I am trying to calculate a few hundred rows of data, solving a system of linear equations using matrices. I am building my matrices based on the row data. I take the inverse of the resultant 3x3 and then multiply twice, once for x's and once for y's. I get 6 variables from this: a,b,c,d,e and f. How can I copy down the example so that is solves for all rows? I am providing the data and then the formulas I am using now. Right now, if I copy down it skips 3 rows or if I copy down with 3 examples it skips 9 rows. I mean I guess I could go an try to insert extra rows into all 300 so I end up with 900 rows, but there has to be an easier way!
I can't figure out how to get my data to populate in here correctly so here is a link: http://codepad.org/qZwua3h9
Note: I split up the matrix rows so you could see them easier, they are not split up on my sheet.
Edit: If anyone can figure out how to paste the example data here I would welcome it so that this post may be of use to someone in the future. I am not sure how long codepad keeps their pastes.
I gave up and came to the conclusion that there is no reasonable amount of effort that will yield the desired results. Not only was the example case only ONE transformation, but the intended case was for 3 transformations - so three times the work. I came up with a Matlab solution in about 15 minutes. I understand that not everyone has access to Matlab though. So, if someone comes up with any reasonable working excel solution, I would welcome the knowledge and mark that answer as the accepted one. Regardless, here is the Matlab script:
M = csvread('pointData.csv');
T1result = zeros(215,6);
T2result = zeros(215,6);
T3result = zeros(215,6);
for i=1:215,
m = [M(i,1) M(i,2) 1; M(i,3) M(i,4) 1; M(i,5) M(i,6) 1];
x = [M(i,7);M(i,9);M(i,11)];
y = [M(i,8);M(i,10);M(i,12)];
xresult = m\x;
yresult = m\y;
T1result(i,:) = [transpose(xresult),transpose(yresult)];
m = [M(i,7) M(i,8) 1; M(i,9) M(i,10) 1; M(i,11) M(i,12) 1];
x = [M(i,13);M(i,15);M(i,17)];
y = [M(i,14);M(i,16);M(i,18)];
xresult = m\x;
yresult = m\y;
T2result(i,:) = [transpose(xresult),transpose(yresult)];
m = [M(i,13) M(i,14) 1; M(i,15) M(i,16) 1; M(i,17) M(i,18) 1];
x = [M(i,19);M(i,21);M(i,23)];
y = [M(i,20);M(i,22);M(i,24)];
xresult = m\x;
yresult = m\y;
T3result(i,:) = [transpose(xresult),transpose(yresult)];
end
LeafId = csvread('extraColumnsForID.csv');
Tresult = [LeafId, T1result, T2result, T3result];
csvwrite('transforms.csv',Tresult);
Related
In the code supplied below I am trying to iterate over 2D numpy array [i][k]
Originally it is a code which was written in Fortran 77 which is older than my grandfather. I am trying to adapt it to python.
(for people interested whatabouts: it is a simple hydraulics transients event solver)
Bear in mind that all variables are introduced in my code which I don't paste here.
H = np.zeros((NS,50))
Q = np.zeros((NS,50))
Here I am assigning the first row values:
for i in range(NS):
H[0][i] = HR-i*R*Q0**2
Q[0][i] = Q0
CVP = .5*Q0**2/H[N]
T = 0
k = 0
TAU = 1
#Interior points:
HP = np.zeros((NS,50))
QP = np.zeros((NS,50))
while T<=Tmax:
T += dt
k += 1
for i in range(1,N):
CP = H[k][i-1]+Q[k][i-1]*(B-R*abs(Q[k][i-1]))
CM = H[k][i+1]-Q[k][i+1]*(B-R*abs(Q[k][i+1]))
HP[k][i-1] = 0.5*(CP+CM)
QP[k][i-1] = (HP[k][i-1]-CM)/B
#Boundary Conditions:
HP[k][0] = HR
QP[k][0] = Q[k][1]+(HP[k][0]-H[k][1]-R*Q[k][1]*abs(Q[k][1]))/B
if T == Tc:
TAU = 0
CV = 0
else:
TAU = (1.-T/Tc)**Em
CV = CVP*TAU**2
CP = H[k][N-1]+Q[k][N-1]*(B-R*abs(Q[k][N-1]))
QP[k][N] = -CV*B+np.sqrt(CV**2*(B**2)+2*CV*CP)
HP[k][N] = CP-B*QP[k][N]
for i in range(NS):
H[k][i] = HP[k][i]
Q[k][i] = QP[k][i]
Remember i is for rows and k is for columns
What I am expecting is that for all k number of columns the values should be calculated until T<=Tmax condition is met. I cannot figure out what my mistake is, I am getting the following errors:
RuntimeWarning: divide by zero encountered in true_divide
CVP = .5*Q0**2/H[N]
RuntimeWarning: invalid value encountered in multiply
QP[N][k] = -CV*B+np.sqrt(CV**2*(B**2)+2*CV*CP)
QP[N][k] = -CV*B+np.sqrt(CV**2*(B**2)+2*CV*CP)
ValueError: setting an array element with a sequence.
