Related vector values via cell array

2023 8 12
1 个回答
回答已采纳
更新时间：2023 8 12
13 次查看（30 天）

0 个投票

Hi guys, I hope someone can help me:
I have a vector. Suppose this:
A=(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)'
Some values of this vector are related through a cell array, in such a way that a value of the vector is the index of a cell. I need to know if a value of the vector "a" is related to another value "b".
B={15, 12, 0, [11, 7], 9, 0, [1, 20, 21], 18, 7, 0, 16, [13, 14], 17, 4, 6};
a=2; b=7
In this case, 2 and 7 are related:
Index Value
2 12
12 13,14
13 17
17 Nan
14 4
4 7
So, a=5 and b=6; and a=4 and b=1 are also related.
Thanks to the whole community

5 个评论
显示 3更早的评论 隐藏 3更早的评论

Matt J
Matt J 2023-8-12
The way you determine that a is related to b is not clear, at least not to me.
Sergio Rojas Blanco
编辑：Sergio Rojas Blanco 2023-8-12
Sorry, it's a little weird
Index 2->12. Now, I take this value (12) like the new index
Index 12->13 and 14. Now, I take one of theses (13, the first) and I follow the sequence
Index 13->17, I follow the sequence
Index 17->Nan. If there isn't any value in the cell or it's zero, the sequence (with 13) would end here. But now, I take the other value (14, the sencond) and I follow the sequence
Index 14->4. I follow the sequence
Index 4->7. I have arrived to "b" starting from "a"
I hope you understand a little better now.
Thanks
Dyuman Joshi
Dyuman Joshi 2023-8-12
By this logic, a=4 and b=1 are related.
B={15, 12, 0, [11, 7], 9, 0, [1, 20, 21], 18, 7, 0, 16, [13, 14], 17, 4, 6};
Idx 4 - 11
Idx 11 - 16
Idx 16 - NaN
Idx 4 - 7
Idx 7 - 1
Related.
You're rigth. I have edited to remove the error
Dyuman Joshi
Dyuman Joshi 2023-8-12
编辑：Dyuman Joshi 2023-8-12
@Sergio Rojas Blanco, please check my answer.

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Dyuman Joshi
Dyuman Joshi 2023-8-12
Ran in:

0 个投票

Ran in:
A=(1:10)';
B={15, 12, 0, [11, 7], 9, 0, [1, 20, 21], 18, 7, 0, 16, [13, 14], 17, 4, 6};
m = max([B{:}]);
z1 = nested(A,B,2,7)
z1 = 1
%8th index is 18 which is greater than the number of elements
%so no value is related to it
z2 = nested(A,B,8,randi(m))
z2 = 0
%3rd index is 0, so no value related to it
z3 = nested(A,B,3,randi(m))
z3 = 0
%Similarly for 6th index
z4 = nested(A,B,6,randi(m))
z4 = 0
z5 = nested(A,B,5,6)
z5 = 1
%As the 1st index in B goes to the 6th index, no value will be related to it as well
z6 = nested(A,B,1,randi(m))
z6 = 0
function z = nested(A,B,a,b)
%Returns 0 if not related, 1 if related
%Initialize z as 0
z=0;
nB=numel(B);
for m=1:numel(a)
%Assign only if the value is within the range of number of elements of B
if a(m)<=nB && a(m)>0
c=B{a(m)};
else
%If the value is not in the range, skip to the next iteration
continue
end
%Check if b is present in c or not
if ismember(b,c)
z=1;
return
else
%If not present, continue the recursion and keep checking
z=nested(A,B,c,b);
end
end
end

2 个评论
显示 无 隐藏 无

Thanks a lot Dyuman. I have gained a lot of time with your code.

请先登录，再进行评论。

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Matrix Indexing 的更多信息

产品

版本

R2023a

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by