I have to programmatically determine the value of the expression:
S = log(x1y1 + x2y2 + x3y3 ...)
Using only the values of:
lxi = log(xi)
lyi = log(yi)
Calculating anti-logs of each of lxi and lyi would probably be impractical and is not desired ...
Is there any way this evaluation can be broken down into a simple summation?
EDIT
I saw a C function somewhere that does the computation in a simple summation:
double log_add(double lx, double ly)
{
double temp,diff,z;
if (lx<ly) {
temp = lx; lx = ly; ly = temp;
}
diff = ly-lx;
z = exp(diff);
return lx+log(1.0+z);
}
The return values are added for each pair of values, and this seems to be giving the correct answer. But I'm not able to figure out how and why it's working!
The direct way is to perform two exponentiations:
ln(x+y) = ln(eln(x) + eln(y))
The log_add function uses a slightly different approach to get the same result with only one:
ln(x+y) = ln((x+y)x/x)
= ln((x+y)/x) + ln(x)
= ln(1 + y/x) + ln(x)
= ln(1 + eln(y/x)) + ln(x)
= ln(1 + eln(y)-ln(x)) + ln(x)
Related
I need to integrate the following function where there is a differentiation term inside. Unfortunately, that term is not easily differentiable.
Is this possible to do something like numerical integration to evaluate this in R?
You can assume 30,50,0.5,1,50,30 for l, tau, a, b, F and P respectively.
UPDATE: What I tried
InnerFunc4 <- function(t,x){digamma(gamma(a*t*(LF-LP)*b)/gamma(a*t))*(x-t)}
InnerIntegral4 <- Vectorize(function(x) { integrate(InnerFunc4, 1, x, x = x)$value})
integrate(InnerIntegral4, 30, 80)$value
It shows the following error:
Error in integrate(InnerFunc4, 1, x, x = x) : non-finite function value
UPDATE2:
InnerFunc4 <- function(t,L){digamma(gamma(a*t*(LF-LP)*b)/gamma(a*t))*(L-t)}
t_lower_bound = 0
t_upper_bound = 30
L_lower_bound = 30
L_upper_bound = 80
step_size = 0.5
integral = 0
t <- t_lower_bound + 0.5*step_size
while (t < t_upper_bound){
L = L_lower_bound + 0.5*step_size
while (L < L_upper_bound){
volume = InnerFunc4(t,L)*step_size**2
integral = integral + volume
L = L + step_size
}
t = t + step_size
}
Since It seems that your problem is only the derivative, you can get rid of it by means of partial integration:
Edit
Not applicable solution for lower integration bound 0.
In my program, I have to find two random values with certain conditions:
i needs to be int range [2...n]
k needs to be in range [i+2...n]
so I did this:
i = rand() % n + 2;
k = rand() % n + (i+2);
But it keeps giving me wrong values like
for n = 7
I get i = 4 and k = 11
or i = 3 and k = 8
How can I fix this?
The exact formula that I use in my other program is:
i = min + (rand() % (int)(max - min + 1))
Look here for other explanation
As the comments say, your range math is off.
You might find it useful to use a function to work the math out consistently each time. e.g.:
int RandInRange(int x0, int x1)
{
if(x1<=x0) return x0;
return rand() % (x1-x0+1) + x0;
}
then call it with what you want:
i = RandInRange(2,n);
k = RandInRange(i+2,n);
I'm trying to take the derivative of an expression:
x = read.csv("export.csv", header=F)$V1
f = expression(-7645/2* log(pi) - 1/2 * sum(log(w+a*x[1:7644]^2)) + (x[2:7645]^2/(w + a*x[1:7644]^2)),'a')
D(f,'a')
x is simply an integer vector, a and w are the variables I'm trying to find by deriving. However, I get the error
"Function '[' is not in Table of Derivatives"
Since this is my first time using R I'm rather clueless what to do now. I'm assuming R has got some problem with my sum function inside of the expression?
After following the advice I now did the following:
y <- x[1:7644]
z <- x[2:7645]
f = expression(-7645/2* log(pi) - 1/2 * sum(log(w+a*y^2)) + (z^2/(w + a*y^2)),'a')
Deriving this gives me the error "sum is not in the table of derivatives". How can I make sure the expression considers each value of y and z?
Another Update:
y <- x[1:7644]
z <- x[2:7645]
f = expression(-7645/2* log(pi) - 1/2 * log(w+a*y^2) + (z^2/(w + a*y^2)))
d = D(f,'a')
uniroot(eval(d),c(0,1000))
I've eliminated the "sum" function and just entered y and z. Now, 2 questions:
a) How can I be sure that this is still the expected behaviour?
b) Uniroot doesn't seem to like "w" and "a" since they're just symbolic. How would I go about fixing this issue? The error I get is "object 'w' not found"
This should work:
Since you have two terms being added f+g, the derivative D(f+g) = D(f) + D(g), so let's separate both like this:
g = expression((z^2/(w + a*y^2)))
f = expression(- 1/2 * log(w+a*y^2))
See that sum() was removed from expression f, because the multiplying constant was moved into the sum() and the D(sum()) = sum(D()). Also the first constant was removed because the derivative is 0.
So:
D(sum(-7645/2* log(pi) - 1/2 * log(w+a*y^2)) + (z^2/(w + a*y^2)) = D( constant + sum(f) + g ) = sum(D(f)) + D(g)
Which should give:
sum(-(1/2 * (y^2/(w + a * y^2)))) + -(z^2 * y^2/(w + a * y^2)^2)
expression takes only a single expr input, not a vector, and it is beyond r abilities to vectorize that.
you can also do this with a for loop:
foo <- c("1+2","3+4","5*6","7/8")
result <- numeric(length(foo))
foo <- parse(text=foo)
for(i in seq_along(foo))
result[i] <- eval(foo[[i]])
I'd like to know the best way to compute autocorrelation function as defined below.
