Here is my dataset:
pressure(time, case, x, y)
>> size(pressure)
ans =
100 1 289 570
How to get a spatial nanmean pressure for x from 30 to 60 and y from 40 to 70 in each time step?
For example: a nanmean value for that particular region for each timestep from time 1 to time 100.
I tried this, "spatial_mean_pressure = nanmean(pressure(:,:,30:60,40:70))" It averaged the pressure in the timeserie. This is not the result I want.
>> size(spatial_mean_pressure)
ans =
1 1 31 31
I like to get the results like this:
>> size(spatial_mean_pressure)
ans =
100 1 1 1
You are trying to get a mean for an entire block of matrix. Therefore, you should apply nanmean twice and not once. Also, apply it along a particular dimension to get the desired result. I think this is what you want.
x=randi(10,[100 1 10 25]);
First take the mean along the third dimension.
mean_x_3=nanmean(x,3);
You would get an answer of size = [100 1 1 25]. Then take the mean along 4th dimension.
mean_x_4=nanmean(mean_x_3,4);
This should give you the desired answer. You can write this in one line as,
mean_x = nanmean(nanmean(x,3),4);
Related
I am trying to calculate the probability of success in 70 (or less) Bernoulli trials out of 100. I wrote it with Matlab. But, I get the probability to be 1 (it can't be 1 since its not success in all 100 trials).
Is my function OK?
syms k
f = nchoosek(100,k)*0.5^k*0.5^(100-k);
F = double(symsum(nchoosek(100,k)*0.5^k*0.5^(100-k),k,0,70));
If it is, how can I get a more accurate resault in Matlab?
Thanks
edit:
I have have a binary vector that represents success/failure in n trials (like tossing a coin 100 times). And I need the error of my sample (the way statistics does it.. but I don't know statistics). So I thought that maybe i will try to calculate "how far am I from being correct in all trials" which should be 1-F in my code. But then 70 successes out of 100 gives me error = 0 which is obviously not true..
edit2: In the example I gave here I need the probability that there are 70 successes in 100 trials.
You do have everything you need to answer this question.
In the formula you have posted, you sum the probabilities from 0 to 70, that is, it will calculate the probability to have 0 or 1 or 2 .. or 70 successes, which means 70 or less successes.
Without the sum, you get the probability to have exactly k successes. The probability to get exactly 70 successes is:
k = 70;
f = nchoosek(100,k)*0.5^k*0.5^(100-k)
Warning: Result may not be exact. Coefficient is greater than 9.007199e+15 and is only
accurate to 15 digits
> In nchoosek (line 92)
f =
2.3171e-05
You receive a warning that the computation of nchoosek(100,70) is not exact (see below for a better way).
To compute the probability to get 70 or less successes, sum over the probabilities to get 0 or 1 or .. 70 successes:
>> f = 0;
>> for k=0:70;
f = f + nchoosek(100,k)*.5^k*.5^(100-k);
end
You will receive a lot of warnings, but you can look at f:
>> f
f =
1.0000
As you see, if rounded to four digits, the probability is 1. We know, however, that it must be slighly less than one´. If we ask Matlab to show more digits:
>> format long
we see that it is not exactly 1:
>> f
f =
0.999983919992352
If you compute 1-f, you will see that the result is not 0 (I switch back to showing less digits):
>> format short
>> 1-f
ans =
1.6080e-05
To get rid of the warnings and to simplify the code for computing the probabilities, Matlab provides several functions to deal with binomial distributions. For the probability to get exactly 70 successes, use
>> binopdf(70,100,.5)
ans =
2.3171e-05
and to get 70 or less successes:
>> format long
>> binocdf(70,100,.5)
ans =
0.999983919992352
Can anybody help me to find out the method to calculate the elements of different sized matrix in Matlab ?
Let say that I have 2 matrices with numbers.
Example:
A=[1 2 3;
4 5 6;
7 8 9]
B=[10 20 30;
40 50 60]
At first,we need to find maximum number in each column.
In this case, Ans=[40 50 60].
And then,we need to find ****coefficient** (k).
Coefficient(k) is equal to 1 divided by quantity of column of matrix A.
In this case, **coefficient (k)=1/3=0.33.
I wanna create matrix C filling with calculation.
Example in MS Excel.
