compare value between different arrays

2013 6 23
2 个回答
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Hi all. I have 2 array that have value. I want to compare each value of each array, but i dont know how to do that. Can somebody help me ? Thanks in advance.

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Azzi Abdelmalek
Azzi Abdelmalek 2013-6-23
If you have
a=[2 3 4]
b=[1 4 0]
What should be the result of comparison?
Lidank Abiel
Lidank Abiel 2013-6-23
a=[39.4029 46.6748 45.3174 26.4645 21.6932];
b=[36.8585 47.6897 46.7784 26.9878 22.2181];
i want to compare each value of array b to array a, that the result of b approach to a. Value of array a is used here as reference.
Iain
Iain 2013-6-24
By "compare", what do you mean? Do you mean:
1. Check that each element of b is greater than, or equal to, or less than the corresponding element of a?
2. Check that each element of b is greater than, or equal to, or less than ANY element of a?
3. Check that each element of b is within a small error (which you haven't defined) of a (corresponding, or an element)
4. Check that "b" is becoming more like "a"
5. Somethign else?
Iain
Iain 2013-6-24
result = b > a; %(b greater than a)
result = b >= a; %(b greater or equal to a)
result = b == a; %(b equal to a)
result = b <= a; %(b less than or equal to a)
result = b < a; %(b less than a)
result is the same length as b, and is 1 (true) when b is that comparison to a, and 0 (false) everywhere else.
all(result) will be 1 (true) if all the elements of b are "that comparison"
any(result) will be 1 (true) if any of the elements of b are "that comparison"
result = b > (5*a); %(b greater than five times a)
Lidank Abiel
Lidank Abiel 2013-6-25
yes, check each element of b less than or equal to a. ( I mean, check b[1] to a[1], b[2] to a[2], etc. If the result are as expected, it will be 1 (true).

