# sorting a matrix of 4 rows by the fourth element and outputting of combinations that make the 4th row element

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Leo Quintanar 2021년 7월 22일
댓글: Leo Quintanar 2021년 7월 22일
My project so far has been to create a vector of numbers:
%x: a vector of numbers
x = 1:4;
and then create a 3 column matrix of the all the possible combinations of the vector x
%n: number of elements to pick (with repetetition) from x. n must be less than or equal to numel(x)
n = 3;
assert(n < numel(x), 'n can''t be greater than numel(x)')
combs = cell(1, n);
[combs{:}] = ndgrid(x);
combs = reshape(cat(n+1, combs{:}), [], n); %a numel(x)^n X n matrix of combinations
and add a 4th column to the matrix which is the energy of the the 3 columns which is equal
length = length(combs);
for x=1:length
combs(x,4) = combs(x, 1)^2+combs(x, 2)^2+combs(x, 3)^2;
E(x,:) = combs(x,4);
end
This produces a matrix similiar to the picture below:
This code works great but I want to re oragnize the rows according to the energy (4th) column to look like this:
So the rows will start with 4th column is 3 and then the next three are 6, and so fourth. I'm not sure how to do this, and would appreciate any help. Just a FYI I plan to change the vector from 4 to 14, this simulation is for Griffth's Quantum mechanics problem 4.2.B, I already know the answer but I figured it would be a fun to see the matrix blow up as well as practice coding. I remember doing something similiar in Java but its been a long time ago.
Thank you,
-Leo Q

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### 채택된 답변

KSSV 2021년 7월 22일
NOTE: Never use a function name as the variable. You have used length as variable, this is problematic.
clc; clear all ;
%x: a vector of numbers
x = 1:4;
%n: number of elements to pick (with repetetition) from x. n must be less than or equal to numel(x)
n = 3;
assert(n < numel(x), 'n can''t be greater than numel(x)')
combs = cell(1, n);
[combs{:}] = ndgrid(x);
combs = reshape(cat(n+1, combs{:}), [], n); %a numel(x)^n X n matrix of combinations
thelength = length(combs);
for x=1:thelength
combs(x,4) = combs(x, 1)^2+combs(x, 2)^2+combs(x, 3)^2;
E(x,:) = combs(x,4);
end
%% Arrange combs
v = combs(:,end) ;
[c,ia,ib] = unique(v) ;
iwant = cell(length(c),1) ;
for i = 1:length(c)
iwant{i} = combs(ib==i,:) ;
end
celldisp(iwant)
iwant{1} = 1 1 1 3 iwant{2} = 2 1 1 6 1 2 1 6 1 1 2 6 iwant{3} = 2 2 1 9 2 1 2 9 1 2 2 9 iwant{4} = 3 1 1 11 1 3 1 11 1 1 3 11 iwant{5} = 2 2 2 12 iwant{6} = 3 2 1 14 2 3 1 14 3 1 2 14 1 3 2 14 2 1 3 14 1 2 3 14 iwant{7} = 3 2 2 17 2 3 2 17 2 2 3 17 iwant{8} = 4 1 1 18 1 4 1 18 1 1 4 18 iwant{9} = 3 3 1 19 3 1 3 19 1 3 3 19 iwant{10} = 4 2 1 21 2 4 1 21 4 1 2 21 1 4 2 21 2 1 4 21 1 2 4 21 iwant{11} = 3 3 2 22 3 2 3 22 2 3 3 22 iwant{12} = 4 2 2 24 2 4 2 24 2 2 4 24 iwant{13} = 4 3 1 26 3 4 1 26 4 1 3 26 1 4 3 26 3 1 4 26 1 3 4 26 iwant{14} = 3 3 3 27 iwant{15} = 4 3 2 29 3 4 2 29 4 2 3 29 2 4 3 29 3 2 4 29 2 3 4 29 iwant{16} = 4 4 1 33 4 1 4 33 1 4 4 33 iwant{17} = 4 3 3 34 3 4 3 34 3 3 4 34 iwant{18} = 4 4 2 36 4 2 4 36 2 4 4 36 iwant{19} = 4 4 3 41 4 3 4 41 3 4 4 41 iwant{20} = 4 4 4 48
##### 댓글 수: 1표시숨기기 없음
Leo Quintanar 2021년 7월 22일
Thanks that worked perfectly, also thank you for your comment about my stupidity regrading my varaible names.

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### 추가 답변(1개)

Scott MacKenzie 2021년 7월 22일
sortrows(combs,4)
##### 댓글 수: 1표시숨기기 없음
Leo Quintanar 2021년 7월 22일
thxs

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