I am struggling with my Matlab homework:
Write a script to do the following:
Generate a matrix called grades of size 8 x 25 that contains random numbers of type double in the range of 1 to 6.
Calculate the mean of matrix rows (mrow), the mean of matrix columns (mcol), and the overall mean (mall) of the matrix grades.
Copy the matrix grades to a new variable, in which you replace the elements in the 5th row and 20th to 23rd column with NaN. Compute the overall mean (mall_2) of this matrix again, i.e., the mean of the remaining values.
I am done with task 2-5, however, task 1 is not correct. I am not sure what I am doing wrong. I assume that it has something to do with the type of number (double), but I was unable to convert it.
We have to submit our homework to the online tool "Matlab Grader". The system says:
Matrix of random numbers : Variable grades has an incorrect value.
Here is my code:
% Generate matrix 'grades' with random numbers in the range 1 to 6
a = 1;
b = 6;
grades = (b-a).*rand(8,25) + a;
% calculate mean values 'mrow', 'mcol', 'mall'
mrow = mean(grades,2)
mcol = mean(grades,1)
mall = mean(grades(:))
% Replace elements with NaN
grades(5,20:23) = NaN
%Calculate mean of elements omitting NaN
mall_2 = mean(grades(:),'omitnan')
I assume your homework validation system is checking that everything in the variable grades is a (random) number in the range 1 to 6, as required by question 1.
However, by the end of your computation there are also 3 NaN values in the grades variable, because you missed this step of question 3:
Copy the matrix grades to a new variable
Instead, you overrode the elements in grades.
If you did this:
grades_mod = grades;
grades_mod(5,20:23) = NaN;
mall_2 = mean(grades_mod(:),'omitnan');
Then grades would retain its original values (no NaNs) and you can calculate mall_2.
Related
For my experiment I have 20 categories which contain 9 pictures each. I want to show these pictures in a pseudo-random sequence where the only constraint to randomness is that one image may not be followed directly by one of the same category.
So I need something similar to
r = randi([1 20],1,180);
just with an added constraint of two numbers not directly following each other. E.g.
14 8 15 15 7 16 6 4 1 8 is not legitimate, whereas
14 8 15 7 15 16 6 4 1 8 would be.
An alternative way I was thinking of was naming the categories A,B,C,...T, have them repeat 9 times and then shuffle the bunch. But there you run into the same problem I think?
I am an absolute Matlab beginner, so any guidance will be welcome.
The following uses modulo operations to make sure each value is different from the previous one:
m = 20; %// number of categories
n = 180; %// desired number of samples
x = [randi(m)-1 randi(m-1, [1 n-1])];
x = mod(cumsum(x), m) + 1;
How the code works
In the third line, the first entry of x is a random value between 0 and m-1. Each subsequent entry represents the change that, modulo m, will give the next value (this is done in the fourth line).
The key is to choose that change between 1 and m-1 (not between 0 and m-1), to assure consecutive values will be different. In other words, given a value, there are m-1 (not m) choices for the next value.
After the modulo operation, 1 is added to to transform the range of resulting values from 0,...,m-1 to 1,...,m.
Test
Take all (n-1) pairs of consecutive entries in the generated x vector and count occurrences of all (m^2) possible combinations of values:
count = accumarray([x(1:end-1); x(2:end)].', 1, [m m]);
imagesc(count)
axis square
colorbar
The following image has been obtained for m=20; n=1e6;. It is seen that all combinations are (more or less) equally likely, except for pairs with repeated values, which never occur.
You could look for the repetitions in an iterative manner and put new set of integers from the same group [1 20] only into those places where repetitions have occurred. We continue to do so until there are no repetitions left -
interval = [1 20]; %// interval from where the random integers are to be chosen
r = randi(interval,1,180); %// create the first batch of numbers
idx = diff(r)==0; %// logical array, where 1s denote repetitions for first batch
while nnz(idx)~=0
idx = diff(r)==0; %// logical array, where 1s denote repetitions for
%// subsequent batches
rN = randi(interval,1,nnz(idx)); %// new set of random integers to be placed
%// at the positions where repetitions have occured
r(find(idx)+1) = rN; %// place ramdom integers at their respective positions
end
How to increment a variable by a infinite set of numbers, in Matlab. I'v a variable which I want to increment till the loop ends by 0.1 every time but through set of range.
I'm currently doing this by: K=K*0.1; %K = 2 initially but I want this same by Matlab's trick of ranged values like [0.1:0.1:9] where 9 is the loop condination.
