I want to create a temp variable in jags, but it doesn't work as it would work in R
for (cid in 1:CAMPAIGN_N) {
for (time in 1:DATE_N){
index <- time * CAMPAIGN_N + cid - 2
positives[index] ~ dbin( k[time]*ctr[cid], tries[index])
}
}
Gives error, because index variable is being defined only once. So I had to write it the following ugly way:
for (cid in 1:CAMPAIGN_N) {
for (time in 1:DATE_N){
positives[time * CAMPAIGN_N + cid - 2]
~ dbin( k[time]*ctr[cid], tries[time * CAMPAIGN_N + cid - 2])
}
}
Is there a way I can create temp variable in jags?
You'd need to let index vary with time and cid.
index[time,cid] <- time * CAMPAIGN_N + cid - 2
positives[index[time,cid]] ~ dbin( k[time]*ctr[cid], tries[index[time,cid]])
Related
I created my code based on this :
http://users.aims.ac.za/~mackay/BUGS/Manual05/Examples1/node29.html
Now I use different seeds to simulate data. It is strange enough that some seeds give me a redefining node dN[1,1] on line 18 error, while others do not. Could someone help please? BTW why is dN[1,1] on line 18 at the first place? How does Jags count lines?
error message :
"
RUNTIME ERROR:
Compilation error on line 18.
Attempt to redefine node dN[1,1]
"
bugsmodel <- "
# Set up data
data{
for(i in 1:N)
{
for(j in 1:bigt)
{
Y[i,j] <- step(obs.t[i] - t[j] + eps)
dN[i, j] <- Y[i, j] * step(t[j + 1] - obs.t[i] - eps) * fail[i]
}
}
}
# Model
model
{
for(i in 1:N){
betax[i,1] <- 0
for(k in 2:(p+1)){
betax[i,k] <- betax[i,k-1] + beta[k-1]*x[i,k-1]
}
}
for(j in 1:bigt) {
for(i in 1:N) {
dN[i, j] ~ dpois(Idt[i, j]) # Likelihood
Idt[i, j] <- Y[i, j] * exp(betax[i,p+1]) * dL0[j] # Intensity
}
dL0[j] ~ dgamma(mu[j], c)
mu[j] <- dL0.star[j] * c # prior mean hazard
}
c <- 0.001
r <- 0.1
for (j in 1 : bigt) {
dL0.star[j] <- r * (t[j + 1] - t[j])
}
for(k in 1:p){
beta[k] ~ dnorm(0.0,0.000001)
}
}"
Here's the link to the question..
http://www.codechef.com/problems/J7
I figured out that 2 edges have to be equal in order to give the maximum volume, and then used x, x, a*x as the lengths of the three edges to write the equations -
4*x + 4*x + 4*a*x = P (perimeter) and,
2*x^2 + 4*(a*x *x) = S (total area of the box)
so from the first equation I got x in terms of P and a, and then substituted it in the second equation and then got a quadratic equation with the unknown being a. and then I used the greater root of a and got x.
But this method seems to be giving the wrong answer! :|
I know that there isn't any logical error in this. Maybe some formatting error?
Here's the main code that I've written :
{
public static void main(String[] args)
{
TheBestBox box = new TheBestBox();
reader = box.new InputReader(System.in);
writer = box.new OutputWriter(System.out);
getAttributes();
writer.flush();
reader.close();
writer.close();
}
public static void getAttributes()
{
t = reader.nextInt(); // t is the number of test cases in the question
for (int i = 0; i < t; i++)
{
p = reader.nextInt(); // p is the perimeter given as input
area = reader.nextInt(); // area of the whole sheet, given as input
a = findRoot(); // the fraction by which the third side differs by the first two
side = (double) p / (4 * (2 + a)); // length of the first and the second sides (equal)
height = a * side; // assuming that the base is a square, the height has to be the side which differs from the other two
// writer.println(side * side * height);
// System.out.printf("%.2f\n", (side * side * height));
writer.println(String.format("%.2f", (side * side * height))); // just printing out the final answer
}
}
public static double findRoot() // the method to find the 2 possible fractions by which the height can differ from the other two sides and return the bigger one of them
{
double a32, b, discriminant, root1, root2;
a32 = 32 * area - p * p;
b = 32 * area - 2 * p * p;
discriminant = Math.sqrt(b * b - 4 * 8 * area * a32);
double temp;
temp = 2 * 8 * area;
root1 = (- b + discriminant) / temp;
root2 = (- b - discriminant) / temp;
return Math.max(root1, root2);
}
}
could someone please help me out with this? Thank You. :)
I also got stuck in this question and realized that can be done by making equation of V(volume) in terms of one side say 'l' and using differentiation to find maximum volume in terms of any one side 'l'.