Looking at your first iteration:
H = np.zeros((NS,50))
Q = np.zeros((NS,50))
for i in range(NS):
H[0][i] = HR-i*R*Q0**2
Q[0][i] = Q0
The shape of H is (NS,50), but when you iterate over a range(NS) you apply that index to the 2nd dimension. Why? Shouldn't it apply to the dimension with size NS?
In numpy arrays have 'C' order by default. Last dimension is inner most. They can have a F (fortran) order, but let's not go there. Thinking of the 2d array as a table, we typically talk of rows and columns, though they don't have a formal definition in numpy.
Lets assume you want to set the first column to these values:
for i in range(NS):
H[i, 0] = HR - i*R*Q0**2
Q[i, 0] = Q0
But we can do the assignment whole rows or columns at a time. I believe new versions of Fortran also have these 'whole-array' functions.
Q[:, 0] = Q0
H[:, 0] = HR - np.arange(NS) * R * Q0**2
One point of caution when translating to Python. Indexing starts with 0; so does ranges and np.arange(...).
H[0][i] is functionally the same as H[0,i]. But when using slices you have to use the H[:,i] format.
I suspect your other iterations have similar problems, but I'll stop here for now.
Regarding the errors:
The first:
RuntimeWarning: divide by zero encountered in true_divide
CVP = .5*Q0**2/H[N]
You initialize H as zeros so it is normal that it complains of division by zero. Maybe you should add a conditional.
The third:
QP[N][k] = -CV*B+np.sqrt(CV**2*(B**2)+2*CV*CP)
ValueError: setting an array element with a sequence.
You define CVP = .5*Q0**2/H[N] and then CV = CVP*TAU**2 which is a sequence. And then you try to assign a derivate form it to QP[N][K] which is an element. You are trying to insert an array to a value.
For the second error I think it might be related to the third. If you could provide more information I would like to try to understand what happens.
Hope this has helped.
I am working on building a report. One figure on the report needs to be expressed in millions. I wrote some basic formatters to handle various types of formatting that need to be consistent throughout the report. To do this, I use lambda functions and string formatting. The two functions are below. One is to round, the other to format.
formatter_round = lambda x: 0 if (x is None or x is bool) else round(x/1000000,1)
formatter_dollar = lambda x: '${:,.1}'.format(0) if (x is None or x == 0) else ('${:,.1}'.format(x) if x >= 0 else '$({:,.1})'.format(abs(x)))
Now comes the problem. See my example below.
I am dealing with two numbers, a = 350000 and b = 850000.
For a everything works as I'd expect. The float isn't necessarily correct when rounded ( not "what I'd expect", but understandable behavior), but the decimal is correct.
a = 350000
formatter_dollar(formatter_round(a))
Out[89]: '$0.3'
a = Decimal(a)
formatter_dollar(formatter_round(a))
Out[91]: '$0.4'
When I run the same example with b, however, this breaks down.
b = 850000
formatter_dollar(formatter_round(b))
Out[93]: '$0.8'
b = Decimal(b)
formatter_dollar(formatter_round(b))
Out[95]: '$0.8'
My question is, how can I properly round and display numbers?
I thought my issue was floating point numbers, and a seemed to confirm that. Then when I ran the same with b, I realized that isn't the case.
Some background...
I am currently building a macro that will estimate the cost of an injection molding tool. These tools have cavities which are filled with plastic. The number of cavities a tool has is the number of parts that will be formed.
So far my program will determine the minimum number of cavities a tool can have based on customer demand. This number is always even. The tool should have an even number of cavities. Given the bounding length and width of a cavity, and setting a limit to how much space the cavities can occupy within the tool, I need my program to calculate the combination of number of cavities along the length and width whose difference is minimized and whose product is equal to the total number of minimum cavities the tool should have.
I am programming my macro is SolidWorks VBA. I first constructed this problem in Excel and used the solver tool. But, I am unable to find a way to reference the Excel Solver Tool in SolidWorks to automate this optimization problem. I am hoping to find a clever set of equations that can solve this specific problem for me. But if someone else has a better idea of what to use, that would be awesome.
Rephrasing in an optimization format...
