Effacer les filtres
Effacer les filtres

Counting First digit of a certain number

5 vues (au cours des 30 derniers jours)
Min
Min le 12 Mar 2024
Commenté : Min le 13 Mar 2024
Hi, I am working with a large number of data to analyze if there are any failures in data which having some hard time manually looking at the data or even with graphs. So I was wondering if there is a way to count the first digit number out of table which then I just need to look into how many failures happened. Thanks in advance...
For example,
t = [0, 0; 0, 1; 0, 1; 1, 1; 1, 0; 0, 0; 0, 1]
t = 7×2
0 0 0 1 0 1 1 1 1 0 0 0 0 1
t2 = [1, 1; 0, 1; 0, 1; 0, 1; 1, 1; 0, 1; 1, 1]
t2 = 7×2
1 1 0 1 0 1 0 1 1 1 0 1 1 1
t3 = [0, 1, 1; 1, 0, 0; 1, 1, 0; 1, 0, 1; 1, 1, 0; 0, 0, 1; 1, 1, 1]
t3 = 7×3
0 1 1 1 0 0 1 1 0 1 0 1 1 1 0 0 0 1 1 1 1
then
T_count = [1, 2]
T_count = 1×2
1 2
T2_count = [3, 1]
T2_count = 1×2
3 1
T3_count = [2, 4, 3]
T3_count = 1×3
2 4 3

Réponse acceptée

Stephen23
Stephen23 le 12 Mar 2024
t1 = [0, 0; 0, 1; 0, 1; 1, 1; 1, 0; 0, 0; 0, 1];
t2 = [1, 1; 0, 1; 0, 1; 0, 1; 1, 1; 0, 1; 1, 1];
t3 = [0, 1, 1; 1, 0, 0; 1, 1, 0; 1, 0, 1; 1, 1, 0; 0, 0, 1; 1, 1, 1];
fh = @(m) sum(diff([0*m(1,:);m],1,1)>0,1);
fh(t1)
ans = 1×2
1 2
fh(t2)
ans = 1×2
3 1
fh(t3)
ans = 1×3
2 4 3
  7 commentaires
Min
Min le 13 Mar 2024
Oh! I see, yep my data was in a form of table since I was working with the timetable data.
I will go ahead and make some changes and see if I see any error. Thank you!
Min
Min le 13 Mar 2024
Hi Stephen, I tried out with some changes and it works great.
Thank you :)

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