I am trying to do the classic method of decomposition of time series. I have somehow managed to get myself to the very last step where I'm supposed to calculate the Trend from the Trend-Cycle series using slope and intercept but I get 2 values instead of one for some reason. Why does slope contain two values?
pkg load io;
pkg load financial;
data = xlsread('exerciseinfo.xlsx','Φύλλο1','A1:B60');
t = data(:,1);
Y = data(:,2);
#Γραφική Παράσταση Δεδομένων
plot(t,Y);
title('Αφίξεις Αυτοκινήτων ανά Μήνα');
xlabel('Μήνας');
ylabel('Αφίξεις');
average = mean(Y);
#ΚΜΟ (5)
M = movmean(Y,5);
plot(t,M);
title('KMO(5)');
xlabel('Μήνας');
ylabel('Αφίξεις');
#Διπλός ΚΜΟ 3χ5
doublema = movmean(M,3);
plot(t,doublema);
title('Διπλός ΚΜΟ 5χ3');
xlabel('Μήνας');
ylabel('Αφίξεις');
# Κεντρικός ΚΜΟ 2χ4
MA4 = movmean(Y,4);
CMA = movmean(MA4,2);
plot(t,CMA);
title('Κεντρικός ΚΜΟ 2χ4');
xlabel('Μήνας');
ylabel('Αφίξεις');
#Βήμα 2 Αποσύνθεσης
Step2 = Y./CMA;
#Βήμα 3 Αποσύνθεσης
Step3 = mean(Step2);
#Βήμα 4 Αποσύνθεσης - Αποεποχικοποιημένη Σειρά
Step4 = Y./Step3;
#Βήμα 5 Αποσύνθεσης - Αφαίρεση τυχαιότητας
MA3 = movmean(Step4,3);
Step5 = movmean(MA3,3);
#Βήμα 6 Αποσύνθεσης
slope = polyfit(t,Step5,1)
I don't speak Greek, so I'll assume that by two "equations" you actually mean two "values", which is what polyfit should give you if you pass it the argument 1. You're fitting a polynomial of degree 1, i.e. an equation of the form y = ax + b, where you fit both a and b. To reconstruct your line then you use y = slope(0)*t + slope(1), where slope(0) is your "slope" and slope(0) your "intercept" (or y = slope(1)*t + slope(0), the docs don't state whether the coefficients are stored with increasing or decreasing degree.)
Related
I'm trying to make a Monte Carlo Simulation with time series and I can't get what I'm doing wrong due to my little knowledge in Stata.
With my ARMA model I'm able to create a time series with 300 observations.
My idea is to do this process 1000 times and in every process I want to save the mean and variance of the 300 observations in a matrix.
This is my ARMA model:
This is my code:
clear all
set more off
set matsize 1000
matrix simulaciones =J(1000,2,0) *To save every simulation of every time series generated
matrix serie = J(300,3,0) *To save each time series
set obs 300 *For the 300 observations in every time series
gen t = _n
tsset t
g y1=0
forvalue j = 1(1)1000{
* Creating a time series
forvalues i = 1(1)300 {
gen e = rnormal(0,1)
replace y1=0 if t==1
replace y1 = 0.7*L1.y1 + e - 0.6*L1.e if t == 2
replace y1 = 0.7*L1.y1 - 0.1*L2.y1 + e - 0.6*L1.e + 0.08*L2.e if t > 2
matrix serie[`i',3] = y1
drop e y1
}
svmat serie
matrix simulaciones[`j',1] = mean(y1)
matrix simulaciones[`j',2] = var(y1)
}
I have no idea how to follow and any idea or recommendation is more than welcomed.
Thanks a lot for your help and time.
I want to calculate the information gain on 20_newsgroup data set.
I am using the code here(also I put a copy of the code down of the question).
As you see the input to the algorithm is X,y
My confusion is that, X is going to be a matrix with documents in rows and features as column. (according to 20_newsgroup it is 11314,1000
in case i only considered 1000 features).
but according to the concept of information gain, it should calculate information gain for each feature.
(So I was expecting to see the code in a way loop through each feature, so the input to the function be a matrix where rows are features and columns are class)
But X is not feature here but X stands for documents, and I can not see the part in the code that take care of this part! ( I mean considering each document, and then going through each feature of that document; like looping through rows but at the same time looping through columns as the features are stored in columns).
I have read this and this and many similar questions but they are not clear in terms of input matrix shape.
this is the code for reading 20_newsgroup:
newsgroup_train = fetch_20newsgroups(subset='train')
X,y = newsgroup_train.data,newsgroup_train.target
cv = CountVectorizer(max_df=0.99,min_df=0.001, max_features=1000,stop_words='english',lowercase=True,analyzer='word')
X_vec = cv.fit_transform(X)
(X_vec.shape) is (11314,1000) which is not features in the 20_newsgroup data set. I am thinking am I calculating Information gain in a incorrect way?
