I am trying to model Kruschke's "filtration-condensation experiment" with pymc 2.3.5. (numpy 1.10.1)
Basicaly there are:
4 groups
each group has 40 individuals
each individual has 64 Bernoulli trials (correct/incorrect)
What I am modeling:
each individual's results are Binomial distribution (e.g. 45 correct out of 64).
my belief about each individual's performance is Beta distribution.
this Beta distribution is influenced by group to which individual belongs (through parameters A=mu*kappa and B=(1-mu)*kappa)
my belief about how strong each group's influence is Gamma distribution (kappa variable)
my belief about each group's average is Beta distribution (mu variable)
The problem:
when I do modeling with "size=" parameters, pymc get's lost
when I seperate each distribution manually (no size=) the pymc does good job
I include the code below:
Data
import numpy as np
import seaborn as sns
import pymc as pm
from pymc.Matplot import plot as mcplot
%matplotlib inline
# Data
ncond = 4
nSubj = 40
trials = 64
N = np.repeat([trials], (ncond * nSubj))
z = np.array([45, 63, 58, 64, 58, 63, 51, 60, 59, 47, 63, 61, 60, 51, 59, 45,
61, 59, 60, 58, 63, 56, 63, 64, 64, 60, 64, 62, 49, 64, 64, 58, 64, 52, 64, 64,
64, 62, 64, 61, 59, 59, 55, 62, 51, 58, 55, 54, 59, 57, 58, 60, 54, 42, 59, 57,
59, 53, 53, 42, 59, 57, 29, 36, 51, 64, 60, 54, 54, 38, 61, 60, 61, 60, 62, 55,
38, 43, 58, 60, 44, 44, 32, 56, 43, 36, 38, 48, 32, 40, 40, 34, 45, 42, 41, 32,
48, 36, 29, 37, 53, 55, 50, 47, 46, 44, 50, 56, 58, 42, 58, 54, 57, 54, 51, 49,
52, 51, 49, 51, 46, 46, 42, 49, 46, 56, 42, 53, 55, 51, 55, 49, 53, 55, 40, 46,
56, 47, 54, 54, 42, 34, 35, 41, 48, 46, 39, 55, 30, 49, 27, 51, 41, 36, 45, 41,
53, 32, 43, 33])
condition = np.repeat([0,1,2,3], nSubj)
Does not work
# modeling
mu = pm.Beta('mu', 1, 1, size=ncond)
kappa = pm.Gamma('gamma', 1, 0.1, size=ncond)
# Prior
theta = pm.Beta('theta', mu[condition] * kappa[condition], (1 - mu[condition]) * kappa[condition], size=len(z))
# likelihood
y = pm.Binomial('y', p=theta, n=N, value=z, observed=True)
# model
model = pm.Model([mu, kappa, theta, y])
mcmc = pm.MCMC(model)
#mcmc.use_step_method(pm.Metropolis, mu)
#mcmc.use_step_method(pm.Metropolis, theta)
#mcmc.assign_step_methods()
mcmc.sample(100000, burn=20000, thin=3)
# outputs never converge and does vary in new simulations
mcplot(mcmc.trace('mu'), common_scale=False)
Works
z1 = z[:40]
z2 = z[40:80]
z3 = z[80:120]
z4 = z[120:]
Nv = N[:40]
mu1 = pm.Beta('mu1', 1, 1)
mu2 = pm.Beta('mu2', 1, 1)
mu3 = pm.Beta('mu3', 1, 1)
mu4 = pm.Beta('mu4', 1, 1)
kappa1 = pm.Gamma('gamma1', 1, 0.1)
kappa2 = pm.Gamma('gamma2', 1, 0.1)
kappa3 = pm.Gamma('gamma3', 1, 0.1)
kappa4 = pm.Gamma('gamma4', 1, 0.1)
# Prior
theta1 = pm.Beta('theta1', mu1 * kappa1, (1 - mu1) * kappa1, size=len(Nv))
