Store data in cell arrays
43 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Hello Everyone,
I know the question below is quite stupid but please help as I am not able o find a solution for it. This is a sample array that I have. I want to check the second column of this. The places where I get 1 should be stored together in cell array. As I start scanning the first 4 values get stored in cell array 1 then as the value in second column is 0 I do not store it . Again when I get 1 I should store but in cell array 2. The number of values that I will get is not fixed.
26.81267 1
26.82500 1
26.83733 1
26.86200 1
37.41933 0
37.40700 0
37.39466 0
37.38233 0
27.08400 1
27.07167 1
27.05933 1
27.04700 1
27.02233 1
0 comentarios
Respuestas (4)
the cyclist
el 17 de Jul. de 2013
Editada: the cyclist
el 17 de Jul. de 2013
If A is your original array, then
C{1} = A(A(:,2)==1,:);
is a one-element cell array that stores the 9 rows with a zero.
If you only wanted to store the first column, then
C{1} = A(A(:,2)==1,1);
instead.
Andrei Bobrov
el 17 de Jul. de 2013
Editada: Andrei Bobrov
el 18 de Jul. de 2013
a =[26.81267 1
26.82500 1
26.83733 1
26.86200 1
37.41933 0
37.40700 0
37.39466 0
37.38233 0
27.08400 1
27.07167 1
27.05933 1
27.04700 1
27.02233 1];
l = [true;diff(a(:,2))~=0];
l(l) = a(l,2);
ll = cumsum(l);
out = accumarray(ll(a(:,2)>0),a(a(:,2)>0,1),[],@(x){x});
OR
l = [true;diff(a(:,2))~=0];
d = diff([find(l);size(a,1)+1]);
out = mat2cell(a(a(:,2)>0,1),d(a(l,2)>0),1);
0 comentarios
Azzi Abdelmalek
el 17 de Jul. de 2013
Editada: Azzi Abdelmalek
el 17 de Jul. de 2013
i2=[1 1 0 0 1 1 1 1 0 0 1 1 0 0 0 1 1 0 1];
a=[i2(1) diff(i2)];
idx1=find(a==1);
idx2=find(a==-1);
if numel(idx2)<numel(idx1)
idx2(end+1)=numel(a)+1;
end
out=arrayfun(@(x1,x2) i2(x1:x2),idx1,idx2-1,'un',0)
5 comentarios
Azzi Abdelmalek
el 17 de Jul. de 2013
It's true, I 've used arrayfun several times, even when it's not the more efficient solution (because I'm not yet good with accumarray), for 'Uniforme output', it's dictated by each case,"I am not guilty!"
Swapnali Gujar
el 17 de Jul. de 2013
You can try below code. This code will give you output as a cell whose first 2 cells will have names as "cell_array1" and "cell_array2", since there are 2 times when 1s appear in your input array in continuation (i.e. first 4 times 1s and next 5 times 1s). Below each column name, you will find respective 1s of each scan.
clc
clear all;
arr = [26.81267 1;
26.82500 1;
26.83733 1;
26.86200 1;
37.41933 0;
37.40700 0;
37.39466 0;
37.38233 0;
27.08400 1;
27.07167 1;
27.05933 1;
27.04700 1;
27.02233 1;
]
size_arr = size(arr);
j=0;
scan = 0;
for i=1:length(arr)
if((arr(i,2)) == 1.000)
j=j+1;
scan=j;
if (i<length(arr)) && (arr((i+1),2) == 1.000)
j = j-1;
end
cell_arr{scan} = strcat('cell_array',num2str(scan)); %creates numbered cell arrays
end
end
%===== Now start storing the values into the cell arrays.
index=1;
columns = length(cell_arr)
k=1;
for i =1:length(arr)
if(((arr(i,2)) == 1.000)) && (k<=columns)
cell_arr(index+1,k) = num2cell(arr(i,2));
index = index+1;
elseif (arr(i,2) == 0.000)&&((arr(i+1,2)) == 1.000)
k = k+1;
index=1;
end
end
%end of file
This will give you output as cell_arr =
'cell_array1' 'cell_array2'
[ 1] [ 1]
[ 1] [ 1]
[ 1] [ 1]
[ 1] [ 1]
[] [ 1]
You may need further processing to access these columns separately as per your need.
0 comentarios
Ver también
Categorías
Más información sobre Matrix Indexing en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!