I have a, hopefully, simple question. Im using Nuke to do a linear animation and I have 2 points.
point1 # frame 1 is (5,90)
point2 # frame 10 is (346,204)
Using a linear interpolation type, I want to fiqure out where the x and y point is at frame 30.
The way i tried is using the slope formula and then finding the y intercept.
m = (204 - 90) / (346 - 5)
m = 114/341 = .3343
then I got the intercept by:
Y = Mx + b
90 = .3343(5) + b
90 = 1.6715 + b
88.3285 = b
so...I got the formula for my line. y = .3343X + 88.3285
Can someone help me figure out where the point is going to be at any given frame?
If you'd please refer to the image attached... you can see image of my graph.
I guess the problem I'm having is relating the time to the coord points.
Thanks
Just consider x as a function of time (t).
Here's some coordinates:
(t, x)
(1, 5)
(10, 346)
and some calculation of the line equation:
x = mt+b
m = (346-5) / (10-1)
m = 341/9
b = 5 - (341/9)*1
b = - 296/9
x = (341t - 296)/9
And using my formula (t -> x) and your formula (x -> y), I can calculate where things are at t=30
t = 30
x = 1103 + 7/9
y = 457.3214
Related
After constructing a graph, when using the detach method to change some tensor value, it is expected that an error pops up when computing the back propagation. However, this is not always the case. In the following two blocks of code: the first one raises an error, while the second one does not. Why does this happen?
x = torch.tensor(3.0, requires_grad=True)
y = x + 1
z = y**2
c = y.detach()
c.zero_()
z.backward(retain_graph=True)
print(x.grad) # errors pop up
x = torch.tensor(3.0, requires_grad=True)
y1 = x+1
y2 = x**2
z = 3*y1 + 4*y2
c = y2.detach()
c.zero_()
z.backward(retain_graph=True)
print(x.grad) # no errors. The printed value is 27
TLDR; In the former example z = y**2, so dz/dy = 2*y, i.e. it's a function of y and requires its values to be unchanged to properly compute the backpropagation, hence the error message when applying the in-place operation. In the latter z = 3*y1 + 4*y2, so dz/dy2 = 4, i.e. y2 values are not needed to compute the gradient, as such its values can be modified freely.
In the former example you have the following computation graph:
x ---> y = x + 1 ---> z = y**2
\
\ ---> c = y.detach().zero_()
Corresponding code:
x = torch.tensor(3.0, requires_grad=True)
y = x + 1
z = y**2
c = y.detach()
c.zero_()
z.backward() # errors pop up
When calling c = y.detach() you effectively detach c from the computation graph, while y remains attached. However, c shares the same data as y. This means when you call the in-place operation c.zero_, you end up affecting y. This is not allowed, because the y is part of a computation graph, and its values will be needed for a potential backpropagation from variable z.
The second scenario corresponds to this layout:
/--> y1 = x + 1 \
x ---> z = 3*y1 + 4*y2
\--> y2 = x**2 /
\
\ ---> c = y2.detach().zero_()
Corresponding code:
x = torch.tensor(3.0, requires_grad=True)
y1 = x + 1
y2 = x**2
z = 3*y1 + 4*y2
c = y2.detach()
c.zero_()
z.backward()
print(x.grad) # no errors. The printed value is 27
Here again, we have the same setup, you detach then in-place modify c and y with zero_.
The only difference is the operation performed on y and y2 (in the 1st and 2nd example respectively).
In the former, you have z = y**2, so the derivative is 2*y, hence the value of y is needed to compute the gradient of that operation.
In the latter example though z(y2) = constant + 4*y2 so the derivative with respect to y2 is just a constant: 4, i.e. it doesn't require the value of y2 to compute its derivative. You can check this by, for instance, defining in 2nd example z with z = 3*y1 + 4*y2**2: it will raise an error.
I'm having problems implementing this exercise from a quantitative economics course.
Here's my code:
N = 50
M = 20
a = 0.1
b = 0.2
c = 0.5
d = 1.0
σ = 0.1
estimates = zeros(M, 5)
for i ∈ 1:M
x₁ = Vector{BigFloat}(randn(N))
x₂ = Vector{BigFloat}(randn(N))
w = Vector{BigFloat}(randn(N))
# Derive y vector (element wise operations)
y = a*x₁ .+ b.*(x₁.^2) .+ c.*x₂ .+ d .+ σ.*w
# Derive X matrix
X = [x₁ x₁ x₂ fill(d, (N, 1)) w]
# Implementation of the formula β = inv(XᵀX)Xᵀy
estimates[i, :] = (X'*X)\X'*y
end
histogram(estimates, layout=5, labels=["a", "b", "c", "d", "σ"])
I get a SingularException(5) error, as the matrix X'X has a determinant of 0 and has no inverse. My question is, where have I gone wrong in this exercise? I heard that a reason the determinant might be zero is floating point inaccuracy, so I made the random variables BigFloats to no avail. I know the mistake I'm making isn't very complicated but I'm lost. Thank you!
Your X should be
X = [x₁ x₁*x₁ x₂ fill(d, (N, 1))]
Explanation
It looks that you are trying to test OLS to estimate the parameters of the model:
y = α₀ + α₁x₁ + α₁₁x₁² + α₂x₂ + ϵ
where α₀, is the intercept of the model, α₁, α₁₁, α₂ are parameters for explanatory variables, and ϵ is the random error with the expected value 0 and variance σ². Hence the structure of X must match your case.
Putting the α₁ twice you introduced co-linearity and got the error.
You also do not want to "estimate" the parameter for ϵ because it represents the randomness.
I have a fairly simple equation, in which I have direct measurements of the variables through time, and two different unknown parameters I need to solve for, but which I know can be considered constants over the time periods I'm studying.
