Reversing a part of matrix

7 vues (au cours des 30 derniers jours)
Bartosz Bagrowski
Bartosz Bagrowski le 29 Avr 2022
Commenté : Star Strider le 30 Avr 2022
Hey, I would like to choose two random values in a matrix and reverse from the first random number to the second one.
i=randsample(10,2)
i1=i(1);
i2=i(2);
let's say we got two random numbers - 3 and 7
M=[10 20 30 40 50 60 70 80 90 100];
I would like to get the matrix as following:
M_new=[10 20 70 60 50 40 30 80 90 100];
Could anyone help me to code it?
The method with the flip doesn't help much.

Réponse acceptée

Star Strider
Star Strider le 29 Avr 2022
A few examples of a robust approach —
M=[10 20 30 40 50 60 70 80 90 100];
i=sort(randsample(10,2))
i = 2×1
5 6
M_new = [M(1:i(1)-1) flip(M(i(1):i(2))) M(i(2)+1:end)]
M_new = 1×10
10 20 30 40 60 50 70 80 90 100
i=sort(randsample(10,2))
i = 2×1
3 9
M_new = [M(1:i(1)-1) flip(M(i(1):i(2))) M(i(2)+1:end)]
M_new = 1×10
10 20 90 80 70 60 50 40 30 100
i=sort(randsample(10,2))
i = 2×1
1 9
M_new = [M(1:i(1)-1) flip(M(i(1):i(2))) M(i(2)+1:end)]
M_new = 1×10
90 80 70 60 50 40 30 20 10 100
i=sort(randsample(10,2))
i = 2×1
2 7
M_new = [M(1:i(1)-1) flip(M(i(1):i(2))) M(i(2)+1:end)]
M_new = 1×10
10 70 60 50 40 30 20 80 90 100
i=sort(randsample(10,2))
i = 2×1
8 9
M_new = [M(1:i(1)-1) flip(M(i(1):i(2))) M(i(2)+1:end)]
M_new = 1×10
10 20 30 40 50 60 70 90 80 100
.
  2 commentaires
Bartosz Bagrowski
Bartosz Bagrowski le 30 Avr 2022
Thank you very much for helping me to solve the problem!
Star Strider
Star Strider le 30 Avr 2022
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.

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Plus de réponses (3)

Voss
Voss le 29 Avr 2022
i=sort(randsample(10,2)) % sort so that i(2) >= i(1)
i = 2×1
3 7
i1=i(1);
i2=i(2);
M=[10 20 30 40 50 60 70 80 90 100];
M_new = M;
M_new(i1:i2) = M(i2:-1:i1)
M_new = 1×10
10 20 70 60 50 40 30 80 90 100

Riccardo Scorretti
Riccardo Scorretti le 29 Avr 2022
A possible way is to pass through an vector of index (= ind in the code hereafter):
M=[10 20 30 40 50 60 70 80 90 100];
t = randsample(10, 2) % I reserve i for the imaginary unit
t = 2×1
8 10
ind = 1 : 10;
ind(t) = t([2 1]);
M = M(ind)
M = 1×10
10 20 30 40 50 60 70 100 90 80

Riccardo Scorretti
Riccardo Scorretti le 29 Avr 2022
Another way (more efficient, I think) is to do a swp by hand:
M=[10 20 30 40 50 60 70 80 90 100];
t = randsample(10, 2) % I reserve i for the imaginary unit
t = 2×1
1 10
tmp_ = M(t(1)) ; M(t(1)) = M(t(2)) ; M(t(2)) = tmp_
M = 1×10
100 20 30 40 50 60 70 80 90 10
  1 commentaire
Voss
Voss le 29 Avr 2022
@Riccardo Scorretti Note that these answers swap the two elements at indices t, but the question asks to reverse the order of all elements between those indices.

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