theta2 = pm.Beta('theta2', mu2 * kappa2, (1 - mu2) * kappa2, size=len(Nv))
theta3 = pm.Beta('theta3', mu3 * kappa3, (1 - mu3) * kappa3, size=len(Nv))
theta4 = pm.Beta('theta4', mu4 * kappa4, (1 - mu4) * kappa4, size=len(Nv))
# likelihood
y1 = pm.Binomial('y1', p=theta1, n=Nv, value=z1, observed=True)
y2 = pm.Binomial('y2', p=theta2, n=Nv, value=z2, observed=True)
y3 = pm.Binomial('y3', p=theta3, n=Nv, value=z3, observed=True)
y4 = pm.Binomial('y4', p=theta4, n=Nv, value=z4, observed=True)
# model
model = pm.Model([mu1, kappa1, theta1, y1, mu2, kappa2, theta2, y2,
mu3, kappa3, theta3, y3, mu4, kappa4, theta4, y4])
mcmc = pm.MCMC(model)
#mcmc.use_step_method(pm.Metropolis, mu)
#mcmc.use_step_method(pm.Metropolis, theta)
#mcmc.assign_step_methods()
mcmc.sample(100000, burn=20000, thin=3)
# outputs converge and are not too much different in every simulation
mcplot(mcmc.trace('mu1'), common_scale=False)
mcplot(mcmc.trace('mu2'), common_scale=False)
mcplot(mcmc.trace('mu3'), common_scale=False)
mcplot(mcmc.trace('mu4'), common_scale=False)
mcmc.summary()
Can someone please explain it to me why mu[condition] and gamma[condition] does not work? :)
I guess that not splitting thetas into different variables is the problem but cannot understand why and maybe there is a way to pass a shape parameter to size= on theta?
First of all, I can confirm that the first version doesn't lead to stable results. What I can't confirm is that the second one is much better; I have seen very different results also with the second code, with values for the first mu parameter varying between 0.17 and 0.9 for different runs.
The convergence problems can be cured by using good starting values for the Markov chain. This can be done by first doing a maximum a posteriori (MAP) estimate, and then starting the Markov chain from there. The MAP step is computationally inexpensive and leads to a converging Markov chain with reproducible results for both variants of your code. For reference and comparison: The values I see for the four mu parameters are around 0.94 / 0.86 / 0.72 and 0.71.
You can do the MAP estimation by inserting the following two lines of code right after the line in which you define your model with "model=pm.Model(...":
map_ = pm.MAP(model)
map_.fit()
This technique is covered in more detail in Cameron Davidson-Pilon's Bayesian Methods for Hackers, together with other helpful topics around PyMC.
Related
I have a LSTM model I am using to predict the unemployment rate from federal reserve filings. It uses glove vectors and vocab2index embedding and the training went as planned. However, upon attempting to feed a word embedding into the model for prediction testing it keeps throwing various errors.
