# 'Logical' function to read 2 variables and output to a new variable

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10B 2015 年 9 月 25 日

Hello Community,
I am having a novice moment here - difficulty in organising my function to do what I want. In short, I am trying to:
1. Read the data in 'a', AND
2. Read the data in 'b'
3. When they agree with the terms (a basic logical test)
4. A numerical count is recorded in the appropriate row/column of 'Imloc'
Example, if a(:,1) & b(:,1) == '1', then in IMloc(1,1) a count is recorded (or nothing if not). If a(:,2) & b(:,1) == '1' a count would be recorded at Imloc(1,2) etc.
Here is what code I have:
function [ Imloc ] = IMLocation( a, b )
% Preallocate Imloc matrix with zeros
Imloc = zeros(1,11);
% Set counter
numout = 0;
% Create storage for output vector
outvec = [];
% Run for the length of
for k=1:length(a,b)
if a(k)==1 && b(k)==1
numout = (numout + 1), 'Imloc',(1:1))
Imloc(outvec(numout))
elseif a(k)==2 && b(k)==1
numout = (numout + 1) 'Imloc',1:2))
Imloc(outvec(numout))
elseif a(k)==3 && b(k)==1
numout = (numout + 1) 'Imloc',1:3))
Imloc(outvec(numout))
% etc. etc. etc.
elseif a(k)==11 && b(k)==1
numout = (numout + 1) 'Imloc',1:11))
Imloc(outvec(numout))
else
end
end
So, can anyone make a suggestion as to what is going wrong as I am not doing very well with this!
Regards,
10B.
##### 3 件のコメント表示非表示 2 件の古いコメント
10B 2015 年 9 月 29 日
Sorry to send a reminder dpb - but did my comment answer your question for this?
Kind regards,
10B.

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### 採用された回答

dpb 2015 年 9 月 29 日
Sorry, I didn't see the previous response. Your dataset isn't complete, but presuming it is, then
d=a(b==1); % only the 1's count...
u=unique(d); % the unique values that are in a
n=hist(d,u); % count for each possible bin
Your header row or whatever is simply the vector u with the 'm' appended if want (altho you'll then have to use a cell array to hold both datatypes together, of course)...
>> cellstr(num2str(u,'%dm')).'
ans =
'1m' '2m' '5m' '6m'
>>
for the above (partial) dataset...
##### 2 件のコメント表示非表示 1 件の古いコメント
dpb 2015 年 9 月 29 日
You're welcome, of course. Was why was checking for sure on what the actual objective was, first...I thought this was likely the question initially, but wasn't positive so figured may as well solve the question asked...
Of course, if the data are complete in the real case, accumarray works wonders here...
>> accumarray(a,b).'
ans =
2 3 0 0 1 1
>>
See the doc for details. This relies on the b vector being [0,1] so the sum is therefore identical to the count of nonzero elements.
As you see, this gives entries in the final array for each element at the index position where as the unique|hist solution only returns those actually found.

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