MMULT / MINVERSE Maximum array size - excel

I am attempting to solve this set of equations:
696x + 405y + 61z = 1385699
618x + 463y + 81z = 1401476
573x + 476y + 113z = 1407438
508x + 537y + 117z = 1418256
473x + 566y + 123z = 1427224
x,y and z are removed in the worksheet.
Using the following formula: E2:E5 {=MMULT(MINVERSE("A1:C5"),"D1:D5")}
But I keep encountering value errors.
Can MINVERSE only be used with a 3x3 matrix?

Since your system of equations is overdetermined you may instead be looking to find the pseudoinverse or least squares estimate. To calculate this, select a 1 x 3 range such as F1:H1 then enter the array formula:
=LINEST(D1:D5,A1:C5,0)
entered by holding down CTRL+SHIFT+ENTER. For the given example, this returns {z,y,x} = {1191.8,1345.3,1103.7}. Note that the results coincide with MINVERSE for square matrices.

Related

How do i Square Root a Function in VBA

I am working on a MonteCarlo simulation model and part of it is to calculate the following formula:
X = Sqr(1-p)Y + Sqr(p)Z,
Where:
Y and Z are randomly obtained values based (idiosyncratic and systematic factors, respectviely) on a standard normal (inv.) distribution, calculated as:
Application.WorksheetFunction.NormInv (Rnd(), mean, sd)
p represents a correlation factor.
My aim is to square root a recalled formula, however when I try the following (inserting the first Sqr), it does not work and gives an error:
Matrix (n, sims) = (R * Sqr(Application.WorksheetFunction.NormInv(Rnd(), mean, sd))) + (Sqr(1 - R) * RandomS(s, x))
where:
R: Correlation factor
RandomS(s,x): generated matrix with Z values.
I don't want to go into too much details about the background and other variables, as the only problem I am getting is with Square Rooting the equation.
Error message I recieve reads:
Run-time error '5':
Invalid procedure call or argument
When I click debug it takes me to the formula, therefore there must be something wrong with the syntax.
Can you help with directly squaring the formula?
Thank you!
Andrew
Square root is simply Sqr.
It works fine in Excel VBA, so for example:
MsgBox Sqr(144)
...returns 12.
Just don't confuse it with the syntax for a worksheet function with is SQRT.
If you're still having an issue with your formula, tit must be with something other than the Square Root function, and I'd suggest you check the values of your variable, and make sure they are properly declared (preferably with Option Explicit at the top of the module).
Also make sure that you're passing Sqr a positive number.
Documentation: Sqr Function
I'm not a math major, but with your formula:
X = Sqr(1-p)Y + Sqr(p)Z,
...you specified how Y and Z are calculated, so calculate them separately to keep it simple:
Dim X as Double, Y as Double, Z as Double
Y = Application.WorksheetFunction.NormInv (Rnd(), mean, sd)
Z = Application.WorksheetFunction.NormInv (Rnd(), mean, sd)
Assuming the comma is not supposed to be in the formula, and having no idea what p is, your final code to calculate X is:
X = Sqr(1-p) * Y + Sqr(p) * Z

Apply function to masked region

I have an image like that:
I have both the mask and the original image. I would like to calculate the colour temperature of ONLY the ducks region.
Right now, I'm iterating through each row and column of the image below and getting pixels where their values are not zero. But I think this isn't the right way to do this. Any suggestions?
What I did was:
xyzImg = cv2.cvtColor(resImage, cv2.COLOR_BGR2XYZ)
x,y,z = cv2.split(xyzImg)
xList=[]
yList=[]
zList=[]
rows=x.shape[0]
cols=x.shape[1]
for i in range(rows):
for j in range(cols):
if (x[i][j]!=0) and (y[i][j]!=0) and (z[i][j]!=0):
xList.append(x[i][j])
yList.append(y[i][j])
zList.append(z[i][j])
xAvg = np.mean(xList)
yAvg = np.mean(yList)
zAvg = np.mean(zList)
xs = xAvg / (xAvg + yAvg + zAvg)
ys = yAvg / (xAvg + yAvg + zAvg)
xyChrome = np.array([xs,ys])
But this is very slow and I don't think its right...
The simplest way would be to use cv2.mean() function.
It takes two arguments src (having 1 to 4 channels) and mask and returns a vector with mean values for individual channels.
Refer to cv2::mask