This is the code for Information gain:
def information_gain(X, y):
def _calIg():
entropy_x_set = 0
entropy_x_not_set = 0
for c in classCnt:
probs = classCnt[c] / float(featureTot)
entropy_x_set = entropy_x_set - probs * np.log(probs)
probs = (classTotCnt[c] - classCnt[c]) / float(tot - featureTot)
entropy_x_not_set = entropy_x_not_set - probs * np.log(probs)
for c in classTotCnt:
if c not in classCnt:
probs = classTotCnt[c] / float(tot - featureTot)
entropy_x_not_set = entropy_x_not_set - probs * np.log(probs)
return entropy_before - ((featureTot / float(tot)) * entropy_x_set
+ ((tot - featureTot) / float(tot)) * entropy_x_not_set)
tot = X.shape[0]
classTotCnt = {}
entropy_before = 0
for i in y:
if i not in classTotCnt:
classTotCnt[i] = 1
else:
classTotCnt[i] = classTotCnt[i] + 1
for c in classTotCnt:
probs = classTotCnt[c] / float(tot)
entropy_before = entropy_before - probs * np.log(probs)
nz = X.T.nonzero()
pre = 0
classCnt = {}
featureTot = 0
information_gain = []
for i in range(0, len(nz[0])):
if (i != 0 and nz[0][i] != pre):
for notappear in range(pre+1, nz[0][i]):
information_gain.append(0)
ig = _calIg()
information_gain.append(ig)
pre = nz[0][i]
classCnt = {}
featureTot = 0
featureTot = featureTot + 1
yclass = y[nz[1][i]]
if yclass not in classCnt:
classCnt[yclass] = 1
else:
classCnt[yclass] = classCnt[yclass] + 1
ig = _calIg()
information_gain.append(ig)
return np.asarray(information_gain)
Well, after going through the code in detail, I learned more about X.T.nonzero().
Actually it is correct that information gain needs to loop through features.
Also it is correct that the matrix scikit-learn give us here is based on doc-features.
But:
in code it uses X.T.nonzero() which technically transform all the nonzero values into array. and then in the next row loop through the length of that array range(0, len(X.T.nonzero()[0]).
Overall, this part X.T.nonzero()[0] is returning all the none zero features to us :)
I was following Siraj Raval's videos on logistic regression using gradient descent :
1) Link to longer video :
https://www.youtube.com/watch?v=XdM6ER7zTLk&t=2686s
2) Link to shorter video :
https://www.youtube.com/watch?v=xRJCOz3AfYY&list=PL2-dafEMk2A7mu0bSksCGMJEmeddU_H4D
In the videos he talks about using gradient descent to reduce the error for a set number of iterations so that the function converges(slope becomes zero).
He also illustrates the process via code. The following are the two main functions from the code :
def step_gradient(b_current, m_current, points, learningRate):
b_gradient = 0
m_gradient = 0
N = float(len(points))
for i in range(0, len(points)):
x = points[i, 0]
y = points[i, 1]
b_gradient += -(2/N) * (y - ((m_current * x) + b_current))
m_gradient += -(2/N) * x * (y - ((m_current * x) + b_current))
new_b = b_current - (learningRate * b_gradient)
new_m = m_current - (learningRate * m_gradient)
return [new_b, new_m]
def gradient_descent_runner(points, starting_b, starting_m, learning_rate, num_iterations):
b = starting_b
m = starting_m
for i in range(num_iterations):
b, m = step_gradient(b, m, array(points), learning_rate)
return [b, m]
#The above functions are called below:
learning_rate = 0.0001
initial_b = 0 # initial y-intercept guess
initial_m = 0 # initial slope guess
num_iterations = 1000
[b, m] = gradient_descent_runner(points, initial_b, initial_m, learning_rate, num_iterations)
# code taken from Siraj Raval's github page
Why does the value of b & m continue to update for all the iterations? After a certain number of iterations, the function will converge, when we find the values of b & m that give slope = 0.
So why do we continue iteration after that point and continue updating b & m ?
This way, aren't we losing the 'correct' b & m values? How is learning rate helping the convergence process if we continue to update values after converging? Thus, why is there no check for convergence, and so how is this actually working?
In practice, most likely you will not reach to slope 0 exactly. Thinking of your loss function as a bowl. If your learning rate is too high, it is possible to overshoot over the lowest point of the bowl. On the contrary, if the learning rate is too low, your learning will become too slow and won't reach the lowest point of the bowl before all iterations are done.
That's why in machine learning, the learning rate is an important hyperparameter to tune.
Actually, once we reach a slope 0; b_gradient and m_gradient will become 0;
thus, for :
new_b = b_current - (learningRate * b_gradient)
new_m = m_current - (learningRate * m_gradient)
new_b and new_m will remain the old correct values; as nothing will be subtracted from them.