Here is the model:
def load_glove_vectors(glove_file= glove_embedding_vectors_text_file):
"""Load the glove word vectors"""
word_vectors = {}
with open(glove_file) as f:
for line in f:
split = line.split()
word_vectors[split[0]] = np.array([float(x) for x in split[1:]])
return word_vectors
def get_emb_matrix(pretrained, word_counts, emb_size = 300):
""" Creates embedding matrix from word vectors"""
vocab_size = len(word_counts) + 2
vocab_to_idx = {}
vocab = ["", "UNK"]
W = np.zeros((vocab_size, emb_size), dtype="float32")
W[0] = np.zeros(emb_size, dtype='float32') # adding a vector for padding
W[1] = np.random.uniform(-0.25, 0.25, emb_size) # adding a vector for unknown words
vocab_to_idx["UNK"] = 1
i = 2
for word in word_counts:
if word in word_vecs:
W[i] = word_vecs[word]
else:
W[i] = np.random.uniform(-0.25,0.25, emb_size)
vocab_to_idx[word] = i
vocab.append(word)
i += 1
return W, np.array(vocab), vocab_to_idx
word_vecs = load_glove_vectors()
pretrained_weights, vocab, vocab2index = get_emb_matrix(word_vecs, counts)
Unfortunately when I feed this array
[array([ 3, 10, 6287, 6, 113, 271, 3, 6639, 104, 5105, 7525,
104, 7526, 9, 23, 9, 10, 11, 24, 7527, 7528, 104,
11, 24, 7529, 7530, 104, 11, 24, 7531, 7530, 104, 11,
24, 7532, 7530, 104, 11, 24, 7533, 7534, 24, 7535, 7536,
104, 7537, 104, 7538, 7539, 7540, 6643, 7541, 7354, 7542, 7543,
7544, 9, 23, 9, 10, 11, 24, 25, 8, 10, 11,
24, 3, 10, 663, 168, 9, 10, 290, 291, 3, 4909,
198, 10, 1478, 169, 15, 4621, 3, 3244, 3, 59, 1967,
113, 59, 520, 198, 25, 5105, 7545, 7546, 7547, 7546, 7548,
7549, 7550, 1874, 10, 7551, 9, 10, 11, 24, 7552, 6287,
7553, 7554, 7555, 24, 7556, 24, 7557, 7558, 7559, 6, 7560,
323, 169, 10, 7561, 1432, 6, 3134, 3, 7562, 6, 7563,
1862, 7144, 741, 3, 3961, 7564, 7565, 520, 7566, 4833, 7567,
7568, 4901, 7569, 7570, 4901, 7571, 1874, 7572, 12, 13, 7573,
10, 7574, 7575, 59, 7576, 59, 638, 1620, 7577, 271, 6488,
59, 7578, 7579, 7580, 7581, 271, 7582, 7583, 24, 669, 5932,
7584, 9, 113, 271, 3764, 3, 5930, 3, 59, 4901, 7585,
793, 7586, 7587, 6, 1482, 520, 7588, 520, 7589, 3246, 7590,
13, 7591])
into torch.LongTensor() I keep getting the following error:
TypeError: can't convert np.ndarray of type numpy.object_. The only supported types are: float64, float32, float16, complex64, complex128, int64, int32, int16, int8, uint8, and bool.
Any ideas on how to remedy? I am fairly new to AI in general, and I am an economist by trade so I am almost certain I have made a boneheaded error.
I am performing an image segmentation with a u-net model.
My mask has classes from 0-50.
I also have a text file dictionary with codes representing each class.
For example -
{1: '1234', 2:'5678', 3:'1245'} etc.
How do I combine when the 2 first string characters are the same so for example above key 1 and 3 are the same because they both start with "12".
How can I do this for all classes?
firstTwoCharDict = {}
for key, value in dictionary.items():
if key == 0:
value == value
firstTwoCharDict[key] = value
else:
value = value[:2]
firstTwoCharDict[key] = value
newDict = {}
for key, value in firstTwoCharDict.items():
if value not in newDict:
newDict[value] = [key]
else:
newDict[value].append(key)
This provides this
{'62': [1, 39],
'90': [2, 5, 9, 20, 32, 42, 47, 72, 88, 91, 95],
'97': [3, 49, 55],
'98': [4, 24, 34, 40, 53, 76, 81, 90, 96],
'31': [6, 17, 30, 48, 83],
'69': [7, 13, 15, 16, 27, 44, 51, 54, 56, 75],
'79': [8, 50],
'71': [10, 19, 22, 35, 61, 63, 65],
'99': [11, 12, 21, 46, 52, 69, 78, 84, 89],
'48': [14, 36, 74],
'60': [18],
'64': [23, 38, 66, 97]
```
Now i have an 2d array with integers, how do I replace them with they keys if the array values are equal to the values in the dict?