My Code:
K=2;
for ii=1:9
K=K*0.1;
end
If I understand correctly:
for K = 2 * 0.1.^(1:9)
%// do something with K
end
You can try using the cumprod command which returns the cumulative product of the elements in a matrix or vector. For your example, something like:
K=cumprod([2 repmat(0.1,1,9)]); % returns a row vector of 9 elements
repmat just creates a row vector of nine elements each set to the value 0.1. The last element in the vector, K(end), will be the product returned by your example. i.e.K = 2*0.1^9;
Hi I have a column of values in Matlab (PDS(:,39)). This column is filtered for various things and there are two seperate flagging columns (PDS(:,[41 81])) that are either 0 for a valid row or -1 for a non-valid row. I am taking the mean of the valid data, and if the mean is above 0, I'd like to make this value non-valid and take the mean again until the mean is below a certain value (0.2 in this instance). Here is my code:
% identify the VALID values
U1 = (PDS(:,81)==0);
F1 = (PDS(:,41)==0);
% only calculate using the valid elements
shearave = mean(PDS(U1&F1,39));
while shearave > 0.2
clear im
% determine the largest shear value overall for filtered and
% non-flagged
[c im] = max(PDS(U1&F1,39));
% make this value a NaN
PDS(im,39)=NaN;
% filter using a specific column and the overall column
PDS(im,41)=-1;
F1 = (PDS(:,41)==0);
% calculate shear ave again using new flagging column - remove the ";" so I can see the average change
shearave = mean(PDS(U1&F1,39))
end
The output that Matlab gives me is:
shearave =
0.3032
shearave =
0.3032
shearave =
0.3032
etc
The loop is not re-evalulating with the new valid data. How do I solve this problem? Do I have to use a break or continue? Or perhaps a different type of loop? Thanks for any help.
You don't need to use a loop, I'd do the following:
sort your data:
m=PDS(U1&F1,39);
[x isort]=sort(m);
Then calculate the cumulative mean of the sorted vector:
y = cumsum(x)./[1:numel(x)]';
Then truncate at 0.2, and retrieve the values needed using the indices found ...
ind=find(y<=0.2);
values_needed=m(isort(ind));
You iteratively replace values in column 39 with NaN. However, mean will not ignore NaN, but instead return NaN as the new average. You can see this with a little experiment:
>> mean([3, 4, 2, NaN, 4, 1])
ans = NaN
Therefore, shearave < 0.2 will never be true.
This post follows a previous question regarding the restructuring of a matrix:
re-formatting a matrix in matlab
An additional problem I face is demonstrated by the following example:
depth = [0:1:20]';
data = rand(1,length(depth))';
d = [depth,data];
d = [d;d(1:20,:);d];
Here I would like to alter this matrix so that each column represents a specific depth and each row represents time, so eventually I will have 3 rows (i.e. days) and 21 columns (i.e. measurement at each depth). However, we cannot reshape this because the number of measurements for a given day are not the same i.e. some are missing. This is known by:
dd = sortrows(d,1);
for i = 1:length(depth);
e(i) = length(dd(dd(:,1)==depth(i),:));
end
From 'e' we find that the number of depth is different for different days. How could I insert a nan into the matrix so that each day has the same depth values? I could find the unique depths first by:
unique(d(:,1))
From this, if a depth (from unique) is missing for a given day I would like to insert the depth to the correct position and insert a nan into the respective location in the column of data. How can this be achieved?
You were thinking correctly that unique may come in handy here. You also need the third output argument, which maps the unique depths onto the positions in the original d vector. have a look at this code - comments explain what I do
% find unique depths and their mapping onto the d array
[depths, ~, j] = unique(d(:,1));
% find the start of every day of measurements
% the assumption here is that the depths for each day are in increasing order
days_data = [1; diff(d(:,1))<0];
% count the number of days
ndays = sum(days_data);
% map every entry in d to the correct day
days_data = cumsum(days_data);
% construct the output array full of nans
dd = nan(numel(depths), ndays);
% assing the existing measurements using linear indices
% Where data does not exist, NaN will remain
dd(sub2ind(size(dd), j, days_data)) = d(:,2)
dd =
0.5115 0.5115 0.5115
0.8194 0.8194 0.8194
0.5803 0.5803 0.5803
0.9404 0.9404 0.9404
0.3269 0.3269 0.3269
0.8546 0.8546 0.8546
0.7854 0.7854 0.7854
0.8086 0.8086 0.8086
0.5485 0.5485 0.5485
0.0663 0.0663 0.0663
0.8422 0.8422 0.8422
0.7958 0.7958 0.7958
0.1347 0.1347 0.1347
0.8326 0.8326 0.8326
0.3549 0.3549 0.3549
0.9585 0.9585 0.9585
0.1125 0.1125 0.1125
0.8541 0.8541 0.8541
0.9872 0.9872 0.9872
0.2892 0.2892 0.2892
0.4692 NaN 0.4692
You may want to transpose the matrix.
