How to tell if there are at least 5 consecutive entries in one 8-by-1 matrix with 8 integers?
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
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 Kommentare
Image Analyst
am 31 Mai 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.
Akzeptierte Antwort
Daniel Shub
am 31 Mai 2013
Bearbeitet: Daniel Shub
am 3 Jun. 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.
Weitere Antworten (4)
Roger Stafford
am 31 Mai 2013
Bearbeitet: Roger Stafford
am 1 Jun. 2013
It can all be put into one line:
any(diff(find([true;diff(unique(x))~=1;true]))>=5)
1 Kommentar
Image Analyst
am 31 Mai 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 Kommentare
Daniel Shub
am 31 Mai 2013
@IA I would be surprised if bwlabel followed by regionprops would win Cody or any type of speed test.
Azzi Abdelmalek
am 31 Mai 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 Kommentare
Azzi Abdelmalek
am 31 Mai 2013
Bearbeitet: Azzi Abdelmalek
am 31 Mai 2013
diff(a) can have several values, 0,-1,1,2,....I have grouped 1 and 0
Daniel Shub
am 31 Mai 2013
Of course, apparently I am not thinking straight, but a = ones(1, 8) still passes your test.
Azzi Abdelmalek
am 31 Mai 2013
Bearbeitet: Azzi Abdelmalek
am 31 Mai 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 Kommentare
Daniel Shub
am 31 Mai 2013
This code doesn't work you need: e=[1 , diff(a)];.
Second, it says a = ones(1, 8) is a straight.
Azzi Abdelmalek
am 31 Mai 2013
Bearbeitet: Azzi Abdelmalek
am 31 Mai 2013
Yes, Look at edited answer (e' instead of e), and this is working for a=ones(8,1)
Siehe auch
Kategorien
Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!