error using conv

I get error when using conv function

dataset=[10 89 87 56 28 96 58 ;1 5 70 59 30 18 25 ;14 20 89 87 100 54 20]

i want to find mid point for example

A=[3 5 6 15 17 21 35 45 46 51 56 57 66 70 71]'

i get answer as

4.0000
    5.5000
   10.5000
   16.0000
   19.0000
   28.0000
   40.0000
   45.5000
   48.5000
   53.5000
   56.5000
   61.5000
   68.0000
   70.5000

using code F_mid = conv(A, [0.5 0.5], 'valid')

when i use this code for my dataset

i get error

Error using ==> conv at 27
A and B must be vectors.

Error in ==> new at 3
F_mid = conv(F, [0.5 0.5], 'valid')

please help

Answer 1

[Walter Roberson]

The code you show that you are using is

F_mid = conv(A, [0.5 0.5], 'valid')

but your error message is for

F_mid = conv(F, [0.5 0.5], 'valid')

which does not occur anywhere in the code you show.

We do not have any information about the shape of F.

[kash]

Error using ==> conv at 27

A and B must be vectors.

Error in ==> new at 3

F_mid = conv(F, [0.5 0.5], 'valid')

[kash]

F is mt dataset

dataset=[10 89 87 56 28 96 58 ;1 5 70 59 30 18 25 ;14 20 89 87 100 54 20]

[Walter Roberson]

And is your "dataset" variable a vector like the error message says is required?

conv() is for 1D convolution. If you want 2D convolution, then use a 2D convolution function such as conv2(), probably with [0.5 0.5].' as your kernel. Or if you want each column to be done independently, then make one conv() call per column.

arrayfun(@(IDX) conv(F(:,IDX),[0.5 0.5]), 1:size(F,2))

[kash]

thanks walter

in ur code

reshape(F, 2, []).'

F = 3 5 6 15 17 21 35 45 46 51 56 57 66 70 71

the output is is

3 5

6 15

17 21

but i need as

3 5

5 6

6 15

i need same for dataset also

[Walter Roberson]

That's a different question and Andrew provided a solution for that.

Note however that your F is *not* a vector or else you would not be having problems passing it to conv(). You have not defined how you want to handle splitting matrices.

I do not see any connection between the two questions, other than that they use the same variable name "F" (for two different purposes.) Unless, that is, you are trying to find a way other than using conv() to find the midpoints.

[kash]

error in using Error using ==> arrayfun

arrayfun(@(IDX) conv(F(:,IDX),[0.5 0.5]), 1:size(F,2))

Non-scalar in Uniform output, at index 1, output 1.

Set 'UniformOutput' to false.

Error in ==> new at 7

arrayfun(@(IDX) conv(F(:,IDX),[0.5 0.5]), 1:size(F,2))

[Walter Roberson]

arrayfun(@(IDX) conv(F(:,IDX),[0.5 0.5],'valid'), 1:size(F,2))

[kash]

walter i get same error for my dataset

[Walter Roberson]

Whatever. Use a "for" loop then. Or use conv2()