf.float32) #caution! shape changed caution caution 1, N+1(problem) is different from N+1,
self.exact = tm.scalar_mul(scalar=np.exp(1), x=tf.sin(math.pi*self.domain))
def error(self):
return np.linalg.norm(self.exact.numpy() - self.approx, 2)
def func (self, y, t):
y = tf.Variable(y)
y = y[0].assign(0.0)
y = y[self.N].assign(0.0)
if self.N**2> 100:
y_dash = tl.matvec(tf.linalg.inv(self.A), tl.matvec(a=tm.negative(self.D), b=y, a_is_sparse=True) + tm.scalar_mul(scalar=math.exp(t), x=self.F)) #caution! shape changed F is (1, N+1) others too
else:
y_dash = tl.matvec(tf.linalg.inv(self.A), tl.matvec(a=tm.negative(self.D), b=y) + tm.scalar_mul(scalar=math.exp(t), x=self.F)) #caution! shape changed F is (1, N+1) others too
y_dash = tf.Variable(y_dash) #!!y_dash performs Hadamard product like multiplication not matrix-like multiplication;returns 2-D
y_dash = y_dash[0].assign(0.0)
y_dash = y_dash[self.N].assign(0.0)
return y_dash
def algo_1(self):
self.approx = tf.contrib.integrate.odeint(
func=self.func,
y0=tf.sin(tm.scalar_mul(scalar=math.pi, x=self.domain)),
t=tf.constant([0.0, 1.0]),
rtol=1e-06,
atol=1e-12,
method='dopri5',
options={"max_num_steps":10**10},
full_output=False,
name=None
).numpy()[1]
def algo_2(self):
self.approx = tf.contrib.integrate.odeint_fixed(
func=self.func,
y0=tf.sin(tm.scalar_mul(scalar=math.pi, x=self.domain)),
t=tf.constant([0.0, 1.0]),
dt=tf.constant([self.H**2], dtype=tf.float32),
method='rk4',
name=None
).numpy()[1]
df = pd.DataFrame(columns=["NumBasis", "Errors"])
Ns = [2**r for r in range(2, 10)]
l =[]
for i in tqdm_notebook(Ns):
heateqn = Heat(i)
heateqn.algo_1()
l.append([i, heateqn.error()])
df.append({"NumBasis":i, "Errors":heateqn.error()}, ignore_index=True)
tf.keras.backend.clear_session()

Related

How to implement custom Keras ordinal loss function with tensor evaluation without disturbing TF>2.0 Model Graph?