Variables
x = number of cavities along width of tool
y = number of cavities along length of tool
z = suggested number of total cavities
Objective Function
Minimize x - y
Such that
x * y = z
x >= 1
y >= 1
x <= y
x is an integer
y is an integer
Example
My macro says that in order to meet demand, our tool needs to have at least 48 cavities. Find the number of cavities along the length and width of the tool such that the difference is minimized and the product is equal to 48. Ideally in this case the macro would return x = 6 and y = 8.
Thanks!
Just to clarify, in the question did you actually mean to Min y-x rather than Min x-y? Otherwise there is a naïve solution taking x = 1 and y = z. Min x - y = 1-z.
I don't program in VBA but here is the idea.
Since x and y are positive integers and there product is z, with x <= y. You can essentially start with x = floor(sqrt(z)) and decrement until x = 1.
For each x, check if there exists an integer y such that x * y = z. If there is, break the loop and that's the pair you are looking for. Otherwise continue until x = 1
If you need any pseudo code so you can translate it into VBA. Here it is
int x, y;
for (x = floor(sqrt(z)); x >= 1; --x)
{
y = z / x;
if (x * y == z)
break;
}
I think you can just test out a few examples. No fancy algorithm is needed.
If you relax the condition to be 2 numbers, x and y, whose product is z and with a minimum difference, then the answer is SQRT(z).
That is not an integer that meets your needs (in general). However, you can then try integers around the square root to see if they divide z. The first one you hit (i.e. minimum difference from SQRT(z)) should have the minimum difference.
If you relax the condition to be |z - x * y| is minimized, then I would recommend testing the numbers around sqrt(z). You need to check two cases -- the floor and ceiling of the square root (and the appropriate other number).
Just in case someone is needs something similar to this in the future, but can't figure out the pseudo-code I went ahead wrote it up. I wasn't sure how to output it as two values so I just threw them together as a string for the user to see.
Option Explicit
Function Factors(ByVal Test As Long) As String
Dim Val As Long
Dim i As Long
Val = Test
i = Int(Sqr(Val))
While Val / i >= 2
If Int(Val / i) * i = Val Then
Factors = i & " & " & Val / i
Exit Function
End If
i = i - 1
Wend
End Function
Here is background information to the problem I am encountering:
1) output is a cell array, each cell contains a matrix of size = 1024 x 1024, type = double
2) labelbout is a cell array which is the identical to output, except that each matrix has been binarized.
3) I am using the function regionprops to extract the mean intensity and centroid values for ROIs (there are multiple ROIs in each image) for each cell of output
4) props is a 5 x 1 struct with 2 fields (centroid and mean intensity)
The problem: I would like to take the mean intensity values for each ROI in every matrix and export to excel. Here is what I have so far:
for i = 1:size(output,2)
props = regionprops(labelboutput{1,i},output{1,i},'MeanIntensity','Centroid');
end
for i = 1:size(output,2)
meanValues = getfield(props(1:length(props),'MeanIntensity'));
end
writetable(struct2table(props), 'advanced_test.xlsx');
There seem to be a few issues:
1) my getfield command is not working and gets the error: "Index exceeds matrix dimensions"
2) when the information is being stored into props, it overwrites the values for each matrix. How do I make props a 5 x n (where n = number of cells in output)?
Please help!!
1) my getfield command is not working and gets the error: "Index exceeds matrix dimensions"
An easier way to get numeric values out of the same field in an array of structs, as an array is: [structArray.fieldName]. In your case this will be:
meanValues = [props.MeanIntensity];
2) when the information is being stored into props, it overwrites the values for each matrix. How do I make props a 5 x n (where n = number of cells in output)?
One option would be to preallocate an empty cell of the necessary dimensions and then fill it in with your regionprops output. Like this:
props = cell(size(output,1),1);
for k = 1:size(output,2)
props{k} = regionprops(labelboutput{1,k},output{1,k},'MeanIntensity','Centroid');
end
for k = 1:size(output,2)
meanValues = [props{k}.MeanIntensity];
end
...
Another option would be to combine your loops so that you can use your matrix data before it is overwritten. Like this:
for i = 1:size(output,2)
props = regionprops(labelboutput{1,i},output{1,i},'MeanIntensity','Centroid');
meanValues = [props.MeanIntensity];
% update this call to place props in non-overlapping parts of your file (e.g. append)
% writetable(struct2table(props), 'advanced_test.xlsx');
end
The bad thing about this second one is it has a file I/O step right inside your loop which can really slow things down; not to mention you will need to curtail your writetable call so it places the resulting table in non-overlapping regions of 'advanced_test.xlsx'.
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