In my code, I can filter a column from exact texts, and it works without problems. However, it is necessary to filter another column with the beginning of a sentence.
The phrases in this column are:
A_2020.092222
A_2020.090787
B_2020.983898
B_2020.209308
So, I need to receive everything that starts with A_20 and B_20.
Thanks in advance
My code:
from bs4 import BeautifulSoup
import pandas as pd
import zipfile, urllib.request, shutil, time, csv, datetime, os, sys, os.path
#location
dt = datetime.datetime.now()
file_csv = "/home/Downloads/source.CSV"
file_csv_new = "/var/www/html/Data/Test.csv"
#open CSV
with open(file_csv, 'r', encoding='CP1251') as file:
reader = csv.reader(file, delimiter=';')
data = list(reader)
#list to dataframe
df = pd.DataFrame(data)
#filter UF
df = df.loc[df[9].isin(['PR','SC','RS'])]
#filter key
# A_ & B_
df = df.loc[df[35].isin(['A_20','B_20'])]
#print (df)
#Empty DataFrame
#Columns: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, ...]
#Index: []
#[0 rows x 119 columns]```
Give the following a try:
lst1 = ['A_2020.092222', 'A_2020.090787 ', 'B_2020.983898', 'B_2020.209308', 'C_2020.209308', 'D_2020.209308']
df = pd.DataFrame(lst1, columns =['Name'])
df.loc[df.Name.str.startswith(('A_20','B_20'))]
For the following function:
func CycleClock(c *ballclock.Clock) int {
for i := 0; i < fiveMinutesPerDay; i++ {
c.TickFive()
}
return 1 + CalculateBallCycle(append([]int{}, c.BallQueue...))
}
where c.BallQueue is defined as []int and CalculateBallCycle is defined as func CalculateBallCycle(s []int) int. I am having a huge performance decrease between the for loop and the return statement.
I wrote the following benchmarks to test. The first benchmarks the entire function, the second benchmarks the for loop, while the third benchmarks the CalculateBallCycle function:
func BenchmarkCycleClock(b *testing.B) {
for i := ballclock.MinBalls; i <= ballclock.MaxBalls; i++ {
j := i
b.Run("BallCount="+strconv.Itoa(i), func(b *testing.B) {
for n := 0; n < b.N; n++ {
c, _ := ballclock.NewClock(j)
CycleClock(c)
}
})
}
}
func BenchmarkCycle24(b *testing.B) {
for i := ballclock.MinBalls; i <= ballclock.MaxBalls; i++ {
j := i
b.Run("BallCount="+strconv.Itoa(i), func(b *testing.B) {
for n := 0; n < b.N; n++ {
c, _ := ballclock.NewClock(j)
for k := 0; k < fiveMinutesPerDay; k++ {
c.TickFive()
}
}
})
}
}
func BenchmarkCalculateBallCycle123(b *testing.B) {
m := []int{8, 62, 42, 87, 108, 35, 17, 6, 22, 75, 116, 112, 39, 119, 52, 60, 30, 88, 56, 36, 38, 26, 51, 31, 55, 120, 33, 99, 111, 24, 45, 21, 23, 34, 43, 41, 67, 65, 66, 85, 82, 89, 9, 25, 109, 47, 40, 0, 83, 46, 73, 13, 12, 63, 15, 90, 121, 2, 69, 53, 28, 72, 97, 3, 4, 94, 106, 61, 96, 18, 80, 74, 44, 84, 107, 98, 93, 103, 5, 91, 32, 76, 20, 68, 81, 95, 29, 27, 86, 104, 7, 64, 113, 78, 105, 58, 118, 117, 50, 70, 10, 101, 110, 19, 1, 115, 102, 71, 79, 57, 77, 122, 48, 114, 54, 37, 59, 49, 100, 11, 14, 92, 16}
for n := 0; n < b.N; n++ {
CalculateBallCycle(m)
}
}
Using 123 balls, this gives the following result:
BenchmarkCycleClock/BallCount=123-8 200 9254136 ns/op
BenchmarkCycle24/BallCount=123-8 200000 7610 ns/op
BenchmarkCalculateBallCycle123-8 3000000 456 ns/op
Looking at this, there is a huge disparity between benchmarks. I would expect that the first benchmark would take roughly ~8000 ns/op since that would be the sum of the parts.