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回答（2 个）

Azzi Abdelmalek
Azzi Abdelmalek 2013-6-23
编辑：Azzi Abdelmalek 2013-6-23

0 个投票

a=[2 3 4]
b=[1 4 0]
comp=a<b

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Lidank Abiel
Lidank Abiel 2013-6-23
a=[39.4029 46.6748 45.3174 26.4645 21.6932];
b=[36.8585 47.6897 46.7784 26.9878 22.2181];
i want to compare each value of array b to array a, that the result of b approach to a. Value of array a is used here as reference.
Azzi Abdelmalek
Azzi Abdelmalek 2013-6-23
Ok, but what should be the result in this case?
Lidank Abiel
Lidank Abiel 2013-6-23
I had the training data, which is a. later, I enter test data, namely b. The b value would I compare it with a. if the value of b close to the value of a, then the data is valid. that's what I want to do. is there a way to make it?
Tom
Tom 2013-6-23
编辑：Tom 2013-6-23
How do you define close?
a=[39.4029 46.6748 45.3174 26.4645 21.6932];
b=[36.8585 47.6897 46.7784 26.9878 22.2181];
tol = 1;
%check that the elements of a and b are within a given tolerance
abs(a-b) <= tol
Lidank Abiel
Lidank Abiel 2013-6-23
ya, like that, but each value from an array sir, not overall ( i mean not (a-b), but maybe like (a1-b1). is it possible ?.
Azzi Abdelmalek
Azzi Abdelmalek 2013-6-23
编辑：Azzi Abdelmalek 2013-6-23
Do you want to compare the first value of a with the first of b, and so on? To be more clear, please post what should be the result for your example!
Lidank Abiel
Lidank Abiel 2013-6-23
编辑：Lidank Abiel 2013-6-23
function deteksi_Callback(hObject, eventdata, handles)
% hObject handle to deteksi (see GCBO)
% eventdata reserved - to be defined in a future version of MATLAB
% handles structure with handles and user data (see GUIDATA)
global capture
cla(handles.axes3);
cla(handles.axes7);
set(handles.status, 'String', '------');
set(handles.nominal, 'String', '------');
set(handles.euc_uji, 'String', '------');
set(handles.euc_sampel, 'String', '------');
toleran=0;
if(isempty(get(handles.toleransi,'String')))
toleran=93;
else
toleran=eval(get(handles.toleransi,'String'));
end
load dataeuc;
[x y]=size(nilaieuc);
jum=0;
citragray=rgb2gray(capture);
citrafilter=medfilt2(citragray,[3 3]);
citraedge=edge(citrafilter,'canny',[0.01 0.11]);
jlh=20;
citraedgethin=bwmorph(citraedge,'thin',jlh);
citraakhir=imcomplement(citraedgethin);
[ximg2 yimg2]=size(citraakhir);
jum=0;
for i=1:ximg2
for j=1:yimg2
if(citraakhir(i,j)==0)
jum=jum+1;
pos(jum,1)=i;
pos(jum,2)=j;
end
end
end
[ximg3 yimg3]=size(pos);
euclid1=0;
for i=1:ximg3-1
euc(i)= sqrt(((pos(i,1)-pos(i+1,1))^2)+((pos(i,2)-pos(i+1,2))^2));
euclid1=euclid1+euc(i);
end
axes(handles.axes3);
imshow(citraakhir);
euclid1
set(handles.euc_uji,'String',euclid1); %set nilai euclidean gambar di static text
bandingkan=0;
for counti=2:x
counti
euclid2=nilaieuc(counti)
besar=0;
kecil=0;
temp_bandingkan=0;
if(euclid1>=euclid2)
besar=euclid1;
kecil=euclid2;
else
besar=euclid2;
kecil=euclid1;
end
temp_bandingkan=(kecil/besar)*100
if(temp_bandingkan>toleran)
%set(handles.euc_uji,'String',toleran);
if(temp_bandingkan>bandingkan)
bandingkan=temp_bandingkan;
posisikoin=counti;
end
end
end
This is it sir. I want to make like this.
Azzi Abdelmalek
Azzi Abdelmalek 2013-6-23
I've asked a simple question: what should be the result for the short example you gave.
Lidank Abiel
Lidank Abiel 2013-6-23
to save the largest value and it is used as a reference in the percentage of proximity or similarity.
the 'toleran = 93' seems like minimal percentage tolerance. and the 'temp_bandingkan=(kecil/besar)*100' is to know the percentage.
Jan
Jan 2013-6-23
The code does not really answer the question, so I ask again: What is the desired result for the values:
a = [39.4029 46.6748 45.3174 26.4645 21.6932];
b = [36.8585 47.6897 46.7784 26.9878 22.2181];
Lidank Abiel
Lidank Abiel 2013-6-24
I want that values used to percentage of proximity.
My plan, i set value for tolerance, example, tolerance = 93. Then, i process image to get the euclidean value. The result above, is value from 2 image. and then, i want to calculate 2 value above ( maybe divided ) if the result approach to value tolerance, so, the image have percentage minimal similarity 90%.
I'm sorry make all of you confused. I'm new in programming. So, I explained it according to my knowledge.
Thanks in advance.
Jan
Jan 2013-6-24
Thanks for the explanantions. Unfortuantely I do not understand this:
i want to calculate 2 value above ( maybe divided ) if the result approach to value tolerance, so, the image have percentage minimal similarity 90%.
I still do not know hwta kind of result you expect for the above mentioned input data. Are you able to calculate it manually or give a definition of the calculations? In the comments to the question you wrote "i want to compare each value of array b to array a". As soon as it is explained, what "campare" means explicitly, I assume the problem can be solved in seconds. So please try to explain this clearly.

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Thorsten
Thorsten 2013-6-24

0 个投票

You can compute two fractions a/b and b/a, and then take the minimum to ensure that the value is not above 100%. Then take the minimum of these values to get "percentage minimal similarity"
min_similarity = min(min([a./b; b./a]))

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