I have a question regarding indexing and loops in MATLAB. I have a vector of length n (named data in the code below). I want to examine this vector 4 elements at a time inside of a for loop. How can I do this? My attempt included below does not work because it will exceed the array dimensions at the end of the loop.
for k = 1:length(data)
index = k:k+3;
cur_data = data(index);
pre_q_data1 = cur_data(1);
pre_q_data2 = cur_data(2);
% Interweaving the data
q = [pre_q_data1; pre_q_data2];
qdata = q(:)';
pre_i_data1 = cur_data(3);
pre_i_data2 = cur_data(4);
i = [pre_i_data1; pre_i_data2];
idata = i(:)';
end
You shouldn't have k go all the way to length(data) if you're planning on indexing up to k+3.
I've also taken the liberty of greatly simplifying your code, but feel free to ignore that!
for k = 1:length(data)-3
% maximum k = length(data)-3, so maximum index = length(data)-3+3=length(data)
index = k:k+3;
cur_data = data(k:k+3);
% Interweaving the data
q = cur_data(1:2); % transpose at end of line here if need be
i = cur_data(3:4); % could just use data(k+2:k+3) and not use cur_data
end
Related
I have a question above removing rows from a matrix. I have the code below which removes the rows I want, but the problem is each time a row is removed it changes the size of the matrix. When the size of the matrix changes, the for loops can no longer run through the original size of the matrix because it has changed. Does anyone know how to get around this? Thanks.
for i = 1:NT
for j = 1:NP
for k = 1:NP
if ContactPartData((i-1)*(NP*(NP-1)) + ((j-1)*NP + k),2) == 0
ContactPartData((i-1)*(NP*(NP-1)) + ((j-1)*NP + k),:) = [];
else
end
end
end
end
For these cases, it is typically easier to record which rows you want to remove, and then remove them all at once at the end. This is more efficient than repeatedly removing a single row. And it solves your problem at the same time!
toremove = false(size(ContactPartData,1),1);
for i = 1:NT
for j = 1:NP
for k = 1:NP
if ContactPartData((i-1)*(NP*(NP-1)) + ((j-1)*NP + k),2) == 0
toremove((i-1)*(NP*(NP-1)) + ((j-1)*NP + k)) = true;
end
end
end
end
ContactPartData(toremove,:) = [];
Of course, in this particular case, the loop is not needed at all:
toremove = ContactPartData(:,2) == 0;
ContactPartData(toremove,:) = [];
Additionally, it might be more efficient to do it the other way around, selecting which rows to preserve (time the code to find out!):
tokeep = ContactPartData(:,2) ~= 0;
ContactPartData = ContactPartData(tokeep,:);
This is what I am trying to do, created a random array to demonstrate:
% all IDs
all_IDS = 1:216000000;
% Array 1
X = round(1550*rand(216000000,1));
Y = round(1550*rand(216000000,1));
Z = round(90*rand(216000000,1));
% Array 2
Xsub = round(1550*rand(160000,1));
Ysub = round(1550*rand(160000,1));
Zsub = round(90*rand(160000,1));
del_val =1;
% required o/p
reqd_op = zeros(1,10);
% boolean indexing
indx =1;
for jj = 1:160000
VID_X = Xsub(jj);
VID_Y = Ysub(jj);
VID_Z = Zsub(jj);
I2 = (X>VID_X-del_val & X<VID_X+del_val)& (Y>VID_Y-del_val & Y<VID_Y+del_val) & (Z>VID_Z-del_val & Z<VID_Z+del_val);
len = numel(all_IDS(I2));
reqd_op(1,indx:indx+len-1) = all_IDS(I2);
indx=indx+len;
end
The above code takes a lot of time as I am dealing with a very large array , Is there a way to eliminate the for loop, meaning, instead of doing Boolean indexing element by element - can I do it for the whole array at once ?
This will run x2.5 faster, anyway, array is too big so it still takes 0.3s per loop, so 160000 loops is like 13 hours on single cpu.
if ~exist('X','var')
% Array 1
X = round(1550*rand(216000000,1,'single'));
Y = round(1550*rand(216000000,1,'single'));
Z = round(90*rand(216000000,1,'single'));
% Array 2
Xsub = round(1550*rand(160000,1,'single'));
Ysub = round(1550*rand(160000,1,'single'));
Zsub = round(90*rand(160000,1,'single'));
end
del_val =single(1);
reqd_op = zeros(1,10,'single');% required o/p
tic
index =1;
for jj = 1:10
VID_X = Xsub(jj);
VID_Y = Ysub(jj);
VID_Z = Zsub(jj);
IdxFinal=[];
Idx1=find(abs(X-VID_X)<del_val); %little better than X>VID_X-del_val & X<VID_X+del_val)
if ~isempty(Idx1)
Idx2 = Idx1(Y(Idx1)>VID_Y-del_val & Y(Idx1)<VID_Y+del_val);
if ~isempty(Idx2)
Idx3= Idx2(Z(Idx2)>VID_Z-del_val & Z(Idx2)<VID_Z+del_val);
IdxFinal=Idx3;
end
end
len = length(IdxFinal);
index=index+len;
if len>0
reqd_op(1,index:index+len-1) = IdxFinal;
end
end
toc
I'm still confused why am not able to know the results of this small algorithm of my array. the array has almost 1000 number 1-D. am trying to find the peak and the index of each peak. I did found the peaks, but I can't find the index of them. Could you please help me out. I want to plot all my values regardless the indexes.
%clear all
%close all
%clc
%// not generally appreciated
%-----------------------------------
%message1.txt.