H4 = ABS((C2-C6)/C9)*0.33+ABS((D2-D6)/D9)*0.33+ABS((E2-E6)/E9)*0.33
I4 = ABS((C3-C6)/C9)*0.33+ABS((D3-D6)/D9)*0.33+ABS((E3-E6)/E9)*0.33
J4 = ABS((C4-C6)/C9)*0.33+ABS((D4-D6)/D9)*0.33+ABS((E4-E6)/E9)*0.33
And then (Like above)
H5 = ABS((C2-C7)/C9)*0.33+ABS((D2-D7)/D9)*0.33+ABS((E2-E7)/E9)*0.33
I5 = ABS((C3-C7)/C9)*0.33+ABS((D3-D7)/D9)*0.33+ABS((E3-E7)/E9)*0.33
J5 = ABS((C4-C7)/C9)*0.33+ABS((D4-D7)/D9)*0.33+ABS((E4-E7)/E9)*0.33
C =
0.34 =|(1-10)|/40*0.33+|(2-20)|/50*0.33+|(3-30)|/60*0.33
0.28 =|(4-10)|/40*0.33+|(5-20)|/50*0.33+|(6-30)|/60*0.33
0.22 =|(7-10)|/40*0.33+|(8-20)|/50*0.33+|(9-30)|/60*0.33
0.95 =|(1-40)|/40*0.33+|(2-50)|/50*0.33+|(3-60)|/60*0.33
0.89 =|(4-40)|/40*0.33+|(5-50)|/50*0.33+|(6-60)|/60*0.33
0.83 =|(7-40)|/40*0.33+|(8-50)|/50*0.33+|(9-60)|/60*0.33
Actually A is a 15x4 matrix and B is a 5x4 matrix.
Perhaps,the matrices dimensions are more than this matrices (variables).
How can i write this in Matlab?
Thanks you!
You can do it like so. Let's assume that A and B are defined as you did before:
A = vec2mat(1:9, 3)
B = vec2mat(10:10:60, 3)
A =
1 2 3
4 5 6
7 8 9
B =
10 20 30
40 50 60
vec2mat will transform a vector into a matrix. You simply specify how many columns you want, and it will automatically determine the right amount of rows to transform the vector into a correctly shaped matrix (thanks #LuisMendo!). Let's also define more things based on your post:
maxCol = max(B); %// Finds maximum of each column in B
coefK = 1 / size(A,2); %// 1 divided by number of columns in A
I am going to assuming that coefK is multiplied by every element in A. You would thus compute your desired matrix as so:
cellMat = arrayfun(#(x) sum(coefK*(bsxfun(#rdivide, ...
abs(bsxfun(#minus, A, B(x,:))), maxCol)), 2), 1:size(B,1), ...
'UniformOutput', false);
outputMatrix = cell2mat(cellMat).'
You thus get:
outputMatrix =
0.3450 0.2833 0.2217
0.9617 0.9000 0.8383
Seems like a bit much to chew right? Let's go through this slowly.
Let's start with the bsxfun(#minus, A, B(x,:)) call. What we are doing is taking the A matrix and subtracting with a particular row in B called x. In our case, x is either 1 or 2. This is equal to the number of rows we have in B. What is cool about bsxfun is that this will subtract every row in A by this row called by B(x,:).
Next, what we need to do is divide every single number in this result by the corresponding columns found in our maximum column, defined as maxCol. As such, we will call another bsxfun that will divide every element in the matrix outputted in the first step by their corresponding column elements in maxCol.
Once we do this, we weight all of the values of each row by coefK (or actually every value in the matrix). In our case, this is 1/3.
After, we then sum over all of the columns to give us our corresponding elements for each column of the output matrix for row x.
As we wish to do this for all of the rows, going from 1, 2, 3, ... up to as many rows as we have in B, we apply arrayfun that will substitute values of x going from 1, 2, 3... up to as many rows in B. For each value of x, we will get a numCol x 1 vector where numCol is the total number of columns shared by A and B. This code will only work if A and B share the same number of columns. I have not placed any error checking here. In this case, we have 3 columns shared between both matrices. We need to use UniformOutput and we set this to false because the output of arrayfun is not a single number, but a vector.
After we do this, this returns each row of the output matrix in a cell array. We need to use cell2mat to transform these cell array elements into a single matrix.
You'll notice that this is the result we want, but it is transposed due to summing along the columns in the second step. As such, simply transpose the result and we get our final answer.
Good luck!
Dedication
This post is dedicated to Luis Mendo and Divakar - The bsxfun masters.