I'm trying to learn theano and decided to implement linear regression (using their Logistic Regression from the tutorial as a template). I'm getting a wierd thing where T.grad doesn't work if my cost function uses .sum(), but does work if my cost function uses .mean(). Code snippet:
(THIS DOESN'T WORK, RESULTS IN A W VECTOR FULL OF NANs):
x = T.matrix('x')
y = T.vector('y')
w = theano.shared(rng.randn(feats), name='w')
b = theano.shared(0., name="b")
# now we do the actual expressions
h = T.dot(x,w) + b # prediction is dot product plus bias
single_error = .5 * ((h - y)**2)
cost = single_error.sum()
gw, gb = T.grad(cost, [w,b])
train = theano.function(inputs=[x,y], outputs=[h, single_error], updates = ((w, w - .1*gw), (b, b - .1*gb)))
predict = theano.function(inputs=[x], outputs=h)
for i in range(training_steps):
pred, err = train(D[0], D[1])
(THIS DOES WORK, PERFECTLY):
x = T.matrix('x')
y = T.vector('y')
w = theano.shared(rng.randn(feats), name='w')
b = theano.shared(0., name="b")
# now we do the actual expressions
h = T.dot(x,w) + b # prediction is dot product plus bias
single_error = .5 * ((h - y)**2)
cost = single_error.mean()
gw, gb = T.grad(cost, [w,b])
train = theano.function(inputs=[x,y], outputs=[h, single_error], updates = ((w, w - .1*gw), (b, b - .1*gb)))
predict = theano.function(inputs=[x], outputs=h)
for i in range(training_steps):
pred, err = train(D[0], D[1])
The only difference is in the cost = single_error.sum() vs single_error.mean(). What I don't understand is that the gradient should be the exact same in both cases (one is just a scaled version of the other). So what gives?
The learning rate (0.1) is way to big. Using mean make it divided by the batch size, so this help. But I'm pretty sure you should make it much smaller. Not just dividing by the batch size (which is equivalent to using mean).
Try a learning rate of 0.001.
Try dividing your gradient descent step size by the number of training examples.
I wanna combine three graphics on one graph. The data from inside of R which is " nottem ". Can someone help me to write code to put a seasonal mean and harmonic (cosine model) and its time series plots together by using different colors? I already wrote model code just don't know how to combine them together to compare.
Code :library(TSA)
nottem
month.=season(nottem)
model=lm(nottem~month.-1)
summary(nottem)
har.=harmonic(nottem,1)
model1=lm(nottem~har.)
summary(model1)
plot(nottem,type="l",ylab="Average monthly temperature at Nottingham castle")
points(y=nottem,x=time(nottem), pch=as.vector(season(nottem)))
Just put your time series inside a matrix:
x = cbind(serie1 = ts(cumsum(rnorm(100)), freq = 12, start = c(2013, 2)),
serie2 = ts(cumsum(rnorm(100)), freq = 12, start = c(2013, 2)))
plot(x)
Or configure the plot region:
par(mfrow = c(2, 1)) # 2 rows, 1 column
serie1 = ts(cumsum(rnorm(100)), freq = 12, start = c(2013, 2))
serie2 = ts(cumsum(rnorm(100)), freq = 12, start = c(2013, 2))
require(zoo)
plot(serie1)
lines(rollapply(serie1, width = 10, FUN = mean), col = 'red')
plot(serie2)
lines(rollapply(serie2, width = 10, FUN = mean), col = 'blue')
hope it helps.
PS.: zoo package is not needed in this example, you could use the filter function.
You can extract the seasonal mean with:
s.mean = tapply(serie, cycle(serie), mean)
# January, assuming serie is monthly data
print(s.mean[1])
This graph is pretty hard to read, because your three sets of values are so similar. Still, if you want to simply want to graph all of these on the sample plot, you can do it pretty easily by using the coefficients generated by your models.
Step 1: Plot the raw data. This comes from your original code.
plot(nottem,type="l",ylab="Average monthly temperature at Nottingham castle")
Step 2: Set up x-values for the mean and cosine plots.
x <- seq(1920, (1940 - 1/12), by=1/12)
Step 3: Plot the seasonal means by repeating the coefficients from the first model.
lines(x=x, y=rep(model$coefficients, 20), col="blue")
Step 4: Calculate the y-values for the cosine function using the coefficients from the second model, and then plot.
y <- model1$coefficients[2] * cos(2 * pi * x) + model1$coefficients[1]
lines(x=x, y=y, col="red")
ggplot variant: If you decide to switch to the popular 'ggplot2' package for your plot, you would do it like so:
x <- seq(1920, (1940 - 1/12), by=1/12)
y.seas.mean <- rep(model$coefficients, 20)
y.har.cos <- model1$coefficients[2] * cos(2 * pi * x) + model1$coefficients[1]
plot_Data <- melt(data.frame(x=x, temp=nottem, seas.mean=y.seas.mean, har.cos=y.har.cos), id="x")
ggplot(plot_Data, aes(x=x, y=value, col=variable)) + geom_line()