Hello, I wanted to ask about how you might concatenate two vectors or matrices under conditions , ie comparing the elements. For example I have a matrix such that:
a =
0
1.0000
2.0000
2.4000
2.5000
3.0000
3.5000
4.8000
5.0000
5.5000
and another:
b =
0 1.0000
2.4000 2.0000
4.8000 3.0000
And I want to have such a result that :
result =
0 1.0000
1.0000 1.0000
2.0000 1.0000
2.4000 2.0000
2.5000 2.0000
3.0000 2.0000
3.5000 2.0000
4.8000 3.0000
5.0000 3.0000
5.5000 3.0000
That is, comparing with the first column of b assign the value 1, 2,3 .. depending on whether the value is greater or is between two numeros.Por example 2.4 is greater and equal to the second element b but less the third and so .... Thanks in advanced.

1 Comment

Hello, first thanks . What I 'm looking for is that when comparing the elements of the two vectors , for example when a (2) = 1.0000 is between the interval b(1) = 0 and b (2) = 2.4 in that row add a column value of 1 as it is between that range. That is, if the condition is met , for example a(2) is greater than or equal b(1) and less than b(2) corresponds to the first interval 1 comparing all elements of the row of the two matrices fulfilled those conditions

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 Accepted Answer

Andrei Bobrov
Andrei Bobrov on 21 May 2016
Edited: Andrei Bobrov on 21 May 2016
a =[ 0
1.0000
2.0000
2.4000
2.5000
3.0000
3.5000
4.8000
5.0000
5.5000];
b =[ 0 1.0000
2.4000 2.0000
4.8000 3.0000];
out = [a, b(cumsum(ismember(a,b(:,1))),2)]
or
[~,ii] = histc(a,[b(:,1);inf]);
out = [a, b(ii,2)]