Assuming by maximum number in each column, you mean columnwise maximum after vertically concatenating A and B, you can try this one-liner -
sum(abs(bsxfun(#rdivide,bsxfun(#minus,permute(A,[3 1 2]),permute(B,[1 3 2])),permute(max(vertcat(A,B)),[1 3 2]))),3)./size(A,2)
Output -
ans =
0.3450 0.2833 0.2217
0.9617 0.9000 0.8383
If by maximum number in each column, you mean columnwise maximum of B, you can try -
sum(abs(bsxfun(#rdivide,bsxfun(#minus,permute(A,[3 1 2]),permute(B,[1 3 2])),permute(max(B),[1 3 2]))),3)./size(A,2)
The output for this case stays the same as the previous case, owing to the values of A and B.
I am trying to find the Extremum of a 3-dim matrix along the 2nd dimension.
I started with
[~,index] = max(abs(mat),[],2), but I don't know how to advance from here. How is the index vector to be used together with the original matrix. Or is there a completely different solution to this problem?
To illustrate the task assume the following matrix:
mat(:,:,1) =
23 8 -4
-1 -26 46
mat(:,:,2) =
5 -27 12
2 -1 18
mat(:,:,3) =
-10 49 39
-13 -46 41
mat(:,:,4) =
30 -24 18
-40 -16 -36
The expected result would then be
ext(:,:,1) =
23
-46
ext(:,:,2) =
-27
18
ext(:,:,3) =
49
-46
ext(:,:,4) =
30
-40
I don't know how to use the index vector with mat to get the desired result ext.
1) If you want to find a maximum just along, let's say, 2d dimension, your variable index will be a matrix having dimensions (N,1,M), where N and M are number of elements of your matrix in the first and third dimensions respectively. In order to remove dummy dimensions, there is function squeeze() exist: index=squeeze(index) After that size(index) gives N,M
2) Depending on your problem, you probably need matlab function ind2sub(). First, you take a slice of your matrix, than find its maximum with linear indexing, and than you can restore your indicies with int2sub(). Here is an example for a 2D matrix:
M = randn(5,5);
[C,I] = max(M(:));
[index1,index2] = ind2sub(size(M),I);
Same method allows to find the absolute maximal element in whole 3D matrix.
Use ndgrid to generate the values along dimensions 1 and 3, and then sub2ind to combine the three indices into a linear index:
[~, jj] = max(abs(mat),[],2); %// jj: returned by max
[ii, ~, kk] = ndgrid(1:size(mat,1),1,1:size(mat,3)); %// ii, kk: all combinations
result = mat(sub2ind(size(mat), ii, jj, kk));
A fancier, one-line alternative:
result = max(complex(mat),[],2);
This works because, acccording to max documentation,
For complex input A, max returns the complex number with the largest complex modulus (magnitude), computed with max(abs(A)).
I'm trying to check the full range of values on the x-axis of a histogram, especially while I expected the full range to be [0, 255], when I used the following command for the histogram [h, bins] = hist(H), I got the following:
h =
221 20 6 4 1 1 2 0 0 1
bins =
Columns 1 through 7
8.2500 24.7500 41.2500 57.7500 74.2500 90.7500 107.2500
Columns 8 through 10
123.7500 140.2500 156.7500
This implies that the maximum range I got here is up to 165.
If we look at the histogram below, we can see that 165 seems to be the maximum number of frequency value. How do I know the maximum value (range) of the x-axis?
I think you need either one of these 3 options:
max(H)
Or
numel(unique(H))
or
numel(H)
I would start at the top till you find the one you need.
How can I interpolate a vector in MATLAB?
For example, I have the following matrix:
M=
1 10
2 20
3 30
4 40
The first column of M denotes the independent parameter of x coordinate while the second column of M denotes the output or y coordinate.
I also have the following input vector:
a =
2.3
2.1
3.5
For each value of a, I wish to determine what the output interpolated result would be. In this case, given a, I wish to return
23
21
35
Here's the answer to the question after the edit, i.e. "how to interpolate"
You want to use interp1
M = [1 10;2 20;3 30;4 40];
a = [2.3;2.1;3.5;1.2];
interpolatedVector = interp1(M(:,1),M(:,2),a)
interpolatedVector =
23
21
35
12
Here's the answer to the question "find the two closest entries in a vector", i.e. the original question before the edit.
x=[1,2,3,4,5]'; %'#
a =3.3;
%# sort the absolute difference
[~,idx] = sort(abs(x-a));
%# find the two closest entries
twoClosestIdx = idx(1:2);
%# turn it into a logical array
%# if linear indices aren't good enough
twoClosestIdxLogical = false(size(x));
twoClosestIdxLogical(twoClosestIdx) = true;
twoClosestIdxLogical =
0
0
1
1
0