I am trying to implement a custom loss function in Tensorflow 2.4 using the Keras backend.
The loss function is a ranking loss; I found the following paper with a somewhat log-likelihood loss: Chen et al. Single-Image Depth Perception in the Wild.
Similarly, I wanted to sample some (in this case 50) points from an image to compare the relative order between ground-truth and predicted depth maps using the NYU-Depth dataset. Being a fan of Numpy, I started working with that but came to the following exception:
ValueError: No gradients provided for any variable: [...]
I have learned that this is caused by the arguments not being filled when calling the loss function but instead, a C function is compiled which is then used later. So while I know the dimensions of my tensors (4, 480, 640, 1), I cannot work with the data as wanted and have to use the keras.backend functions on top so that in the end (if I understood correctly), there is supposed to be a path between the input tensors from the TF graph and the output tensor, which has to provide a gradient.
So my question now is: Is this a feasible loss function within keras?
I have already tried a few ideas and different approaches with different variations of my original code, which was something like:
def ranking_loss_function(y_true, y_pred):
# Chen et al. loss
y_true_np = K.eval(y_true)
y_pred_np = K.eval(y_pred)
if y_true_np.shape[0] != None:
num_sample_points = 50
total_samples = num_sample_points ** 2
err_list = [0 for x in range(y_true_np.shape[0])]
for i in range(y_true_np.shape[0]):
sample_points = create_random_samples(y_true, y_pred, num_sample_points)
for x1, y1 in sample_points:
for x2, y2 in sample_points:
if y_true[i][x1][y1] > y_true[i][x2][y2]:
#image_relation_true = 1
err_list[i] += np.log(1 + np.exp(-1 * y_pred[i][x1][y1] + y_pred[i][x2][y2]))
elif y_true[i][x1][y1] < y_true[i][x2][y2]:
#image_relation_true = -1
err_list[i] += np.log(1 + np.exp(y_pred[i][x1][y1] - y_pred[i][x2][y2]))
else:
#image_relation_true = 0
err_list[i] += np.square(y_pred[i][x1][y1] - y_pred[i][x2][y2])
err_list = np.divide(err_list, total_samples)
return K.constant(err_list)
As you can probably tell, the main idea was to first create the sample points and then based on the existing relation between them in y_true/y_pred continue with the corresponding computation from the cited paper.
Can anyone help me and provide some more helpful information or tips on how to correctly implement this loss using keras.backend functions? Trying to include the ordinal relation information really confused me compared to standard regression losses.
EDIT: Just in case this causes confusion: create_random_samples() just creates 50 random sample points (x, y) coordinate pairs based on the shape[1] and shape[2] of y_true (image width and height)
EDIT(2): After finding this variation on GitHub, I have tried out a variation using only TF functions to retrieve data from the tensors and compute the output. The adjusted and probably more correct version still throws the same exception though:
def ranking_loss_function(y_true, y_pred):
#In the Wild ranking loss
y_true_np = K.eval(y_true)
y_pred_np = K.eval(y_pred)
if y_true_np.shape[0] != None:
num_sample_points = 50
total_samples = num_sample_points ** 2
bs = y_true_np.shape[0]
w = y_true_np.shape[1]
h = y_true_np.shape[2]
total_samples = total_samples * bs
num_pairs = tf.constant([total_samples], dtype=tf.float32)
output = tf.Variable(0.0)
for i in range(bs):
sample_points = create_random_samples(y_true, y_pred, num_sample_points)
for x1, y1 in sample_points:
for x2, y2 in sample_points:
y_true_sq = tf.squeeze(y_true)
y_pred_sq = tf.squeeze(y_pred)
d1_t = tf.slice(y_true_sq, [i, x1, y1], [1, 1, 1])
d2_t = tf.slice(y_true_sq, [i, x2, y2], [1, 1, 1])
d1_p = tf.slice(y_pred_sq, [i, x1, y1], [1, 1, 1])
d2_p = tf.slice(y_pred_sq, [i, x2, y2], [1, 1, 1])
d1_t_sq = tf.squeeze(d1_t)
d2_t_sq = tf.squeeze(d2_t)
d1_p_sq = tf.squeeze(d1_p)
d2_p_sq = tf.squeeze(d2_p)
if d1_t_sq > d2_t_sq:
# --> Image relation = 1
output.assign_add(tf.math.log(1 + tf.math.exp(-1 * d1_p_sq + d2_p_sq)))
elif d1_t_sq < d2_t_sq:
# --> Image relation = -1
output.assign_add(tf.math.log(1 + tf.math.exp(d1_p_sq - d2_p_sq)))
else:
output.assign_add(tf.math.square(d1_p_sq - d2_p_sq))
return output/num_pairs
EDIT(3): This is the code for create_random_samples():
(FYI: Because it was weird to get the shape from y_true in this case, I first proceeded to hard-code it here as I know it for the dataset which I am currently using.)
def create_random_samples(y_true, y_pred, num_points=50):
y_true_shape = (4, 480, 640, 1)
y_pred_shape = (4, 480, 640, 1)
if y_true_shape[0] != None:
num_samples = num_points
population = [(x, y) for x in range(y_true_shape[1]) for y in range(y_true_shape[2])]
sample_points = random.sample(population, num_samples)
return sample_points

