I have the following code:
for i = 1450:9740:89910
n = i+495;
range = ['B',num2str(i),':','H',num2str(n)];
iter = xlsread('BrokenDisplacements.xlsx' , range);
displ = iter;
displ = [displ; iter];
end
Which takes values from an Excel file from a number of ranges I want and outputs them as matricies. However, this code just uses the final value of displ and creates the total matrix from there. I would like to total these outputs (displ) into one large matrix saving values along the way, how would I go about doing this?
Since you know the size of the block of data you are reading, you can make your code much more efficient as follows:
firstVals = 1450:9740:89910;
displ = zeros((firstVals(end) - firstVals(1) + 1 + 496), 7);
for ii = firstVals
n = ii + 495;
range = sprintf('B%d:H%d', ii, ii+495);
displ((ii:ii+495)-firstVals(1)+1,:) = xlsread('BrokenDiplacements.xlsx', range);
end
Couple of points:
I prefer not to use i as a variable since it is built in as sqrt(-1) - if you later execute code that assumes that to be true, you're in trouble
I am not assuming that the last value of ii is 89910 - by first assigning the value to a vector, then finding the last value in the vector, I sidestep that question
I assign all space in iter at once - otherwise, as it grows, Matlab keeps having to move the array around which can slow things down a lot
I used sprintf to generate the string representing the range - I think it's more readable but it's a question of style
I assign the return value of xlsread directly to a block in displ that is the right size
I hope this helps.
How about this:
displ=[];
for i = 1450:9740:89910
n = i+495;
range = ['B',num2str(i),':','H',num2str(n)];
iter = xlsread('BrokenDisplacements.xlsx' , range);
displ = [displ; iter];
end
Related
Suppose we are running an infinite for loop in MATLAB, and we want to store the iterative values in a vector. How can we declare the vector without knowing the size of it?
z=??
for i=1:inf
z(i,1)=i;
if(condition)%%condition is met then break out of the loop
break;
end;
end;
Please note first that this is bad practise, and you should preallocate where possible.
That being said, using the end keyword is the best option for extending arrays by a single element:
z = [];
for ii = 1:x
z(end+1, 1) = ii; % Index to the (end+1)th position, extending the array
end
You can also concatenate results from previous iterations, this tends to be slower since you have the assignment variable on both sides of the equals operator
z = [];
for ii = 1:x
z = [z; ii];
end
Sadar commented that directly indexing out of bounds (as other answers are suggesting) is depreciated by MathWorks, I'm not sure on a source for this.
If your condition computation is separate from the output computation, you could get the required size first
k = 0;
while ~condition
condition = true; % evaluate the condition here
k = k + 1;
end
z = zeros( k, 1 ); % now we can pre-allocate
for ii = 1:k
z(ii) = ii; % assign values
end
Depending on your use case you might not know the actual number of iterations and therefore vector elements, but you might know the maximum possible number of iterations. As said before, resizing a vector in each loop iteration could be a real performance bottleneck, you might consider something like this:
maxNumIterations = 12345;
myVector = zeros(maxNumIterations, 1);
for n = 1:maxNumIterations
myVector(n) = someFunctionReturningTheDesiredValue(n);
if(condition)
vecLength = n;
break;
end
end
% Resize the vector to the length that has actually been filled
myVector = myVector(1:vecLength);
By the way, I'd give you the advice to NOT getting used to use i as an index in Matlab programs as this will mask the imaginary unit i. I ran into some nasty bugs in complex calculations inside loops by doing so, so I would advise to just take n or any other letter of your choice as your go-to loop index variable name even if you are not dealing with complex values in your functions ;)
You can just declare an empty matrix with
z = []
This will create a 0x0 matrix which will resize when you write data to it.
In your case it will grow to a vector ix1.
Keep in mind that this is much slower than initializing your vector beforehand with the zeros(dim,dim) function.
So if there is any way to figure out the max value of i you should initialize it withz = zeros(i,1)
cheers,
Simon
You can initialize z to be an empty array, it'll expand automatically during looping ...something like:
z = [];
for i = 1:Inf
z(i) = i;
if (condition)
break;
end
end
However this looks nasty (and throws a warning: Warning: FOR loop index is too large. Truncating to 9223372036854775807), I would do here a while (true) or the condition itself and increment manually.
z = [];
i = 0;
while !condition
i=i+1;
z[i]=i;
end
And/or if your example is really what you need at the end, replace the re-creation of the array with something like:
while !condition
i=i+1;
end
z = 1:i;
As mentioned in various times in this thread the resizing of an array is very processing intensive, and could take a lot of time.
