How to tell if there are at least 5 consecutive entries in one 8-by-1 matrix with 8 integers?
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Hi:
I am looking for a way to determine if there are AT LEAST 5 consecutive values in one 8-by-1 matrix with 8 integers. The 8 integers do NOT have to be unique. And I prefer this to be short and loop-free.
The consecutive values have to be in a "straight", meaning that [5 1 3 4 2 6 8 7] would work because it has at least 5 (actually 8) consecutive values in a straight. The straight is 1 2 3 4 5 6 7 8.
But [1 2 3 4 10 9 8 7] would not because it only has 4 values in each of the straight. The first straight is 1 2 3 4. The second straight is 7 8 9 10.
Thank you very much for your help (I'm writing this for my TexasHoldEm function for fun)
10 comentarios
Image Analyst
el 31 de Mayo de 2013
[5 1 3 4 2 6 8 7] is 1 by 8, not 8 by 1 like you said, and my code assumes. Which is it??? [5; 1; 3; 4; 2; 6; 8; 7] would be 8 by 1.
Respuesta aceptada
Daniel Shub
el 31 de Mayo de 2013
Editada: Daniel Shub
el 3 de Jun. de 2013
As people are giving answers, I am pretty confident that
not(isempty(strfind(diff(sort(unique(x))), ones(1, 4))))
works. I am less confident that
not(all(diff(sort(unique(x)), 4)))
works, but if it does, it is much cooler as I rarely use the second argument to diff. After further thinking the second approach does not work. It fails in all sorts of unique ways, for example 1,3,5,7,9.
Más respuestas (4)
Roger Stafford
el 31 de Mayo de 2013
Editada: Roger Stafford
el 1 de Jun. de 2013
It can all be put into one line:
any(diff(find([true;diff(unique(x))~=1;true]))>=5)
1 comentario
Image Analyst
el 31 de Mayo de 2013
How about
data = [9; 1; 2; 3; 4; 5; 6; 9] % Sample data.
diffData = diff(data)
countOf1s = sum(diffData==1)+1
atLeast5 = countOf1s >= 5
7 comentarios
Daniel Shub
el 31 de Mayo de 2013
@IA I would be surprised if bwlabel followed by regionprops would win Cody or any type of speed test.
Azzi Abdelmalek
el 31 de Mayo de 2013
a=[3 2 3 4 5 5 7 6 8];
e=[1 diff(a)];
e(e==0)=1;
idx=strfind(e,[true,true,true,true]) % it exist if idx~=0
4 comentarios
Azzi Abdelmalek
el 31 de Mayo de 2013
Editada: Azzi Abdelmalek
el 31 de Mayo de 2013
diff(a) can have several values, 0,-1,1,2,....I have grouped 1 and 0
Daniel Shub
el 31 de Mayo de 2013
Of course, apparently I am not thinking straight, but a = ones(1, 8) still passes your test.
Azzi Abdelmalek
el 31 de Mayo de 2013
Editada: Azzi Abdelmalek
el 31 de Mayo de 2013
a=[1; 2; 3; 5; 6; 7; 9; 10]
a=sort(a)
e=[1 ;diff(a)];
e(e==0)=1;
idx=~isempty(strfind(e',[true,true,true,true])) % it exist if idx=1
2 comentarios
Daniel Shub
el 31 de Mayo de 2013
This code doesn't work you need: e=[1 , diff(a)];.
Second, it says a = ones(1, 8) is a straight.
Azzi Abdelmalek
el 31 de Mayo de 2013
Editada: Azzi Abdelmalek
el 31 de Mayo de 2013
Yes, Look at edited answer (e' instead of e), and this is working for a=ones(8,1)
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