tensorflow nested map_fn concat two tensors

say I have two tensors:
a=Tensor("zeros_3:0", shape=(2, 4, 5), dtype=float32)
b=Tensor("ones_3:0", shape=(2, 3, 5), dtype=float32)
how can I concat each element along axis 2 to get a new tensor shaped (2,3,4,10), using nested map_fn or other tf functions?
here is my for loop version
concat_list = []
for i in range(a.get_shape()[1]):
for j in range(b.get_shape()[1]):
concat_list.append(tf.concat([a[:, i, :], b[:, j, :]], axis=1))
there is a similar question using "new unit dimension", but I don't know how to use tf.concat with the "new unit dimension".
You can use tf.tile and tf.expand_dims with tf.concat. An example:
import tensorflow as tf
a = tf.random_normal(shape=(2,4,5),dtype=tf.float32)
b = tf.random_normal(shape=(2,3,5),dtype=tf.float32)
# your code
concat_list = []
for i in range(a.get_shape()[1]):
for j in range(b.get_shape()[1]):
concat_list.append(tf.concat([a[:, i, :], b[:, j, :]], axis=1))
# Application method
A = tf.tile(tf.expand_dims(a,axis=1),[1,b.shape[1],1,1])
B = tf.tile(tf.expand_dims(b,axis=2),[1,1,a.shape[1],1])
result = tf.concat([A,B],axis=-1)
with tf.Session() as sess:
concat_list_val,result_val = sess.run([concat_list,result])
print(concat_list_val[-1])
print(result_val.shape)
print(result_val[:,-1,-1,:])
# your result
[[ 1.0459949 1.5562199 -0.04387079 0.17898582 -1.9795663 0.988437
-0.40415847 0.8865694 -1.4764767 -0.8417388 ]
[-0.3542176 -0.3281141 0.01491702 0.91899025 -1.0651684 0.12315683
0.6555444 -0.80451876 -1.3260773 0.33680603]]
# Application result shape
(2, 3, 4, 10)
# Application result
[[ 1.0459949 1.5562199 -0.04387079 0.17898582 -1.9795663 0.988437
-0.40415847 0.8865694 -1.4764767 -0.8417388 ]
[-0.3542176 -0.3281141 0.01491702 0.91899025 -1.0651684 0.12315683
0.6555444 -0.80451876 -1.3260773 0.33680603]]
Performance
You can use follow code to compare speed.
import datetime
...
with tf.Session() as sess:
start = datetime.datetime.now()
print('#' * 60)
for i in range(10000):
result_val = sess.run(result)
end = datetime.datetime.now()
print('cost time(seconds) : %.2f' % ((end - start).total_seconds()))
start = datetime.datetime.now()
print('#' * 60)
for i in range(10000):
concat_list_val = sess.run(concat_list)
end = datetime.datetime.now()
print('cost time(seconds) : %.2f' % ((end - start).total_seconds()))
The vectorization method 10000 iterations takes 1.48s and the loop 10000 iterations takes 5.76s when a.shape=(2,4,5) and b.shape=(2,3,5) on my 8GB GPU memory. But the vectorization method takes 3.28s and the loop time is 317.23s when a.shape=(20,40,5) and b.shape=(20,40,5).
The vectorization method will be significantly faster than the tf.map_fn() and python loop.

