In this page https://courses.cit.cornell.edu/bionb441/CA/forest.m
I found a code named "Forest Fire"
I am trying to figure out how this code works for educational purposes.
Here are the rules:
Cells can be in 3 different states. State=0 is empty, state=1 is burning and state=2 is forest.
If one or more of the 4 neighbors of a cell is burning and it is forest (state=2) then the new state is burning (state=1).
A cell which is burning (state=1) becomes empty (state=0).
There is a low probablity (0.000005) of a forest cell (state=2) starting to burn on its own (from lightning).
There is a low probability (say, 0.01) of an empty cell becoming forest to simulate growth.
what it is not very clear how it works is...
sum = (veg(1:n,[n 1:n-1])==1) + (veg(1:n,[2:n 1])==1) + ...
(veg([n 1:n-1], 1:n)==1) + (veg([2:n 1],1:n)==1) ;
veg = 2*(veg==2) - ((veg==2) & (sum> 0 | (rand(n,n)< Plightning))) + ...
2*((veg==0) & rand(n,n)< Pgrowth) ;
There is no problem in running the code, it just I am confused what are these vectors (sum and veg). Especially what makes (veg(1:n,[n 1:n-1])==1).
What I see is that, both are matrixes and veg is the data of the plot (matriz with 0's 1's and 2's).
I really appreciate any help you can provide.
Binary comparison operators on a matrix and a scalar return the matrix of elements of that binary comparison with the scalar and the corresponding element of the original matrix.
sum is a matrix in which each cell contains the number of adjacent cells in the corresponding matrix veg that are on fire (==1).
(veg(1:n,[n 1:n-1])==1) is a matrix of logical 1s and 0s (I don't know if the data type is static or dynamic) in which each cell equals 1 when the cell to the left of the corresponding one in veg is on fire (==1).
https://courses.cit.cornell.edu/bionb441/CA/
Look at the URL, go back up the tree to see the source.
The rule:
Cells can be in 3 different states. State=0 is empty, state=1 is burning and state=2 is forest.
If one or more of the 4 neighbors if a cell is burning and it is forest (state=2) then the new state is burning (state=1).
There is a low probablity (say 0.000005) of a forest cell (state=2) starting to burn on its own (from lightning).
A cell which is burning (state=1) becomes empty (state=0).
There is a low probability (say, 0.01) of an empty cell becoming forest to simulate growth.
The array is considered to be toroidly connected, so that fire which burns to left side will start fires on the right. The top and bottom are similarly connected.
The update code:
sum = (veg(1:n,[n 1:n-1])==1) + (veg(1:n,[2:n 1])==1) + ...
(veg([n 1:n-1], 1:n)==1) + (veg([2:n 1],1:n)==1) ;
veg = ...
2*(veg==2) - ((veg==2) & (sum> 0 | (rand(n,n)< Plightning))) + ...
2*((veg==0) & rand(n,n)< Pgrowth) ;
Note that the toroidal connection is implemented by the ordering of subscripts.
Related
I am trying to understand what the VADER does for analysis of sentences.
Why is the hyper-parameter Alpha set to 15 here? I understand that the it is unstable when left unbound, but why 15?
def normalize(score, alpha=15):
"""
Normalize the score to be between -1 and 1 using an alpha that
approximates the max expected value
"""
norm_score = score/math.sqrt((score*score) + alpha)
return norm_score
Vader's normalization equation is which is the equation for
I have read the paper of the research for Vader from here:http://comp.social.gatech.edu/papers/icwsm14.vader.hutto.pdf
Unfortunately, I could not find any reason why such a formula and 15 as the value for alpha was chosen but the experiments and the graph show that as x grows which is the sum of sentiments' scores grow the value becomes closer to -1 or 1 which indicates that as number of words grow the score tends more towards -1 or 1. Which means that Vader works better with short documents or tweets compared to long documents.
My bold claim is that the Octave implementation of the cubic spline, as implemented in interp1(..., "spline") differs from the "natural cubic spline" algorithm outlined in, e.g., Wolfram's Mathworld. I have written my own implementation of the latter and compared it to the output of the interp1(..., "spline") function, with the following results:
I discovered that when I try the same comparison with 4 points, the solutions also differ, and, moreover, the Octave solution is identical to fitting a single cubic polynomial to all four points (and not actually producing a piecewise spline for the three intervals).
I also tried to look under the hood at Octave's implementation of splines, and found it was too obtuse to read and understand in 5 minutes.
I know that there are a few options for boundary conditions one can choose ("natural" vs "clamped") when implementing a cubic spline. My implementation uses "natural" boundary conditions (in which the second derivative of the two exterior points is set to zero).
If Octave's cubic spline is indeed different to the standard cubic spline, then what actually is it?
EDIT:
The second order differences of the two solutions shown in the Comparison plot above are plotted here:
Firstly, there appear to be only two cubic polynomials in Octave's case: one that is fit over the first two intervals, and one that is fit over the last two intervals. Secondly, they are clearly not using "natural" splines, since the second derivatives at the extremes do not tend to zero.
Also, I think the fact that the second order difference for my implementation at the middle (i.e. 3rd) point is zero is just a coincidence, and not demanded by the algorithm. Repeating this test for a different set of points will confirm/refute this.
