How to have a vector that is obtained by discretizing a and b for each i.

For example, when i=1, a(1)=1, b(1)=24; c=1 0.5 2 2.5 up to 24. Then we move on to i=2

Answer 1

[Chuguang Pan]

you can use arrayfun function to process array element.

a=[1 2 3 4 5 6 7 8 9 10 11 12];
b=[24 23 22 21 20 19 18 17 16 15 14 13];

stepHalfFun = @(start,stop) start:.5:stop;
res = arrayfun(stepHalfFun,a,b,'UniformOutput',false)

res = 1×12 cell array
  Columns 1 through 10

    {1×47 double}    {1×43 double}    {1×39 double}    {1×35 double}    {1×31 double}    {1×27 double}    {1×23 double}    {1×19 double}    {1×15 double}    {1×11 double}

  Columns 11 through 12

    {1×7 double}    {[12 12.5000 13]}

[albert Kinda]

THANKS

Another question how to transform it in matrix with 12 line ?.

cell 1*1 (1*47) is first line for matrix

cell 1*2(1*43) is second line

up to cell 1*12({[12 12.5000 13]}) is 12th line

[VBBV]

a=[1 2 3 4 5 6 7 8 9 10 11 12];
b=[24 23 22 21 20 19 18 17 16 15 14 13];

stepHalfFun = @(start,stop) start:.5:stop;
res = arrayfun(stepHalfFun,a,b,'UniformOutput',false).'  % transpose

res = 12×1 cell array
    {[      1 1.5000 2 2.5000 3 3.5000 4 4.5000 5 5.5000 6 6.5000 7 7.5000 8 8.5000 9 9.5000 10 10.5000 11 11.5000 12 12.5000 13 13.5000 14 14.5000 15 15.5000 16 … ] (1×47 double)}
    {[    2 2.5000 3 3.5000 4 4.5000 5 5.5000 6 6.5000 7 7.5000 8 8.5000 9 9.5000 10 10.5000 11 11.5000 12 12.5000 13 13.5000 14 14.5000 15 15.5000 16 16.5000 17 … ] (1×43 double)}
    {[  3 3.5000 4 4.5000 5 5.5000 6 6.5000 7 7.5000 8 8.5000 9 9.5000 10 10.5000 11 11.5000 12 12.5000 13 13.5000 14 14.5000 15 15.5000 16 16.5000 17 17.5000 18 … ] (1×39 double)}
    {[4 4.5000 5 5.5000 6 6.5000 7 7.5000 8 8.5000 9 9.5000 10 10.5000 11 11.5000 12 12.5000 13 13.5000 14 14.5000 15 15.5000 16 16.5000 17 17.5000 18 18.5000 19 … ] (1×35 double)}
    {[               5 5.5000 6 6.5000 7 7.5000 8 8.5000 9 9.5000 10 10.5000 11 11.5000 12 12.5000 13 13.5000 14 14.5000 15 15.5000 16 16.5000 17 17.5000 18 18.5000 19 19.5000 20]}
    {[                                   6 6.5000 7 7.5000 8 8.5000 9 9.5000 10 10.5000 11 11.5000 12 12.5000 13 13.5000 14 14.5000 15 15.5000 16 16.5000 17 17.5000 18 18.5000 19]}
    {[                                                       7 7.5000 8 8.5000 9 9.5000 10 10.5000 11 11.5000 12 12.5000 13 13.5000 14 14.5000 15 15.5000 16 16.5000 17 17.5000 18]}
    {[                                                                           8 8.5000 9 9.5000 10 10.5000 11 11.5000 12 12.5000 13 13.5000 14 14.5000 15 15.5000 16 16.5000 17]}
    {[                                                                                               9 9.5000 10 10.5000 11 11.5000 12 12.5000 13 13.5000 14 14.5000 15 15.5000 16]}
    {[                                                                                                                   10 10.5000 11 11.5000 12 12.5000 13 13.5000 14 14.5000 15]}
    {[                                                                                                                                         11 11.5000 12 12.5000 13 13.5000 14]}
    {[                                                                                                                                                               12 12.5000 13]}

[Steven Lord]

Another question how to transform it in matrix with 12 line ?.

In MATLAB, arrays in MATLAB cannot be "jagged" -- all the rows have to have the same number of columns. [There is an exception for arrays of Java objects under certain circumstances, I think.] So you can't have a matrix where one row has 2 elements and one has 1 element. You could pad it with NaN values or find some other padding value.

A = [1 2; 3 NaN] % works

A = 2×2

1 2 3 NaN

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Note that without the NaN, this wouldn't work. I left this commented out so I could run the rest of the code in this answer.

% B = [1 2; 3] % would not work

One way to pad the data would be with paddata. Let's take some sample data.

C = {[1 2], 3, 4:6}

C = 1×3 cell array
    {[1 2]}    {[3]}    {[4 5 6]}

Determine the maximum length of the data stored in cells in C.

lengthOfVectorsInCell = cellfun(@numel, C)

lengthOfVectorsInCell = 1×3

2 1 3

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maxLength = max(lengthOfVectorsInCell)

maxLength = 3

Create the padding function that will pad each cell to that maximum length, filling in the newly added elements with NaN.

paddingFunction = @(x) paddata(x, maxLength, FillValue = NaN);

Apply the padding function to the cell array.

paddedC = cellfun(paddingFunction, C, UniformOutput = false)

paddedC = 1×3 cell array
    {[1 2 NaN]}    {[3 NaN NaN]}    {[4 5 6]}

Now that the vectors in the cell array are the same length, they can be concatenated.

D = vertcat(paddedC{:})

D = 3×3

1 2 NaN 3 NaN NaN 4 5 6

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