It's not entirely clear from your question what your data looks like exactly, but the following might help you towards an answer.
Suppose you have a column vector
day1 = 1:21';
and, initially, all the values are NaN
day1(:) = NaN
Suppose next that you have a 2d array of measurements, in which the first column represents depths, and the second the measurements at those depths. For example
msrmnts = [1,2;2,3;4,5;6,7] % etc
then the assignment
day1(msrmnts(:,1)) = msrmnts(:,2)
will set values in only those rows of day1 whose indices are found in the first column of msrmnts. This second statement uses Matlab's capabilities for using one array as a set of indices into another array, for example
d([9 7 8 12 4]) = 1:5
would set elements [9 7 8 12 4] of d to the values 1:5. Note that the indices of the elements do not need to be in order. You could even insert the same value several times into the index array, eg [4 4 5 6 3 4] though it's not terribly useful.
I am currently trying to code up a function to assign probabilities to a collection of vectors using a histogram count. This is essentially a counting exercise, but requires some finesse to be able to achieve efficiently. I will illustrate with an example:
Say that I have a matrix X = [x1, x2....xM] with N rows and M columns. Here, X represents a collection of M, N-dimensional vectors. IN other words, each of the columns of X is an N-dimensional vector.
As an example, we can generate such an X for M = 10000 vectors and N = 5 dimensions using:
X = randint(5,10000)
This will produce a 5 x 10000 matrix of 0s and 1s, where each column is represents a 5 dimensional vector of 1s and 0s.
I would like to assign a probability to each of these vectors through a basic histogram count. The steps are simple: first find the unique columns of X; second, count the number of times each unique column occurs. The probability of a particular occurrence is then the #of times this column was in X / total number of columns in X.
Returning to the example above, I can do the first step using the unique function in MATLAB as follows:
UniqueXs = unique(X','rows')'
The code above will return UniqueXs, a matrix with N rows that only contains the unique columns of X. Note that the transposes are due to weird MATLAB input requirements.
However, I am unable to find a good way to count the number of times each of the columns in UniqueX is in X. So I'm wondering if anyone has any suggestions?
Broadly speaking, I can think of two ways of achieving the counting step. The first way would be to use the find function, though I think this may be slow since find is an elementwise operation. The second way would be to call unique recursively as it can also provide the index of one of the unique columns in X. This should allow us to remove that column from X and redo unique on the resulting X and keep counting.
Ideally, I think that unique might already be doing some counting so the most efficient way would probably be to work without the built-in functions.
Here are two solutions, one assumes all values are either 0's or 1's (just like the example in your description), the other does not. Both codes should be very fast (more so the one with binary values), even on large data.
1) only zeros and ones
%# random vectors of 0's and 1's
x = randi([0 1], [5 10000]); %# RANDINT is deprecated, use RANDI instead
%# convert each column to a binary string
str = num2str(x', repmat('%d',[1 size(x,1)])); %'
%# convert binary representation to decimal number
num = (str-'0') * (2.^(size(s,2)-1:-1:0))'; %'# num = bin2dec(str);
%# count frequency of how many each number occurs
count = accumarray(num+1,1); %# num+1 since it starts at zero
%# assign probability based on count
prob = count(num+1)./sum(count);
2) any positive integer
%# random vectors with values 0:MAX_NUM
x = randi([0 999], [5 10000]);
%# format vectors as strings (zero-filled to a constant length)
nDigits = ceil(log10( max(x(:)) ));
frmt = repmat(['%0' num2str(nDigits) 'd'], [1 size(x,1)]);
str = cellstr(num2str(x',frmt)); %'
%# find unique strings, and convert them to group indices
[G,GN] = grp2idx(str);
%# count frequency of occurrence
count = accumarray(G,1);
%# assign probability based on count
prob = count(G)./sum(count);
Now we can see for example how many times each "unique vector" occurred:
>> table = sortrows([GN num2cell(count)])
table =
'000064850843749' [1] # original vector is: [0 64 850 843 749]
'000130170550598' [1] # and so on..
'000181606710020' [1]
'000220492735249' [1]
'000275871573376' [1]
'000525617682120' [1]
'000572482660558' [1]
'000601910301952' [1]
...
Note that in my example with random data, the vector space becomes very sparse (as you increase the maximum possible value), thus I wouldn't be surprised if all counts were equal to 1...