Here is the github repository.
EDIT:
I discovered that the result from the benchmark and the result from the running program are widely different. I took what #yazgazan found and modified the benchmark function in main.go mimic somewhat the BenchmarkCalculateBallCycle123 from main_test.go:
func Benchmark() {
for i := ballclock.MinBalls; i <= ballclock.MaxBalls; i++ {
if i != 123 {
continue
}
start := time.Now()
t := CalculateBallCycle([]int{8, 62, 42, 87, 108, 35, 17, 6, 22, 75, 116, 112, 39, 119, 52, 60, 30, 88, 56, 36, 38, 26, 51, 31, 55, 120, 33, 99, 111, 24, 45, 21, 23, 34, 43, 41, 67, 65, 66, 85, 82, 89, 9, 25, 109, 47, 40, 0, 83, 46, 73, 13, 12, 63, 15, 90, 121, 2, 69, 53, 28, 72, 97, 3, 4, 94, 106, 61, 96, 18, 80, 74, 44, 84, 107, 98, 93, 103, 5, 91, 32, 76, 20, 68, 81, 95, 29, 27, 86, 104, 7, 64, 113, 78, 105, 58, 118, 117, 50, 70, 10, 101, 110, 19, 1, 115, 102, 71, 79, 57, 77, 122, 48, 114, 54, 37, 59, 49, 100, 11, 14, 92, 16})
duration := time.Since(start)
fmt.Printf("Ballclock with %v balls took %s;\n", i, duration)
}
}
This gave the output of:
Ballclock with 123 balls took 11.86748ms;
As you can see, the total time was 11.86 ms, all of which was spent in the CalculateBallCycle function. What would cause the benchmark to run in 456 ns/op while the running program runs in around 11867480 ms/op?
You wrote that CalcualteBallCycle() modifies the slice by design.
I can't speak to correctness of that approach, but it is why benchmark time of BenchmarkCalculateBallCycle123 is so different.
On first run it does the expected thing but on subsequent runs it does something completely different, because you're passing different data as input.
Benchmark this modified code:
func BenchmarkCalculateBallCycle123v2(b *testing.B) {
m := []int{8, 62, 42, 87, 108, 35, 17, 6, 22, 75, 116, 112, 39, 119, 52, 60, 30, 88, 56, 36, 38, 26, 51, 31, 55, 120, 33, 99, 111, 24, 45, 21, 23, 34, 43, 41, 67, 65, 66, 85, 82, 89, 9, 25, 109, 47, 40, 0, 83, 46, 73, 13, 12, 63, 15, 90, 121, 2, 69, 53, 28, 72, 97, 3, 4, 94, 106, 61, 96, 18, 80, 74, 44, 84, 107, 98, 93, 103, 5, 91, 32, 76, 20, 68, 81, 95, 29, 27, 86, 104, 7, 64, 113, 78, 105, 58, 118, 117, 50, 70, 10, 101, 110, 19, 1, 115, 102, 71, 79, 57, 77, 122, 48, 114, 54, 37, 59, 49, 100, 11, 14, 92, 16}
for n := 0; n < b.N; n++ {
tmp := append([]int{}, m...)
CalculateBallCycle(tmp)
}
}
This works-around this behavior by making a copy of m, so that CalculateBallCycle modifies a local copy.
