sort matrix by absolute value

2021 3 29
2 个回答
回答已采纳
更新时间：2021 3 30
7 次查看（30 天）

0 个投票

How do I sort a matrix by absolute value and display its original value with +,- sign?

请先登录，再回答此问题。

 采纳的回答

DGM
DGM 2021-3-29
编辑：DGM 2021-3-29

2 个投票

Consider the case for a vector:
A=randi(19,[10 1])-10
[~,idx]=sort(abs(A))
Asorted=A(idx)
For other arrays, you'll have to decide how exactly you want to sort it, and do the indexing accordingly.

4 个评论
显示 2更早的评论 隐藏 2更早的评论

Poorva
Poorva 2021-3-30
clc
clear
% generating the table of rows and columns
row = input('Enter number of rows: ');
col = input('Enter number of columns: ');
C = zeros(row ,col);
%taking values from user of matrix
for s = 1:row
for l = 1:col
str = ['Enter element in row ' num2str(s) ', col ' num2str(l) ': '];
C(s,l) = input(str);
end
end
%taking absolute values of all elements
D=abs(C);
%sorting the rows in descending order
[T,I]=sort(D,2,'descend');
This is my code. Please help me with this, Matrix C has the original input from user in +,- sign, D has the absolute value of C, T gives the sorted matrix but all in + sign ,but I want the sorted result with both +,- sign as the input given by user.
e.g C= [ 1 -8
-4 7]
D=[ 1 8
4 7]
T=[8 1
7 4]
i want the result as res= [-8 1
7 -4]
DGM
DGM 2021-3-30
If you want to sort descending on dim2, my method becomes more cumbersome because of how the index array is organized as row vector indices instead of array indices.
% you could do this without a loop
% but this is less obfuscated
[T,idx]=sort(abs(C),2,'descend');
for r=1:row % because we're sorting on dim2
T(r,:)=C(r,idx(r,:));
end
T
If you're using a sufficiently recent version, you can just use Steven's suggestion from his answer, which is much more concise.
T=sort(C,2,'descend','comparisonmethod','abs')
For your purposes these two methods are probably both fine, but they do handle duplicate values differently.
Poorva
Poorva 2021-3-30
Thank you ....I got it now and the code is working well.
Poorva
Poorva 2021-3-30
[y, idx2] = sort(D,2,'descend');
sx = size(C);
index = sub2ind(sx, repmat((1:sx(1)).', 1, sx(2)), idx2);
y2 = C(index);
res=array2table(y2)

请先登录，再进行评论。

更多回答（1 个）

Steven Lord
Steven Lord 2021-3-30
Ran in:

1 个投票

Ran in:
Specify the 'ComparisonMethod' name-value pair argument.
rng default
A = randi([-10 10], 5, 5)
A = 5×5
7 -8 -7 -8 3 9 -5 10 -2 -10 -8 1 10 9 7 9 10 0 6 9 3 10 6 10 4
[values, locations] = sort(A(:), 'ComparisonMethod', 'abs');
[values.'; locations.']
ans = 2×25
0 1 -2 3 3 4 -5 6 6 7 7 -7 -8 -8 -8 9 9 9 9 10 10 10 10 10 -10 14 8 17 5 21 25 7 15 19 1 23 11 3 6 16 2 4 18 24 9 10 12 13 20 22

2 个评论
显示 无 隐藏 无

DGM
DGM 2021-3-30
Do you know about when that option was introduced? I'm guessing it was some time around 2018, but release notes are vague and I only know it's not in R2015b.
Poorva
Poorva 2021-3-30
thank you steven ....my code is working.

请先登录，再进行评论。

类别

在 帮助中心 和 File Exchange 中查找有关 Shifting and Sorting Matrices 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by