Indexing n-th element with increasing offset.
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Good day
I'm trying to find a whey how it would be possible to index a n-th element with increasing offset in a arrey of zeros (expectation - 1 0 1 0 0 1 0 0 0 1 0 0 0 0 1…). I found a whey, but its something like from first grade. Maybe someone would share some ideas, or give a hint of how it would be possible to reduce repeating typing…
x=1:1:100;
g=cumsum(x)<100;
g_idx=find(g);
X=zeros(size(x));
X(g_idx(1))=1;
X(cumsum(g_idx(1:2)))=1;
X(cumsum(g_idx(1:3)))=1;
X(cumsum(g_idx(1:4)))=1;
X(cumsum(g_idx(1:5)))=1;
X(cumsum(g_idx(1:6)))=1;
X(cumsum(g_idx(1:7)))=1;
X(cumsum(g_idx(1:8)))=1;
X(cumsum(g_idx(1:9)))=1;
X(cumsum(g_idx(1:10)))=1;
X(cumsum(g_idx(1:11)))=1;
X(cumsum(g_idx(1:12)))=1;
X(cumsum(g_idx(1:13)))=1
2 个评论
Mario Malic
2020-8-6
编辑：Mario Malic
2020-8-6
Pascal's triangle, binomial theorem? You can check on Wiki, there's a formula for those numbers on it.https://wikimedia.org/api/rest_v1/media/math/render/svg/23050fcb53d6083d9e42043bebf2863fa9746043
Andrew
2020-8-6
采纳的回答
Fangjun Jiang
2020-8-6
%%
a=1:13;
b=cumsum(a);
X=zeros(1,100);
X(b)=1
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