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How to create a random array from an array

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arash rad
arash rad 2023 年 2 月 16 日
コメント済み: Steven Lord 2023 年 2 月 16 日
Hello every one
I have a 1x450 array and I want to choose 300 elements of this array randomly and create a new array with random elements
I tried randi and randperm but these two didn't give me the answer I want
What should I use instead
Thanks in advance
  2 件のコメント
Dyuman Joshi
Dyuman Joshi 2023 年 2 月 16 日
What did you try?
Do you want 300 different indices? or 300 indices with repetitions works?
Jan
Jan 2023 年 2 月 16 日
randi creates indices with repetitions, randperm without. If both methods are not working for you, the problem is hidden inside "didn't give me the answer I want". The readers cannot guess, what you have exactly tried and what you want. So please post your code and explain, what you want to change.

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Arif Hoq
Arif Hoq 2023 年 2 月 16 日
try this:
A=randi(500,1,450);
B=A(randperm(numel(A),300));% choose 300 elements randomly
  2 件のコメント
arash rad
arash rad 2023 年 2 月 16 日
I try this but sth I don't want is that from the 1x450 array no repeated element the new array
for example array index(20) in matrix A only comes one time in B but it sometimes happen more than one
Steven Lord
Steven Lord 2023 年 2 月 16 日
As Jan said, randi draws with replacement and so may generate the same number more than once. randperm draws without replacement and so won't.
rng default % for reproducibility
x = randi(10, 1, 5)
x = 1×5
9 10 2 10 7
y = randperm(10, 5)
y = 1×5
7 3 1 9 10
The number 10 does appear in x twice. If we use the analogy of a deck of cards, in x I selected five cards, replacing each one back in the deck after I've noted what it was. In y I dealt out a hand of five cards without putting any back on the deck.
Note that using y to index into another array may produce the same number repeatedly in that result if the array you're indexing into contains that number more than once. In the deck analogy:
deck = repmat(1:13, 4, 1)
deck = 4×13
1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 3 4 5 6 7 8 9 10 11 12 13
suits = repmat(["club"; "diamond"; "heart"; "spade"], 1, 13)
suits = 4×13 string array
"club" "club" "club" "club" "club" "club" "club" "club" "club" "club" "club" "club" "club" "diamond" "diamond" "diamond" "diamond" "diamond" "diamond" "diamond" "diamond" "diamond" "diamond" "diamond" "diamond" "diamond" "heart" "heart" "heart" "heart" "heart" "heart" "heart" "heart" "heart" "heart" "heart" "heart" "heart" "spade" "spade" "spade" "spade" "spade" "spade" "spade" "spade" "spade" "spade" "spade" "spade" "spade"
cards = deck(y)
cards = 1×5
2 1 1 3 3
cardsuits = suits(y)
cardsuits = 1×5 string array
"heart" "heart" "club" "club" "diamond"
The first 1 in y corresponds to the Ace of hearts and the second corresponds to the Ace of clubs. If you only wanted to draw at most one of each rank of card (at most one Ace or at most one Queen, for example, regardless of suit) then that would require a little modification.

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