If processing time is not an issue:
Then something like #Wolfie mentioned would be good enough. In each iteration the array length will be increased and that is that:
z = [];
for ii = 1:x
%z = [z; ii];
z(end+1) = ii % Best way
end
If processing time is an issue:
If the processing time is a large factor, and you want it to run as smooth as possible, then you need to preallocating.If you have a rough idea of the maximum number of iterations that will run then you can use #PluginPenguin's suggestion. But there could still be a change of hitting that preset limit, which will break (or severely slow down) the program.
My suggestion:
If your loop is running infinitely until you stop it, you could do occasional resizing. Essentially extending the size as you go, but only doing it once in a while. For example every 100 loops:
z = zeros(100,1);
for i=1:inf
z(i,1)=i;
fprintf("%d,\t%d\n",i,length(z)); % See it working
if i+1 >= length(z) %The array as run out of space
%z = [z; zeros(100,1)]; % Extend this array (note the semi-colon)
z((length(z)+100),1) = 0; % Seems twice as fast as the commented method
end
if(condition)%%condition is met then break out of the loop
break;
end;
end
This means that the loop can run forever, the array will increase with it, but only every once in a while. This means that the processing time hit will be minimal.
Edit:
As #Cris kindly mentioned MATLAB already does what I proposed internally. This makes two of my comments completely wrong. So the best will be to follow what #Wolfie and #Cris said with:
z(end+1) = i
Hope this helps!
I have acceleration (10240x31) data that I want to filter by replacing every data point that exceeds the threshold value of 4 times standard derivation of each column with the mean value of the two adjacent data points.
First, I wanted to replace every data point with a zero, if it exceeds the maximum value. This is my loop:
for w = 1:31
Sigma(w) = std(zacceleration(:,w));
zacceleration(zacceleration<(-4*Sigma(w))) = 0;
zacceleration(zacceleration>(4*Sigma(w))) = 0;
end
That code works if w is just one number, for example:
w = 1;
But when w changes every iteration, the filtered data only contains the values that don't exceed the threshold value of the last dataset, Sigma(31).
So, I guess that I overwrite my data or something like that but I cant seem to find a solution.
Can anybody please give me a hint?
Thank you in advance and best regards.
I think I got it now.
Sigma = std(zacceleration);
for a = 1:10240;
for b = 1:31;
if zacceleration(a,b)<(-4*Sigma(b))
zacceleration(a,b) = 0;
end
if zacceleration(a,b)>(4*Sigma(b))
zacceleration(a,b) = 0;
end
end
end
I am trying to find the max value and its location. Following is the example of the programme,
fname = dir('*.mat');
nfiles = length(fname);
vals = cell(nfiles,1);
phen = cell(nfiles,1);
for i = 1:nfiles
vals{i} = load(fname(i).name);
phen{i} = (vals{i}.phen);
[M, position] = max(phen{i},[],3);
clear vals
end
After the program is executed, all the position is showing 1. There are total 15 files and M is taking the values of the last file.
How to overcome this prpoblem? Any help will be appreciated
I am not sure I understand your question.
However, at every iteration you are computing the max value and position and overwriting them in the next iteration (i.e. not storing them anywhere). So at the end of the loop M and position would correspond to the last entry phen{nfiles}.
Each time you run through your for loop, you are overwriting M with the max from the most recently loaded phen from the dimension of 3. Since your data is only two dimensional, you probably should be using a dimension of 1 or 2 instead of 3. Because you are using 3, max is returning 1 to position. Fix the dimension issue and position should then be the correct value.
What you could do is make M and position the size of nfiles. So instead of
[M, position] = max(phen{i},[],3);
do
%create M and positions arrays here
%ex. M(nfiles) = 0; or a smaller value if your values are negative
%do the same for positions
[M(i), positions(i)] = max(phen{i},[],1); %1 or 2 correction here here!
then after your for loop
...
end
[maxM, maxMposition] = max(M);
position = positions(maxMposition);
I am refering to an example like this
I have a function to analize the elements of a vector, 'input'. If these elements have a special property I store their values in a vector, 'output'.
The problem is that at the begging I don´t know the number of elements it will need to store in 'output'so I don´t know its size.