So, equations are like this :-
P = 4(l+b+h);
S = 2(l*b+b*h+l*h);
V = l*b*h;
so equation in l for V = (l^3) - (l^2)P/4 + lS/2 -------equ(1)
After differentiation we get:-
d(V)/d(l) = 3*(l^2) - l*P/2 + S/2;
to get max V we need to equate above equation to zero(0) and get the value of l.
So, solutions to a quadratic equation will be:-
l = ( P + sqrt((P^2)-24S) ) / 24;
so substitute this l in equation(1) to get max volume.
I need to parse comma separated groups(enclosed in brackets) that may have internal groups inside the groups. It should only separate the outside groups.
I have a function that does this:
function lpeg.commaSplit(arg)
local P,C,V,sep = lpeg.P, lpeg.C, lpeg.V, lpeg.P(",")
local p = P{
"S";
S = lpeg.T_WSpace * C(V"Element") * (lpeg.T_WSpace * sep * lpeg.T_WSpace * C(V"Element"))^0 * lpeg.T_WSpace,
Element = (V"Group")^0 * (1 - lpeg.T_Group - sep)^0 * (V"Group" * (1 - lpeg.T_Group - sep)^0)^0 * (1 - sep)^0,
Group = lpeg.T_LGroup * ((1 - lpeg.T_Group) + V"Group")^0 * lpeg.T_RGroup
}^-1
return lpeg.match(lpeg.Ct(p), arg)
end
But the problem is to remove the extra brackets that may enclose the group.
Here is a test string:
[[a,b,[c,d]],[e,[f,g]]]
should parse to
[a,b,[c,d] & [e,[f,g]]
Notice the internal groups are left alone. A simple removal of the extra brackets on the end does not work since you'll end up with a string like a,b,[c,d]],[e,[f,g].
Any ideas how to modify the lpeg grammar to allow for the outside groups?
As I am not expert in making grammars in LPeg, I found this exercise interesting to do...
I couldn't manage to use your grammar, so I went ahead and made my own, with smaller chunks easier to understand and where I could put the captures I needed.
I think I got a decent empirical result. It works on your test case, I don't know if groups can be more deeply nested, etc. The post-processing of the capture is a bit ad hoc...
require"lpeg"
-- Guesswork...
lpeg.T_WSpace = lpeg.P" "^0
lpeg.T_LGroup = lpeg.P"["
lpeg.T_RGroup = lpeg.P"]"
lpeg.T_Group = lpeg.S"[]"
function lpeg.commaSplit(arg)
local P, C, Ct, V, sep = lpeg.P, lpeg.C, lpeg.Ct, lpeg.V, lpeg.P","
local grammar =
{
"S";
S = lpeg.T_WSpace * V"Group" * lpeg.T_WSpace,
Group = Ct(lpeg.T_LGroup * C(V"Units") * lpeg.T_RGroup),
Units = V"Unit" *
(lpeg.T_WSpace * sep * lpeg.T_WSpace * V"Unit")^0,
Unit = V"Element" + V"Group",
Element = (1 - sep - lpeg.T_Group)^1,
}
return lpeg.match(Ct(P(grammar)^-1), arg)
end
local test = "[[a,b,[c,d]],[e,[f,g]]]"
local res = lpeg.commaSplit(test)
print(dumpObject(res))
print(res[1], res[1][1], res[1][2])
local groups = res[1]
local finalResult = {}
for n, v in ipairs(groups) do
if type(v) == 'table' then
finalResult[#finalResult+1] = "[" .. v[1] .. "]"
end
end
print(dumpObject(finalResult))
dumpObject is just a table dump of my own. The output of this code is as follows:
local T =
{
{
"[a,b,[c,d]],[e,[f,g]]",
{
"a,b,[c,d]",
{
"c,d"
}
},
{
"e,[f,g]",
{
"f,g"
}
}
}
}
table: 0037ED48 [a,b,[c,d]],[e,[f,g]] table: 0037ED70
local T =
{
"[a,b,[c,d]]",
"[e,[f,g]]"
}
Personally, I wouldn't pollute the lpeg table with my stuff, but I kept your style here.