Pairwise distance between a set of Matrices in Keras/Tensorflow

I want to calculate pairwise distance between a set of Tensor (e.g 4 Tensor). Each matrix is 2D Tensor. I don't know how to do this in vectorize format. I wrote following sudo-code to determine what I need:
E.shape => [4,30,30]
sum = 0
for i in range(4):
for j in range(4):
res = calculate_distance(E[i],E[j]) # E[i] is one the 30*30 Tensor
sum = sum + reduce_sum(res)
Here is my last try:
x_ = tf.expand_dims(E, 0)
y_ = tf.expand_dims(E, 1)
s = x_ - y_
P = tf.reduce_sum(tf.norm(s, axis=[-2, -1]))
This code works But I don't know how do this in a Batch. For instance when E.shape is [BATCH_SIZE * 4 * 30 * 30] my code doesn't work and Out Of Memory will happen. How can I do this efficiently?
Edit: After a day, I find a solution. it's not perfect but works:
res = tf.map_fn(lambda x: tf.map_fn(lambda y: tf.map_fn(lambda z: tf.norm(z - x), x), x), E)
res = tf.reduce_mean(tf.square(res))
Your solution with expand_dims should be okay if your batch size is not too large. However, given that your original pseudo code loops over range(4), you should probably expand axes 1 and 2, instead of 0 and 1.
You can check the shape of the tensors to ensure that you're specifying the correct axes. For example,
batch_size = 8
E_np = np.random.rand(batch_size, 4, 30, 30)
E = K.variable(E_np) # shape=(8, 4, 30, 30)
x_ = K.expand_dims(E, 1)
y_ = K.expand_dims(E, 2)
s = x_ - y_ # shape=(8, 4, 4, 30, 30)
distances = tf.norm(s, axis=[-2, -1]) # shape=(8, 4, 4)
P = K.sum(distances, axis=[-2, -1]) # shape=(8,)
Now P will be the sum of pairwise distances between the 4 matrices for each of the 8 samples.
You can also verify that the values in P is the same as what would be computed in your pseudo code:
answer = []
for batch_idx in range(batch_size):
s = 0
for i in range(4):
for j in range(4):
a = E_np[batch_idx, i]
b = E_np[batch_idx, j]
s += np.sqrt(np.trace(np.dot(a - b, (a - b).T)))
answer.append(s)
print(answer)
[149.45960605637578, 147.2815068236368, 144.97487402393705, 146.04866735065312, 144.25537059201062, 148.9300986019226, 146.61229889228133, 149.34259789169045]
print(K.eval(P).tolist())
[149.4595947265625, 147.281494140625, 144.97488403320312, 146.04867553710938, 144.25537109375, 148.9300994873047, 146.6123046875, 149.34259033203125]
Tensorflow allows to compute the Frobenius norm via tf.norm function. In case of 2D matrices, it's equivalent to 1-norm.
The following solution isn't vectorized and assumes that the first dimension in E is known statically:
E = tf.random_normal(shape=[5, 3, 3], dtype=tf.float32)
F = tf.split(E, E.shape[0])
total = tf.reduce_sum([tf.norm(tensor=(lhs-rhs), ord=1, axis=(-2, -1)) for lhs in F for rhs in F])
Update:
An optimized vectorized version of the same code:
E = tf.random_normal(shape=[1024, 4, 30, 30], dtype=tf.float32)
lhs = tf.expand_dims(E, axis=1)
rhs = tf.expand_dims(E, axis=2)
total = tf.reduce_sum(tf.norm(tensor=(lhs - rhs), ord=1, axis=(-2, -1)))
Memory concerns: upon evaluating this code,
tf.contrib.memory_stats.MaxBytesInUse() reports that the peak memory consumption is 73729792 = 74Mb, which indicates relatively moderate overhead (the raw lhs-rhs tensor is 59Mb). Your OOM is most likely caused by the duplication of BATCH_SIZE dimension when you compute s = x_ - y_, because your batch size is much larger than the number of matrices (1024 vs 4).

Row-wise Histogram

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)

Avoiding optimization pitfalls when modeling an ordinal predicted variable in PyMC3