Different end conditions explains the difference between your implementation and Octave's. Octave uses the not-a-knot condition (depending on input)
See help spline
To explain your observations: the third derivative is continuous at the 2nd and (n-1)th break due to the not-a-knot condition, so that's why Octave's second derivative looks like it has less 'breaks', because it is a continuous straight line over the first two and last two segments. If you look at the third derivative, you can see the effect more clearly - the 3rd derivative is discontinuous only at the 3rd break (the middle)
x = 1:5;
y = rand(1,5);
xx = linspace(1,5);
pp = interp1(x, y, 'spline', 'pp');
yy = ppval(pp, xx);
dyy = ppval(ppder(pp, 3), xx);
plot(xx, yy, xx, dyy);
Also the pp data structure looks like this
pp =
scalar structure containing the fields:
form = pp
breaks =
1 2 3 4 5
coefs =
0.427823 -1.767499 1.994444 0.240388
0.427823 -0.484030 -0.257085 0.895156
-0.442232 0.799439 0.058324 0.581864
-0.442232 -0.527258 0.330506 0.997395
pieces = 4
order = 4
dim = 1
orient = first
I'd like to know how to plot power series (whose variable is x), but I don't even know where to start with.
I know it might not be possible plot infinite series, but it'd do as well plotting the sum of the first n terms.
Gnuplot has a sum function, which can be used inside the using statement to sum up several columns or terms. Together with the special file name + you can implement power series.
Consider the exponention function, which has a power series
\sum_{n=0}^\infty x^n/n!
So, we define a term as
term(x, n) = x**n/n!
Now we can plot the power series up to the n=5 term with
set xrange [0:4]
term(x, n) = x**n/n!
set samples 20
plot '+' using 1:(sum [n=0:5] term($1, n))
To plot the results when using 2 to 7 terms and compare it with the actual exp function, use
term(x, n) = x**n/n!
set xrange [-2:2]
set samples 41
set key left
plot exp(x), for [i=1:6] '+' using 1:(sum[t=0:i] term($1, t)) title sprintf('%d terms', i)
The easiest way that I can think of is to generate a file that has a column of x-values and a column of f(x) values, then just plot the table like you would any other data. A power series is continuous, so you can just connect the dots and have a fairly accurate representation (provided your dots are close enough together). Also, when evaluating f(x), you just sum up the first N terms (where N is big enough). Big enough means that the sum of the rest of the terms is smaller than whatever error you allow. (*If you want 3 good digits, then N needs to be large enough that the remaining sum is smaller than .001.)
You can pull out a calc II textbook to determine how to bound the error on the tail of the sum. A lot of calc classes briefly cover it, but students tend to feel like the error estimates are pointless (I know because I've taught the course a few times.) As an example, if you have an alternating series (whose terms are decreasing in absolute value), then the absolute value of the first term you omit (don't sum) is an upperbound on your error.
*This statement is not 100% true, it is slightly over simplified, but is correct for most practical purposes.
I have a 3d cylinder chart that I am having some problems with. I want to effectively sort the cylinders with the highest value at the back and the lowest value at the front. Otherwise the tallest valuest cover the smallest values.
I have tried sorting both a-z and z-a but I really need it to be dynamic based on the values. I have also tried sorting the values by the actual value field. both a-z and z-a but this seems to return completely random results.
the data in the database (example) looks like. I use a parameter to separate by supplier.
Date catgeory_Type cost supplier
01/01/2013 apple $5 abc
01/01/2013 pear $10 def
01/01/2013 bannana $15 cgi
01/02/2013 apple $7 etc
01/02/2013 pear $12 etc
01/02/2013 banana $18 etc
I believe I need some form of expression that sorts the values based on cost. as both a-z and z-a in the instance would provide cylinders that blocked other cylinders.
I have tried sorting the series group by :=Sum(Fields!cost.Value, "DataSet1") and =Fields!cost.Value but this seems to return random results.
I would be happy even if I could achieve a custom sort such as sort by "bannana, pear, apple" although for some "suppliers" this would still cause me an issue.
edit 1: strangely enough this works with a line chart but not a 3d cylinder
edit 2: example
attached is an example. I want the tallest cylinders at the back. but methods mentioned above do not work
In chart area properties -> 3D-options , Enable,
series clustering
Choose this option to cluster series groups. When multiple series for
bar or column charts are clustered, they are displayed along two
distinct rows in the chart area. If series are not clustered, their
corresponding data points are displayed adjacent to each other in one
row. This option is applicable only to bar and column charts.
Also try changing the Rotation & Inclination degrees, to get a better look.
Decrease wall thickness also.
I would like SAS to print the probability of my binary dependent variable occurring (“Calliphoridae” a particular fly family being present (1) or not (0), at a specific instance for my continuous independent variable (“degree_index” that was recorded from .055 to 2.89, but can be continuously recorded past 2.89 and always increases as time goes on) using Proc GENMOD. How do I change my code to print the probability, for example, that Calliphoridae is present at degree_index=.1?
My example code is:
proc genmod data=thesis descending ;
class Body_number ;
model Calliphoridae = degree_index / dist=binomial link=logit ;
repeated subject=Body_number/ type=cs;
estimate 'degreeindex=.1' intercept 1 degree_index 0 /exp;
estimate 'degree_index=.2' intercept 1 degree_index .1 /exp;run;
I get an output for the contrast estimate results as mean estimate at degree_index=.1 is ..99; degree_index=.2 is .98.
I think that it is correctly modeling the probability...I just didn't include the square of
the degree-day index. If you do, it allows the probability to increase and decrease. I
realized this when I did the probability by hand
(e^-1.1307x+.2119)/(1+e^-1.1307x+.2119) to verify that this really was modeling
probability when y=1 for the mean estimates at specific x values...and then I realized that it is
fitting a regression line and cannot increase and decrease because there is only
one x value. http://www.stat.sc.edu/~hansont/stat704/chapter14a.pdf