For i=1,2,... I would like to compute the i-th autocorrelation function acf.
This is the sum, from k = 1 to n-i, of +1 if v(k) = v(k+i) or -1 if v(k) is different from v(k+i), where n is the length of a vector.
For example:
if v<-c(0,1,1,0,0) and i = 2. Then
acf(v) = (-1) + (-1) + (-1) = -3
Thanks!
What about using R-help? There you should have found the acf function.
v = c(1,1,0,0,1,0,1,0,1)
acf(v,plot=F) -> acf_v
acf_v[2]
I created a function to do to it but still looking for a short and efficient way to do it.
Here is the function:
> v<-c(0,1,1,1,0,1,1)
> acf_bit <-function(vec,lag) {
+ m<-length(vec)
+ t<-0
+ for (k in 1:(m-lag)) {
+ if (v[k]==v[k+lag]) {t<-t+1}
+ else {t<-t-1}
+ }
+ return(t)
+ }
> acf_bit(v,2)
[1] -1
How do I map numbers, linearly, between a and b to go between c and d.
That is, I want numbers between 2 and 6 to map to numbers between 10 and 20... but I need the generalized case.
My brain is fried.
If your number X falls between A and B, and you would like Y to fall between C and D, you can apply the following linear transform:
Y = (X-A)/(B-A) * (D-C) + C
That should give you what you want, although your question is a little ambiguous, since you could also map the interval in the reverse direction. Just watch out for division by zero and you should be OK.
Divide to get the ratio between the sizes of the two ranges, then subtract the starting value of your inital range, multiply by the ratio and add the starting value of your second range. In other words,
R = (20 - 10) / (6 - 2)
y = (x - 2) * R + 10
This evenly spreads the numbers from the first range in the second range.
It would be nice to have this functionality in the java.lang.Math class, as this is such a widely required function and is available in other languages.
Here is a simple implementation:
final static double EPSILON = 1e-12;
public static double map(double valueCoord1,
double startCoord1, double endCoord1,
double startCoord2, double endCoord2) {
if (Math.abs(endCoord1 - startCoord1) < EPSILON) {
throw new ArithmeticException("/ 0");
}
double offset = startCoord2;
double ratio = (endCoord2 - startCoord2) / (endCoord1 - startCoord1);
return ratio * (valueCoord1 - startCoord1) + offset;
}
I am putting this code here as a reference for future myself and may be it will help someone.
As an aside, this is the same problem as the classic convert celcius to farenheit where you want to map a number range that equates 0 - 100 (C) to 32 - 212 (F).
https://rosettacode.org/wiki/Map_range
[a1, a2] => [b1, b2]
if s in range of [a1, a2]
then t which will be in range of [b1, b2]
t= b1 + ((s- a1) * (b2-b1))/ (a2-a1)
In addition to #PeterAllenWebb answer, if you would like to reverse back the result use the following:
reverseX = (B-A)*(Y-C)/(D-C) + A
Each unit interval on the first range takes up (d-c)/(b-a) "space" on the second range.
Pseudo:
var interval = (d-c)/(b-a)
for n = 0 to (b - a)
print c + n*interval
How you handle the rounding is up to you.
if your range from [a to b] and you want to map it in [c to d] where x is the value you want to map
use this formula (linear mapping)
double R = (d-c)/(b-a)
double y = c+(x*R)+R
return(y)
Where X is the number to map from A-B to C-D, and Y is the result:
Take the linear interpolation formula, lerp(a,b,m)=a+(m*(b-a)), and put C and D in place of a and b to get Y=C+(m*(D-C)). Then, in place of m, put (X-A)/(B-A) to get Y=C+(((X-A)/(B-A))*(D-C)). This is an okay map function, but it can be simplified. Take the (D-C) piece, and put it inside the dividend to get Y=C+(((X-A)*(D-C))/(B-A)). This gives us another piece we can simplify, (X-A)*(D-C), which equates to (X*D)-(X*C)-(A*D)+(A*C). Pop that in, and you get Y=C+(((X*D)-(X*C)-(A*D)+(A*C))/(B-A)). The next thing you need to do is add in the +C bit. To do that, you multiply C by (B-A) to get ((B*C)-(A*C)), and move it into the dividend to get Y=(((X*D)-(X*C)-(A*D)+(A*C)+(B*C)-(A*C))/(B-A)). This is redundant, containing both a +(A*C) and a -(A*C), which cancel each other out. Remove them, and you get a final result of: Y=((X*D)-(X*C)-(A*D)+(B*C))/(B-A)
TL;DR: The standard map function, Y=C+(((X-A)/(B-A))*(D-C)), can be simplified down to Y=((X*D)-(X*C)-(A*D)+(B*C))/(B-A)
int srcMin = 2, srcMax = 6;
int tgtMin = 10, tgtMax = 20;
int nb = srcMax - srcMin;
int range = tgtMax - tgtMin;
float rate = (float) range / (float) nb;
println(srcMin + " > " + tgtMin);
float stepF = tgtMin;
for (int i = 1; i < nb; i++)
{
stepF += rate;
println((srcMin + i) + " > " + (int) (stepF + 0.5) + " (" + stepF + ")");
}
println(srcMax + " > " + tgtMax);
With checks on divide by zero, of course.