Distance to a straight line in standard form

For a 3D straight line expressed in the standard form
a1*x + b1*y + c1*z + d1 = 0
a2*x + b2*y + c2*z + d2 = 0
and a given point x0,y0,z0
what is the distance from the point to the straight line?
Distance from point P0 to parametric line L(t) = Base + t * Dir is
Dist = Length(CrossProduct(Dir, P0 - Base)) / Length(Dir)
To find direction vector:
Dir = CrossProduct((a1,b1,c1), (a2,b2,c2))
To get some arbitrary base point, solve equation system with 2 equations and three unknowns (find arbitrary solution):
a1*x + b1*y + c1*z + d1 = 0
a2*x + b2*y + c2*z + d2 = 0
Check minors consisting of a and b, a and c, b and c coefficients. When minor is non-zero, corresponding variable might be taken as free one. For example, if a1 * b2 - b1 * a2 <> 0, choose variable z as free - make it zero or another value and solve system for two unknowns x and y.
(I omitted extra cases of parallel or coinciding planes)

gam in mgcv R with big number of covariates

I would like to know if there is another way to write the function:
gam(VariableResponse ~ s(CovariateName1) + s(CovariateName2) + ... + s(CovariateName100),
family = gaussian(link = identity), data = MyData)
in mgcv package without typing 100 covariates' name as above?
Supposing that in MyData I have only VariableResponse in column 1, CovariateName1 in column 2, etc.
Many thank!
Yes, use the brute force approach to generate a formula by pasting together the covariate names with the strings 's(' and ')' and then collapsing the whole things with ' + '. The convert the resultant string to a formula and pass that to gam(). You may need to fix issues with the formula's environment if gam() can't find the variable you name as it is going to do some NSE on the formula to identify which terms need smooths estimating and hence need to be replaced by a basis expansion.
library(mgcv)
set.seed(2) ## simulate some data...
df <- gamSim(1, n=400, dist = "normal", scale = 2)
> names(df)
[1] "y" "x0" "x1" "x2" "x3" "f" "f0" "f1" "f2" "f3"
We'll ignore the last 5 of those columns for the purposes of this example
df <- df[1:5]
Make the formula
fm <- paste('s(', names(df[ -1 ]), ')', sep = "", collapse = ' + ')
fm <- as.formula(paste('y ~', fm))
Now fit the model
m <- gam(fm, data = df)
> m
Family: gaussian
Link function: identity
Formula:
y ~ s(x0) + s(x1) + s(x2) + s(x3)
Estimated degrees of freedom:
2.5 2.4 7.7 1.0 total = 14.6
GCV score: 4.050519
You do have to be careful about fitting GAMs this way however; concurvity (the nonlinear counterpart to multicolinearlity in linear models) can cause catastrophically bad estimates of smooth functions.

Color Histogram

I'm trying to calculate histogram for an image. I'm using the following formula to calculate the bin
%bin = red*(N^2) + green*(N^1) + blue;
I have to implement the following Matlab functions.
[row, col, noChannels] = size(rgbImage);
hsvImage = rgb2hsv(rgbImage); % Ranges from 0 to 1.
H = zeros(4,4,4);
for col = 1 : columns
for row = 1 : rows
hBin = floor(hsvImage(row, column, 1) * 15);
sBin = floor(hsvImage(row, column, 2) * 4);
vBin = floor(hsvImage(row, column, 3) * 4);
F(hBin, sBin, vBin) = hBin, sBin, vBin + 1;
end
end
When I run the code I get the following error message "Subscript indices must either be real positive integers or logical."
As I am new to Matlab and Image processing, I'm not sure if the problem is with implementing the algorithm or a syntax error.
There are 3 problems with your code. (Four if you count that you changed from H to F your accumulator vector, but I'll assume that's a typo.)
First one, your variable bin can be zero at any moment if the values of a giving pixel are low. And F(0) is not a valid index for a vector or matrix. This is why you are getting that error.
You can solve easily by doing F(bin+1) and keep in mind that your F vector will have your values shifted one position over.
Second error, you are assigning the value bin + 1 to your accumulator vector F, which is not what you want, you want to add 1 every time a pixel in that range is found, what you should do is F(bin+1) = F(bin+1) + 1;. This way the values of F will be increasing all the time.
Third error is simpler, you forgot to implement your bin = red*(N^2) + green*(N^1) + blue; equation

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