I have a loop, inside I go around the vector, 'input' through an index. When I consider special some element of this vector capture the values of 'input' and It be stored in a vector 'ouput' through a sentence like this:
For i=1:N %Where N denotes the number of elements of 'input'
...
output(j) = input(i);
...
end
The problem is that I get an Error if I don´t previously "declare" 'output'. I don´t like to "declare" 'output' before reach the loop as output = input, because it store values from input in which I am not interested and I should think some way to remove all values I stored it that don´t are relevant to me.
Does anyone illuminate me about this issue?
Thank you.
How complicated is the logic in the for loop?
If it's simple, something like this would work:
output = input ( logic==true )
Alternatively, if the logic is complicated and you're dealing with big vectors, I would preallocate a vector that stores whether to save an element or not. Here is some example code:
N = length(input); %Where N denotes the number of elements of 'input'
saveInput = zeros(1,N); % create a vector of 0s
for i=1:N
...
if (input meets criteria)
saveInput(i) = 1;
end
end
output = input( saveInput==1 ); %only save elements worth saving
The trivial solution is:
% if input(i) meets your conditions
output = [output; input(i)]
Though I don't know if this has good performance or not
If N is not too big so that it would cause you memory problems, you can pre-assign output to a vector of the same size as input, and remove all useless elements at the end of the loop.
output = NaN(N,1);
for i=1:N
...
output(i) = input(i);
...
end
output(isnan(output)) = [];
There are two alternatives
If output would be too big if it was assigned the size of N, or if you didn't know the upper limit of the size of output, you can do the following
lengthOutput = 100;
output = NaN(lengthOutput,1);
counter = 1;
for i=1:N
...
output(counter) = input(i);
counter = counter + 1;
if counter > lengthOutput
%# append output if necessary by doubling its size
output = [output;NaN(lengthOutput,1)];
lengthOutput = length(output);
end
end
%# remove unused entries
output(counter:end) = [];
Finally, if N is small, it is perfectly fine to call
output = [];
for i=1:N
...
output = [output;input(i)];
...
end
Note that performance degrades dramatically if N becomes large (say >1000).
I need to calculate the mean, standard deviation, and other values for a number of variables and I was wondering how to use a loop to my advantage. I have 5 electrodes of data. So to calculate the mean of each I do this:
mean_ch1 = mean(ch1);
mean_ch2 = mean(ch2);
mean_ch3 = mean(ch3);
mean_ch4 = mean(ch4);
mean_ch5 = mean(ch5);
What I want is to be able to condense that code into a line or so. The code I tried does not work:
for i = 1:5
mean_ch(i) = mean(ch(i));
end
I know this code is wrong but it conveys the idea of what I'm trying to accomplish. I want to end up with 5 separate variables that are named by the loop or a cell array with all 5 variables within it allowing for easy recall. I know there must be a way to write this code I'm just not sure how to accomplish it.
You have a few options for how you can do this:
You can put all your channel data into one large matrix first, then compute the mean of the rows or columns using the function MEAN. For example, if each chX variable is an N-by-1 array, you can do the following:
chArray = [ch1 ch2 ch3 ch4 ch5]; %# Make an N-by-5 matrix
meanArray = mean(chArray); %# Take the mean of each column
You can put all your channel data into a cell array first, then compute the mean of each cell using the function CELLFUN:
meanArray = cellfun(#mean,{ch1,ch2,ch3,ch4,ch5});
This would work even if each chX array is a different length from one another.
You can use EVAL to generate the separate variables for each channel mean:
for iChannel = 1:5
varName = ['ch' int2str(iChannel)]; %# Create the name string
eval(['mean_' varName ' = mean(' varName ');']);
end
If it's always exactly 5 channels, you can do
ch = {ch1, ch2, ch3, ch4, ch5}
for j = 1:5
mean_ch(j) = mean(ch{j});
end
A more complicated way would be
for j = 1:nchannels
mean_ch(j) = eval(['mean(ch' num2str(j) ')']);
end
Apart from gnovice's answer. You could use structures and dynamic field names to accomplish your task. First I assume that your channel data variables are all in the format ch* and are the only variables in your MATLAB workspace. The you could do something like the following
%# Move the channel data into a structure with fields ch1, ch2, ....
%# This could be done by saving and reloading the workspace
save('channelData.mat','ch*');
chanData = load('channelData.mat');
%# Next you can then loop through the structure calculating the mean for each channel
flds = fieldnames(chanData); %# get the fieldnames stored in the structure
for i=1:length(flds)
mean_ch(i) = mean(chanData.(flds{i});
end