The running time becomes more like the others:
BenchmarkCalculateBallCycle123-8 3000000 500 ns/op
BenchmarkCalculateBallCycle123v2-8 100 10483347 ns/op
In your CycleClock function, you are copying the c.BallQueue slice. You can improve performance significantly by using CalculateBallCycle(c.BallQueue) instead (assuming CalculateBallCycle doesn't modify the slice)
For example:
func Sum(values []int) int {
sum := 0
for _, v := range values {
sum += v
}
return sum
}
func BenchmarkNoCopy(b *testing.B) {
for n := 0; n < b.N; n++ {
Sum(m)
}
}
func BenchmarkWithCopy(b *testing.B) {
for n := 0; n < b.N; n++ {
Sum(append([]int{}, m...))
}
}
// BenchmarkNoCopy-4 20000000 73.5 ns/op
// BenchmarkWithCopy-4 5000000 306 ns/op
// PASS
There is a subtle bug in your tests.
Both methods BenchmarkCycleClock and BenchmarkCycle24 run the benchmark in a for loop, passing a closure to b.Run. Inside of those closures you initialize the clocks using the loop variable i like this:ballclock.NewClock(i).
The problem is, that all instances of your anonymous function share the same variable. And, by the time the function is run by the test runner, the loop will be finished, and all of the clocks will be initialized using the same value: ballclock.MaxBalls.
You can fix this using a local variable:
for i := ballclock.MinBalls; i <= ballclock.MaxBalls; i++ {
i := i
b.Run("BallCount="+strconv.Itoa(i), func(b *testing.B) {
for n := 0; n < b.N; n++ {
c, _ := ballclock.NewClock(i)
CycleClock(c)
}
})
}
The line i := i stores a copy of the current value of i (different for each instance of your anonymous function).
I have a signal ts which has rougly mean 40 and applied fft on that with code
ts = array([25, 40, 30, 40, 29, 48, 36, 32, 34, 38, 15, 33, 40, 32, 41, 25, 37,49, 41, 35, 23, 22, 36, 44, 28, 36, 32, 37, 39, 51])
index = fftshift(fftfreq(len(ts)))
ft_ts =fftshift(fft(ts))
output
ft_ts = array([ -76.00000000 +8.34887715e-14j, -57.72501110 +1.17054586e+01j,
7.69492662 +9.79582336e+00j, -29.11145618 -7.22493645e+00j,
14.92140414 +4.58471353e+01j, -26.00000000 -4.67653718e+01j,
-39.61803399 -2.83601821e+01j, -11.34044003 +8.66215368e+00j,
23.68703939 +1.57391882e+01j, -64.88854382 -2.44499549e+01j,
50.00000000 -3.98371686e+01j, 4.09382150 -6.27663403e+00j,
-37.38196601 -3.06708342e+01j, 35.97162964 +1.31929223e+01j,
18.69662985 -2.20453671e+00j, 1048.00000000 +0.00000000e+00j,
18.69662985 +2.20453671e+00j, 35.97162964 -1.31929223e+01j,
-37.38196601 +3.06708342e+01j, 4.09382150 +6.27663403e+00j,
50.00000000 +3.98371686e+01j, -64.88854382 +2.44499549e+01j,
23.68703939 -1.57391882e+01j, -11.34044003 -8.66215368e+00j,
-39.61803399 +2.83601821e+01j, -26.00000000 +4.67653718e+01j,
14.92140414 -4.58471353e+01j, -29.11145618 +7.22493645e+00j,
7.69492662 -9.79582336e+00j, -57.72501110 -1.17054586e+01j])
at 0 frequency ft_ts has value of 1048. Shouldn't that be the mean of my original signal ts which is 40 ? What happened here ?
Many thanks
The FFT is not normalized, so the first term should be the sum, not the mean.
For example, see the definition here
and you can see, that when k=0, the exponential term is 1, and you'll just get the sum of x_n.
This is why the first item in fft(np.ones(10)) is 10, not 1. 1 is the mean (since it's an array of ones), and 10 is the sum.