%-----------------------------------
% t=linspace(0,tmax,length(x)); %get all numbers
% t1_n=0:0.05:tmax;
x=load('ww.txt');
tmax= length(x) ;
tt= 0:tmax -1;
x4 = x(1:5:end);
t1_n = 1:5:tt;
x1_n_ref=0;
k=0;
for i=1:length(x4)
if x4(i)>170
if x1_n_ref-x4(i)<0
x1_n_ref=x4(i);
alpha=1;
elseif alpha==1 && x1_n_ref-x4(i)>0
k=k+1;
peak(k)=x1_n_ref; // This is my peak value. but I also want to know the index of it. which will represent the time.
%peak_time(k) = t1_n(i); // this is my issue.
alpha=2;
end
else
x1_n_ref=0;
end
end
%----------------------
figure(1)
% plot(t,x,'k','linewidth',2)
hold on
% subplot(2,1,1)
grid
plot( x4,'b'); % ,tt,x,'k'
legend('down-sampling by 5');
Here is you error:
tmax= length(x) ;
tt= 0:tmax -1;
x4 = x(1:5:end);
t1_n = 1:5:tt; % <---
tt is an array containing numbers 0 through tmax-1. Defining t1_n as t1_n = 1:5:tt will not create an array, but an empty matrix. Why? Expression t1_n = 1:5:tt will use only the first value of array tt, hence reduce to t1_n = 1:5:tt = 1:5:0 = <empty matrix>. Naturally, when you later on try to access t1_n as if it were an array (peak_time(k) = t1_n(i)), you'll get an error.
You probably want to exchange t1_n = 1:5:tt with
t1_n = 1:5:tmax;
You need to index the tt array correctly.
you can use
t1_n = tt(1:5:end); % note that this will give a zero based index, rather than a 1 based index, due to t1_n starting at 0. you can use t1_n = 1:tmax if you want 1 based (matlab style)
you can also cut down the code a little, there are some variables that dont seem to be used, or may not be necessary -- including the t1_n variable:
x=load('ww.txt');
tmax= length(x);
x4 = x(1:5:end);
xmin = 170
% now change the code
maxnopeaks = round(tmax/2);
peaks(maxnopeaks)=0; % preallocate the peaks for speed
index(maxnopeaks)=0; % preallocate index for speed
i = 0;
for n = 2 : tmax-1
if x(n) > xmin
if x(n) >= x(n-1) & x(n) >= x(n+1)
i = i+1;
peaks(i) = t(n);
index(i) = n;
end
end
end
% now trim the excess values (if any)
peaks = peaks(1:i);
index = index(1:i);
Is it possible in matlab/octave to use the sort function to sort an array based on the relative frequency of their elements?
For example the array
m= [4,4,4,10,10,10,4,4,5]
should result in this array:
[5,10,10,10,4,4,4,4,4]
5 is the less frequent element and is on the top while 4 is the most frequent and it's on bottom.
Should one use the indices provided by histcount?
The following code first calculates how often each element occurs and then uses runLengthDecode to expand the unique elements.
m = [4,4,4,10,10,10,4,4,5];
u_m = unique(m);
elem_count = histc(m,u_m);
[elem_count, idx] = sort(elem_count);
m_sorted = runLengthDecode(elem_count, u_m(idx));
The definition of runLengthDecode is copied from this answer:
For MATLAB R2015a+:
function V = runLengthDecode(runLengths, values)
if nargin<2
values = 1:numel(runLengths);
end
V = repelem(values, runLengths);
end
For versions before R2015a:
function V = runLengthDecode(runLengths, values)
%// Actual computation using column vectors
V = cumsum(accumarray(cumsum([1; runLengths(:)]), 1));
V = V(1:end-1);
%// In case of second argument
if nargin>1
V = reshape(values(V),[],1);
end
%// If original was a row vector, transpose
if size(runLengths,2)>1
V = V.'; %'
end
end
One way would be to use accumarray to find the count of each number (I suspect you can use histcounts(m,max(m))) but then you have to clear all the 0s).
m = [4,4,4,10,10,10,4,4,5];
[~,~,subs]=unique(m);
freq = accumarray(subs,subs,[],#numel);
[~,i2] = sort(freq(subs),'descend');
m(i2)
By combinging my approach with that of m.s. you can get a simpler solution:
m = [4,4,4,10,10,10,4,4,5];
[U,~,i1]=unique(m);
freq= histc(m,U);
[~,i2] = sort(freq(i1),'descend');
m(i2)
You could count the number of repetitions with bsxfun, sort that, and apply that sorting to m:
[~, ind] = sort(sum(bsxfun(#eq,m,m.')));
result = m(ind);
A similar question has been asked, but still I'm looking for a solution.
In MATLAB, I have an array of states s:
s = {'Indiana', 'Texas', 'Alabama'}
Time is a column vector: [120 30 20 40 50]'
Tornadoes is a column vector: [5 5 3 5 5]'
And I need to for loop through this array s for the following code below while placing each string in s in the first line.
index = strcmpi(States,s)
Time = Time(index)
Tornadoes = Tornadoes(index)
h = scatter(Time,Tornadoes)
So how can I write the code to push each state in s to generate a plot for each plot.
Could it be as simple as this?
for ii = 1:numel(s)
index = strcmpi(States, s{ii})
Time = Time(index)
Tornadoes = Tornadoes(index)
figure % make sure you start a new figure each time...
h = scatter(Time,Tornadoes)
title(['Tornadoes in ' s{ii}])
end
If you are wanting to loop through each entry in s, you could do
j = length(s)
for i = 1:j
x = Time(i)
y = Tornadoes(i)
h = scatter(x, y)
end