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Is it possible to express an exponential in terms of a quadratic?
I am trying to see whether or not an exponential or a linear fit is the best fit for my data. I am going to be extrapolating the...
12 years 前 | 1 个回答 | 0
1
个回答提问
Does matlab have the ability to perform a 3 parameter weibull analysis
I have managed to perform a two parameter weibull analysis on my data which provides me with the shape parameter and the scale p...
13 years 前 | 0 个回答 | 0
0
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Weibull 0 time failures
I'm having a lot of trouble trying to analyse the following times to failure, due to their being dead-on-arrival failures. The o...
13 years 前 | 1 个回答 | 0
1
个回答已回答
Problem with appending to assignin
Solved: A(j,1) = regexprep(TXTFilenameMAINTENANCE(j,1),'[\W'',\d]','') s.(A{j,1}) = polyfit (X,Y,1) then to append: ...
Problem with appending to assignin
Solved: A(j,1) = regexprep(TXTFilenameMAINTENANCE(j,1),'[\W'',\d]','') s.(A{j,1}) = polyfit (X,Y,1) then to append: ...
13 years 前 | 0
提问
Problem with appending to assignin
Hi, I have this line of code: assignin('base',char(regexprep(TXTFilenameMAINTENANCE(j,1),'[^\w'']','')),polyfit (X,Y,1)); ...
13 years 前 | 2 个回答 | 0
2
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Trouble with k means
I'm having trouble tweaking the k means options as shown here: http://www.mathworks.co.uk/help/toolbox/stats/kmeans.html I'm...
13 years 前 | 2 个回答 | 0
2
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Identifying duplicates and deleting duplicates
v1 =[300 350 350 350 360 370 380 380 380 390 395 405 450 450 450 450 450 450 500 500 ...
13 years 前 | 3 个回答 | 0
3
个回答提问
Graph analysis question
I have this data: X = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 With Y = 0.5 1 1.5 2 2.5 3 3.5 4 1 1.5...
13 years 前 | 3 个回答 | 0
3
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Selecting part of a string
I have a list of file names in a variable called file_name e.g: [a.txt b.txt c.txt d.txt] I would like to create a new v...
13 years 前 | 1 个回答 | 0
1
个回答已回答
Calling a variable from a variable
Just done it using eval e.g a = eval((char(V(1,1)))); >> a (1,1); thanks for the help guys, Matt
Calling a variable from a variable
Just done it using eval e.g a = eval((char(V(1,1)))); >> a (1,1); thanks for the help guys, Matt
13 years 前 | 0
| 已接受
提问
Calling a variable from a variable
e.g With variables: a b c d V = who; Now: V (2) = b b(1) = 3000 (for example) I would like to call var...
13 years 前 | 4 个回答 | 0
4
个回答已回答
What should be a very simple bit of code is causing me lot's of trouble, X = find(I0_data == I0_unique(i,1));
X = find(I0_data (:,1) == I0_unique(i,1));
What should be a very simple bit of code is causing me lot's of trouble, X = find(I0_data == I0_unique(i,1));
X = find(I0_data (:,1) == I0_unique(i,1));
13 years 前 | 0
| 已接受
已回答
What should be a very simple bit of code is causing me lot's of trouble, X = find(I0_data == I0_unique(i,1));
Right I think I have had a little breakthrough. The problem is here: X = find(I0_data == I0_unique(i,1)); Specificall...
What should be a very simple bit of code is causing me lot's of trouble, X = find(I0_data == I0_unique(i,1));
Right I think I have had a little breakthrough. The problem is here: X = find(I0_data == I0_unique(i,1)); Specificall...
13 years 前 | 0
已回答
What should be a very simple bit of code is causing me lot's of trouble, X = find(I0_data == I0_unique(i,1));
here is the full code: I0_unique = unique(I0_data (1:31136))'; [a,b] = size(I0_unique); for i = 1:a; X = find(I0_da...
What should be a very simple bit of code is causing me lot's of trouble, X = find(I0_data == I0_unique(i,1));
here is the full code: I0_unique = unique(I0_data (1:31136))'; [a,b] = size(I0_unique); for i = 1:a; X = find(I0_da...
13 years 前 | 0
提问
What should be a very simple bit of code is causing me lot's of trouble, X = find(I0_data == I0_unique(i,1));
I believe that the code is correct as it has worked for me on many occasions, but I am having trouble with this data set and for...
13 years 前 | 3 个回答 | 0