8 Comments

Hello, first thanks . What I 'm looking for is that when comparing the elements of the two vectors , for example when a (2) = 1.0000 is between the interval b(1) = 0 and b (2) = 2.4 in that row add a column value of 1 as it is between that range. That is, if the condition is met , for example a(2) is greater than or equal b(1) and less than b(2) corresponds to the first interval 1 comparing all elements of the row of the two matrices fulfilled those conditions
Andrei Bobrov
Andrei Bobrov on 21 May 2016
Edited: Andrei Bobrov on 21 May 2016
see after word 'or'
Hello again, I do not know why when I will run the process with larger matrices (length (b) = 25 and length (a = 684) ) makes me : out = [a, b ( cumSum ( IsMember (a, b ( : , 1) ) ) , 2) ] Index Exceeds matrix dimensions .
Why?
I do not know, I do not see your data
my data would be the nmat matrix (Interested in column 6 are the data shown in column 1 of the Bach matrix.) to which I want to add the column compared with the Bach matrix are as follows :
nmat =
0 0.9979 3.0000 62.0000 75.0000 0 0.5987
0.5000 0.2479 1.0000 78.0000 75.0000 0.3000 0.1487
0.5000 0.2479 2.0000 78.0000 75.0000 0.3000 0.1487
0.7500 0.2479 1.0000 79.0000 75.0000 0.4500 0.1487
0.7500 0.2479 2.0000 79.0000 75.0000 0.4500 0.1487
1.0000 0.4979 1.0000 81.0000 75.0000 0.6000 0.2988
1.0000 0.4979 2.0000 81.0000 75.0000 0.6000 0.2988
1.5000 0.4979 1.0000 72.0000 75.0000 0.9000 0.2988
1.5000 0.4979 2.0000 72.0000 75.0000 0.9000 0.2988
1.5000 0.4979 3.0000 54.0000 75.0000 0.9000 0.2988
2.0000 0.4979 1.0000 71.0000 75.0000 1.2000 0.2988
2.0000 0.4979 2.0000 71.0000 75.0000 1.2000 0.2988
2.0000 0.9979 3.0000 55.0000 75.0000 1.2000 0.5987
2.5000 0.2479 1.0000 76.0000 75.0000 1.5000 0.1487
2.5000 0.2479 2.0000 76.0000 75.0000 1.5000 0.1487
2.7500 0.2479 1.0000 78.0000 75.0000 1.6500 0.1487
2.7500 0.2479 2.0000 78.0000 75.0000 1.6500 0.1487
3.0000 0.4979 1.0000 79.0000 75.0000 1.8000 0.2988
3.0000 0.4979 2.0000 79.0000 75.0000 1.8000 0.2988
3.5000 0.4979 1.0000 71.0000 75.0000 2.1000 0.2988
3.5000 0.4979 2.0000 71.0000 75.0000 2.1000 0.2988
3.5000 0.4979 3.0000 52.0000 75.0000 2.1000 0.2988
4.0000 0.4979 1.0000 69.0000 75.0000 2.4000 0.2988
4.0000 0.9979 3.0000 54.0000 75.0000 2.4000 0.5988
4.5000 0.2479 1.0000 74.0000 75.0000 2.7000 0.1488
4.7500 0.2479 1.0000 76.0000 75.0000 2.8500 0.1488
5.0000 0.4979 1.0000 78.0000 75.0000 3.0000 0.2988
5.5000 0.4979 1.0000 69.0000 75.0000 3.3000 0.2988
5.5000 0.4979 3.0000 50.0000 75.0000 3.3000 0.2988
6.0000 0.2479 1.0000 67.0000 75.0000 3.6000 0.1488
6.0000 0.4979 3.0000 59.0000 75.0000 3.6000 0.2988
6.2500 0.2479 1.0000 78.0000 75.0000 3.7500 0.1488
6.5000 0.2479 1.0000 76.0000 75.0000 3.9000 0.1488
6.5000 0.4979 3.0000 55.0000 75.0000 3.9000 0.2987
6.7500 0.2479 1.0000 74.0000 75.0000 4.0500 0.1487
7.0000 0.2479 1.0000 73.0000 75.0000 4.2000 0.1487
7.0000 0.4979 3.0000 57.0000 75.0000 4.2000 0.2987
7.2500 0.2479 1.0000 83.0000 75.0000 4.3500 0.1487
7.5000 0.2479 1.0000 81.0000 75.0000 4.5000 0.1487
7.5000 0.4979 3.0000 45.0000 75.0000 4.5000 0.2987
7.7500 0.2479 1.0000 79.0000 75.0000 4.6500 0.1487
8.0000 0.2479 1.0000 78.0000 62.0000 4.8000 0.1487
8.0000 0.4979 3.0000 50.0000 75.0000 4.8000 0.2987
8.2500 0.2479 1.0000 76.0000 62.0000 4.9500 0.1487
8.5000 0.2479 1.0000 74.0000 62.0000 5.1000 0.1487
8.5000 0.4979 3.0000 62.0000 75.0000 5.1000 0.2987
8.7500 0.2479 1.0000 76.0000 62.0000 5.2500 0.1487
9.0000 0.2479 1.0000 78.0000 62.0000 5.4000 0.1487
9.0000 0.4979 3.0000 61.0000 75.0000 5.4000 0.2987
9.2500 0.2479 1.0000 76.0000 62.0000 5.5500 0.1487
9.5000 0.2479 1.0000 78.0000 62.0000 5.7000 0.1487
9.5000 0.4979 3.0000 59.0000 75.0000 5.7000 0.2987
9.7500 0.2479 1.0000 80.0000 62.0000 5.8500 0.1487
10.0000 0.2479 3.0000 57.0000 75.0000 6.0000 0.1487
10.0000 0.9979 1.0000 81.0000 62.0000 6.0000 0.5987
10.2500 0.2479 3.0000 56.0000 75.0000 6.1500 0.1487
10.5000 0.2479 2.0000 73.0000 75.0000 6.3000 0.1487
10.5000 0.2479 3.0000 57.0000 75.0000 6.3000 0.1487
10.7500 0.2479 2.0000 74.0000 75.0000 6.4500 0.1487
10.7500 0.2479 3.0000 59.0000 75.0000 6.4500 0.1487
11.0000 0.2479 3.0000 61.0000 75.0000 6.6000 0.1487
11.0000 0.4979 2.0000 76.0000 75.0000 6.6000 0.2987
11.0000 1.9979 1.0000 76.0000 62.0000 6.6000 1.1987
11.2500 0.2479 3.0000 62.0000 75.0000 6.7500 0.1487
11.5000 0.2479 3.0000 61.0000 75.0000 6.9000 0.1487
11.5000 0.4979 2.0000 67.0000 75.0000 6.9000 0.2987
y Bach:
Bach =
0 1.0000
2.4000 2.0000
4.8000 3.0000
Using the second method :
[ ~ , ii ] = histc ( nmat [ Bach ( : , 1) ; inf ] ) ;
out = [ nmat , Bach (ii, 2)]
Error using horzcat Dimensions of arrays being concatenated are not consistent .
[~,ii] = histc(nmat(:,6), [Bach(:,1);inf]);
out = [nmat, Bach(ii,2)];
or (R2015a and later)
out = [nmat, discretize(nmat(:,6),[Bach(:,1);inf],Bach(:,2))];
And if I work, many thanks! and apology for the inconvenience

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