Azzera filtri
Azzera filtri

Why won't this for loop move to the next variable?

2 visualizzazioni (ultimi 30 giorni)
Cary
Cary il 30 Giu 2015
Commentato: Star Strider il 30 Giu 2015
On the first pass, everything is fine. i = 49, j = 1, k = 1112. On the second pass, i and j move to the next variables (50 and 18), but k stays at 1112. It does the same thing on the third and fourth pass. Can someone please point out my error? My sincerest gratitude for reading.
for i = 49:52
expiry=find(vifDate==expDate(i));
for j = [1; 18; 43; 63]
vifCls1 = vifCls(j:expiry);
for k = [1112; 1129; 1154; 1174]
consolFut1 = consolFut(k:expDateIdx(i),i);
arbVIF=vifCls1-consolFut1;
end
end
end

Risposte (2)

Thorsten
Thorsten il 30 Giu 2015
Modificato: Thorsten il 30 Giu 2015
I you code only i cycles to the values 49,50, ..., while j and k are always a 4x1 vector [1; 18; 43; 63] and k = [1112; 1129; 1154; 1174]. You have to use row vectors in the for loop:
for j = [1 18 43 63]
% some code
for k = [1112 1129 1154 1174]

Star Strider
Star Strider il 30 Giu 2015
Your loops probably are incrementing, but because of the way you created your indices, you’re just not seeing it.
Consider:
ix = 1;
for i = 49:52
for j = [1; 18; 43; 63]'
for k = [1112; 1129; 1154; 1174]'
Q(ix,:) = [i j k];
ix = i + 1;
end
end
end
figure(1)
stem3(Q(:,1), Q(:,2), Q(:,3))
grid on
xlabel('i')
ylabel('j')
zlabel('k')
The plot shows that all the loops are incrementing.
  2 Commenti
Thorsten
Thorsten il 30 Giu 2015
Modificato: Thorsten il 30 Giu 2015
Your example works because you use row vectors for the indices (or transposed column vectors, to be more precise). Cary's code does not work, because she uses column vectors; it is not a matter of "not seeing it".
Star Strider
Star Strider il 30 Giu 2015
The column vectors quite definitely do not work (that I found surprising).
I transposed them to demonstrate that even if the loops would increment, it will be difficult to see the result because of the inefficient ways the loops are indexed.

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