I am trying to model an ordinal predicted variable using PyMC3 based on the approach in chapter 23 of Doing Bayesian Data Analysis. I would like to determine a good starting value using find_MAP, but am receiving an optimization error.
The model:
import pymc3 as pm
import numpy as np
import theano
import theano.tensor as tt
# Some helper functions
def cdf(x, location=0, scale=1):
epsilon = np.array(1e-32, dtype=theano.config.floatX)
location = tt.cast(location, theano.config.floatX)
scale = tt.cast(scale, theano.config.floatX)
div = tt.sqrt(2 * scale ** 2 + epsilon)
div = tt.cast(div, theano.config.floatX)
erf_arg = (x - location) / div
return .5 * (1 + tt.erf(erf_arg + epsilon))
def percent_to_thresh(idx, vect):
return 5 * tt.sum(vect[:idx + 1]) + 1.5
def full_thresh(thresh):
idxs = tt.arange(thresh.shape[0] - 1)
thresh_mod, updates = theano.scan(fn=percent_to_thresh,
sequences=[idxs],
non_sequences=[thresh])
return tt.concatenate([[-1 * np.inf, 1.5], thresh_mod, [6.5, np.inf]])
def compute_ps(thresh, location, scale):
f_thresh = full_thresh(thresh)
return cdf(f_thresh[1:], location, scale) - cdf(f_thresh[:-1], location, scale)
# Generate data
real_ps = [0.05, 0.05, 0.1, 0.1, 0.2, 0.3, 0.2]
data = np.random.choice(7, size=1000, p=real_ps)
# Run model
with pm.Model() as model:
mu = pm.Normal('mu', mu=4, sd=3)
sigma = pm.Uniform('sigma', lower=0.1, upper=70)
thresh = pm.Dirichlet('thresh', a=np.ones(5))
cat_p = compute_ps(thresh, mu, sigma)
results = pm.Categorical('results', p=cat_p, observed=data)
with model:
start = pm.find_MAP()
trace = pm.sample(2000, start=start)
When running this, I receive the following error:
Applied interval-transform to sigma and added transformed sigma_interval_ to model.
Applied stickbreaking-transform to thresh and added transformed thresh_stickbreaking_ to model.
Traceback (most recent call last):
File "cm_net_log.v1-for_so.py", line 53, in <module>
start = pm.find_MAP()
File "/usr/local/lib/python3.5/site-packages/pymc3/tuning/starting.py", line 133, in find_MAP
specific_errors)
ValueError: Optimization error: max, logp or dlogp at max have non-finite values. Some values may be outside of distribution support. max: {'thresh_stickbreaking_': array([-1.04298465, -0.48661088, -0.84326554, -0.44833646]), 'sigma_interval_': array(-2.220446049250313e-16), 'mu': array(7.68422528308479)} logp: array(-3506.530143064723) dlogp: array([ 1.61013190e-06, nan, -6.73994118e-06,
-6.93873894e-06, 6.03358122e-06, 3.18954680e-06])Check that 1) you don't have hierarchical parameters, these will lead to points with infinite density. 2) your distribution logp's are properly specified. Specific issues:
My questions:
How can I determine why dlogp is nan at certain points?
Is there a different way that I can express this model to avoid dlogp being nan?
Also worth noting:
This model runs fine if I don't find_MAP and use a Metropolis sampler. However, I'd like to have the flexibility of using other samplers as this model becomes more complex.
I have a suspicion that the issue is due to the relationship between the thresholds and the normal distribution, but I don't know how to disentangle them for the optimization.
Regarding question 2: I expressed the model for the ordinal predicted variable (single group) differently; I used the Theano #as_op decorator for a function that calculates probabilities for the outcomes. That also explains why I cannot use find_MAP() or gradient based samplers: Theano cannot calculate a gradient for the custom function. (http://pymc-devs.github.io/pymc3/notebooks/getting_started.html#Arbitrary-deterministics)
# Number of outcomes
nYlevels = df.Y.cat.categories.size
thresh = [k + .5 for k in range(1, nYlevels)]
thresh_obs = np.ma.asarray(thresh)
thresh_obs[1:-1] = np.ma.masked
#as_op(itypes=[tt.dvector, tt.dscalar, tt.dscalar], otypes=[tt.dvector])
def outcome_probabilities(theta, mu, sigma):
out = np.empty(nYlevels)
n = norm(loc=mu, scale=sigma)
out[0] = n.cdf(theta[0])
out[1] = np.max([0, n.cdf(theta[1]) - n.cdf(theta[0])])
out[2] = np.max([0, n.cdf(theta[2]) - n.cdf(theta[1])])
out[3] = np.max([0, n.cdf(theta[3]) - n.cdf(theta[2])])
out[4] = np.max([0, n.cdf(theta[4]) - n.cdf(theta[3])])
out[5] = np.max([0, n.cdf(theta[5]) - n.cdf(theta[4])])
out[6] = 1 - n.cdf(theta[5])
return out
with pm.Model() as ordinal_model_single:
theta = pm.Normal('theta', mu=thresh, tau=np.repeat(.5**2, len(thresh)),
shape=len(thresh), observed=thresh_obs, testval=thresh[1:-1])
mu = pm.Normal('mu', mu=nYlevels/2.0, tau=1.0/(nYlevels**2))
sigma = pm.Uniform('sigma', nYlevels/1000.0, nYlevels*10.0)
pr = outcome_probabilities(theta, mu, sigma)
y = pm.Categorical('y', pr, observed=df.Y.cat.codes.as_matrix())
http://nbviewer.jupyter.org/github/JWarmenhoven/DBDA-python/blob/master/Notebooks/Chapter%2023.ipynb