I hope this will be useful (or will be a starting point to make you to advance).
I have two list like this :
def a = [100,200,300]
def b = [30,60,90]
I want the Groovier way of manipulating the a like this :
1) First element of a should be changed to a[0]-2*b[0]
2)Second element of a should be changed to a[1]-4*b[1]
3)Third element of a should be changed to a[2]-8*b[2]
(provided that both a and b will be of same length of 3)
If the list changed to map like this, lets say:
def a1 = [100:30, 200:60, 300:90]
how one could do the same above operation in this case.
Thanks in advance.
For List, I'd go with:
def result = []
a.eachWithIndex{ item, index ->
result << item - ((2**index) * b[index])
}
For Map it's a bit easier, but still requires an external state:
int i = 1
def result = a.collect { k, v -> k - ((2**i++) * v) }
A pity, Groovy doesn't have an analog for zip, in this case - something like zipWithIndex or collectWithIndex.
Using collect
In response to Victor in the comments, you can do this using a collect
def a = [100,200,300]
def b = [30,60,90]
// Introduce a list `c` of the multiplier
def c = (1..a.size()).collect { 2**it }
// Transpose these lists together, and calculate
[a,b,c].transpose().collect { x, y, z ->
x - y * z
}
Using inject
You can also use inject, passing in a map of multiplier and result, then fetching the result out at the end:
def result = [a,b].transpose().inject( [ mult:2, result:[] ] ) { acc, vals ->
acc.result << vals.with { av, bv -> av - ( acc.mult * bv ) }
acc.mult *= 2
acc
}.result
And similarly, you can use inject for the map:
def result = a1.inject( [ mult:2, result:[] ] ) { acc, key, val ->
acc.result << key - ( acc.mult * val )
acc.mult *= 2
acc
}.result
Using inject has the advantage that you don't need external variables declared, but has the disadvantage of being harder to read the code (and as Victor points out in the comments, this makes static analysis of the code hard to impossible for IDEs and groovypp)
def a1 = [100:30, 200:60, 300:90]
a1.eachWithIndex{item,index ->
println item.key-((2**(index+1))*item.value)
i++
}
This is a geometry question.
I have a line between two points A and B and want separate it into k equal parts. I need the coordinates of the points that partition the line between A and B.
Any help is highly appreciated.
Thanks a lot!
You just need a weighted average of A and B.
C(t) = A * (1-t) + B * t
or, in 2-D
Cx = Ax * (1-t) + Bx * t
Cy = Ay * (1-t) + By * t
When t=0, you get A.
When t=1, you get B.
When t=.25, you a point 25% of the way from A to B
etc
So, to divide the line into k equal parts, make a loop and find C, for t=0/k, t=1/k, t=2/k, ... , t=k/k
for(int i=0;i<38;i++)
{
Points[i].x = m_Pos.x * (1 - (i/38.0)) + m_To.x * (i / 38.0);
Points[i].y = m_Pos.y * (1 - (i/38.0)) + m_To.y * (i / 38.0);
if(i == 0 || i == 37 || i == 19) dbg_msg("CLight","(%d)\nPos(%f,%f)\nTo(%f,%f)\nPoint(%f,%f)",i,m_Pos.x,m_Pos.y,m_To.x,m_To.y,Points[i].x,Points[i].y);
}
prints:
[4c7cba40][CLight]: (0)
Pos(3376.000000,1808.000000)
To(3400.851563,1726.714111)
Point(3376.000000,1808.000000)
[4c7cba40][CLight]: (19)
Pos(3376.000000,1808.000000)
To(3400.851563,1726.714111)
Point(3388.425781,1767.357056)
[4c7cba40][CLight]: (37)
Pos(3376.000000,1808.000000)
To(3400.851563,1726.714111)
Point(3400.851563,1726.714111)
which looks fine but then my program doesn't work :D.
but your method works so thanks