How to sort one array based on another of a different size

2 Ansichten (letzte 30 Tage)
rbrbrb
rbrbrb am 23 Jun. 2020
Bearbeitet: Stephen23 am 24 Jun. 2020
Hello,
I would like to sort one cell array A according to the order define by a separate cell array B of a smaller size. For example:
A = [a a b c c d d d d e e]
B = [d a c b e]
After sorting, array A would be:
A = [d d d d a a c c b e e]
Thank you for the help.
  1 Kommentar
Rik
Rik am 23 Jun. 2020
I would write code that replaced every value in your top array by the index in your second array. Then you can sort and use indexing to get the original values back.

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Stephen23
Stephen23 am 24 Jun. 2020
Bearbeitet: Stephen23 am 24 Jun. 2020
As Rik wrote, the MATLAB solution is to use ismember, e.g.:
>> A = {'a','a','b','c','c','d','d','d','d','e','e'};
>> B = {'d','a','c','b','e'};
>> [X,Y] = ismember(A,B);
>> [~,Z] = sort(Y(X));
>> C = A(Z)
C =
'd' 'd' 'd' 'd' 'a' 'a' 'c' 'c' 'b' 'e' 'e'

Weitere Antworten (1)

Aakash Mehta
Aakash Mehta am 23 Jun. 2020
Bearbeitet: Aakash Mehta am 23 Jun. 2020
A = ['a' 'a' 'b' 'c' 'c' 'd' 'd' 'd' 'd' 'e' 'e']
B = ['d' 'a' 'c' 'b' 'e']
sizeA = length(A)
sizeB = length(B)
You can use the container.Map to track the frequency of the characters. This approach will also work when A is not sorted.
M = containers.Map
for i=1:sizeA
if isKey(M,A(i))==1
M(A(i)) = M(A(i)) + 1;
else
M(A(i)) = 1;
end
end
You will get the map M, which contains the characters as a key and frequency as their value.
Now using following code you can get the desired output.
str = "";
for i=1:sizeB
if isKey(M,B(i))==1
for j=1:M(B(i))
str = strcat(str,B(i));
end
end
end
  1 Kommentar
Rik
Rik am 23 Jun. 2020
I suspect this can be vectorized with ismember. I suspect that the mapping you are proposing has a lot of unnecessary overhead.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Shifting and Sorting Matrices finden Sie in Help Center und File Exchange

Produkte


Version

R2019b

Community Treasure Hunt

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

Start Hunting!

Translated by