Both of these "constants" have fairly narrow ranges of variability in nature. In principle, it seems like some kind of optimization procedure/function should be able to do this easily, by finding the pair of values that minimizes the standard deviation of each of the constant values across the time series.
However, I am new to optimization and parameter fitting. Any help figuring out how to use r code to find the pair (or pairs) of values in this situation would be greatly appreciated.
Below is a simplified form of the equation I'm dealing with:
A * x + B * z - B * d = c + e
A and B are the constants I need to solve for.
Possible real-world values of A are 0.4-0.8
Possible real-world values of B are 0.85-0.99
To create a reasonable mock data set, assuming perfect measurements of all variables, and known values of A and B:
### Generate mock data
### Variables all have a daily cycle and are strongly autocorrelated,
# and so can be approximated via sin function,
# with unique noise added to each to simulate variability:
# Variability for each variable
n <- 1000 # number of data points
t <- seq(0,4*pi,length.out = 1000)
a <- 3
b <- 2
x.unif <- runif(n)
z.norm <- rnorm(n)
c.unif <- runif(n)
d.norm <- rnorm(n)
d.unif <- runif(n)
e.norm <- rnorm(n)
amp <- 1
# Create reasonable values of mock variable data for all variables except e;
# I will calculate from known fixed values for A and B.
x <- a*sin(b*t)+x.unif*amp + 10 # uniform error
z <- a*sin(b*t)+z.norm*amp + 10 # Gaussian/normal error
c <- ((a*sin(b*t)+c.unif*amp) + 10)/4
d <- ((a*sin(b*t)+d.norm*amp)+(a*sin(b*t)+d.unif*amp)+10)/2
# Put vectors in dataframe
dat <- data.frame("t" = t, "x" = x, "z" = z, "c" = c, "d" = d)
# Equation: A*x + B*z - B*d = c + e
# Solve for e:
# e = A*x + B*z - B*d - c
# Specify "true" values for A and B:
A = 0.6
B = 0.9
# Solve for e:
dat <- dat %>%
mutate(e = A*x + B*z - B*d - c)
# Gather data for easy visualizing of results for e:
dat_gathered <- dat %>%
gather(-t, value = "value", key = "key")
# Plot all variables
ggplot(dat_gathered, aes(x = t, y = value, color = key)) + geom_line()
# Add small error (to simulate measurement error) to all variables except A and B:
dat <- dat %>%
mutate(x_j = x + rnorm(x, sd=0.02)/(1/x)) %>%
mutate(z_j = z + rnorm(z, sd=0.02)/(1/z)) %>%
mutate(c_j = c + rnorm(c, sd=0.02)/(1/c)) %>%
mutate(d_j = d + rnorm(d, sd=0.02)/(1/d)) %>%
mutate(e_j = e + rnorm(e, sd=0.02)/(1/e))
The variables in dat with the _j suffix represent real world data (since they have measurement error added). Knowing the constraint that:
A is within 0.4-0.8
B is within 0.85-0.99
Is it possible to use the noisy "_j" data to optimize for the pair of constant values that minimize deviation of A and B across the entire time series?
A little bit of algebra and setting this up as a linear regression problem with no intercept seems to work fine:
m1 <- lm(e_j+c_j ~ 0 + x_j + I(z_j-d_j), data=dat)
coef(m1) ## A =0.6032, B = 0.8916
It doesn't do anything to constrain the solution, though.
time <- 1:12
y <- c(0,0,0,0,0,0,0,12,34,69,100,100)
mdl <- gcFitModel(time,y,control = grofit.control(fit.opt = "m", model.type = "gompertz"))
I get my parameters from the above mdl to fit the gompertz equation
y <- A* exp(-exp(mu * e * (lambda - time)/mu + 1))
mu <- 36.162016
lambda <- 7.9800164
A <- 100
time <- 1:12
time here is in a step of 15 days. For e.g time = 1 implies mid-day of a 15 day period, time 2 implies mid day of the next 15 days period, time 3 implies implies the mid-day of the next 15 days period and so on.
I fitted the following function:
e <- exp(1)
y <- 100 * exp(-exp(mu * e * (lambda - time)/mu + 1))
plot(time,y)
The lambda controls the movement along the x-axis.
I am looking to modify this curve so that I get more data points by converting the ids into weeks i.e instead of mid-point of every 15 days, I want to get y for every 7 days. How can I do this?
Your code doesn't include any fitting routines, so I am assuming this is a question about plotting.
Here is an example of a plot with 100 points between time = 1 and time = 12.
time <- seq(1, 12, length.out = 100);
mu <- 34.55844
y <- 100 * exp(-exp(mu * e * (3 - time)/mu + 1))
plot(time,y)
I have not needed to do things like this for years and was also never good at it. Below is my graph :
Looking at my artistic numbering on the graph :
I have the X and Y Values : X = 7282, Y = 235
I have the X and Y values : X = 8178, Y = 173
I have the X but not the Y : X = 7882, Y = ?
I need to calculate Y, and im sure it is pretty simple, but I cant seem to figure it out. Ive googled a lot, but all my calculations never work(i.e the new Y point is never on the line, always above or below), so im clearly missing something.
Can anyone help with the formula of how to calculate the new Y value ?
Thanks!
Try this:
Y = (Ymax - Ymin)/(Xmax - Xmin) * K + Ymin;
If slope goes up: K = X - Xmin, another way K = Xmax - X.
Since you know the coordinate values of 2 points
m = ((y2 - y1)/(x2 - x1))
y3 = m*x3 + c
where c is a constant
to calculate c use x = x1 and y = y1
c = y-m*x
apply this c in the above equation we get y3
so for this case
m = (235 - 173)/(7282 - 8178)
c = 8178 - m*173
y3 = m*7882 + c