extracting subsequences of binary string

2013 8 20
3 个回答
回答已采纳
更新时间：2013 10 19
5 次查看（30 天）

0 个投票

as would be the code for the following string have the next subsequences ?
STRING
1(1), 0(2), 1(3), 1(4), 0(5), 0(6), 1(7), 0(8), 0(9), 1(10), 1(11), 1(12), 1(13), 0(14), 0(15), 0(16), 1(17), 1(18), 1(19), 0(20)
SUBSEQUENCES
01: 1(01), 0(02), 1(03), 1(04) -> [1,0,1,1],
02: 1(01), 1(03), 0(05), 1(07) -> [1,1,0,1],
03: 1(01), 1(04), 1(07), 1(10) -> [1,1,1,1],
04: 1(01), 0(05), 0(09), 1(13) -> [1,0,0,1],
05: 1(01), 0(06), 1(11), 0(16) -> [1,0,1,0],
06: 1(01), 1(07), 1(13), 1(19) -> [1,1,1,1],
07: 0(02), 1(03), 1(04), 0(05) -> [0,1,1,0],
08: 0(02), 1(04), 0(06), 0(08) -> [0,1,0,0],
09: 0(02), 0(05), 0(08), 1(11) -> [0,0,0,1],
10: 0(02), 0(06), 1(10), 0(14) -> [0,0,1,0],
11: 0(02), 1(07), 1(12), 1(17) -> [0,1,1,1],
12: 0(02), 0(08), 0(14), 0(20) -> [0,0,0,0],
13: 1(03), 1(04), 0(05), 0(06) -> [1,1,0,0],
14: 1(03), 0(05), 1(07), 0(09) -> [1,0,1,0],
15: 1(03), 0(06), 0(09), 1(12) -> [1,0,0,1],
16: 1(03), 1(07), 1(11), 0(15) -> [1,1,1,0],
17: 1(03), 0(08), 1(13), 1(18) -> [1,0,1,1],
18: 1(04), 0(05), 0(06), 1(07) -> [1,0,0,1],
19: 1(04), 0(06), 0(08), 1(10) -> [1,0,0,1],
20: 1(04), 1(07), 1(10), 1(13) -> [1,1,1,1],
21: 1(04), 0(08), 1(12), 0(16) -> [1,0,1,0],
22: 1(04), 0(09), 0(14), 1(19) -> [1,0,0,1],
23: 0(05), 0(06), 1(07), 0(08) -> [0,0,1,0],
24: 0(05), 1(07), 0(09), 1(11) -> [0,1,0,1],
25: 0(05), 0(08), 1(11), 0(14) -> [0,0,1,0],
26: 0(05), 0(09), 1(13), 1(17) -> [0,0,1,1],
27: 0(05), 1(10), 0(15), 0(20) -> [0,1,0,0],
28: 0(06), 1(07), 0(08), 0(09) -> [0,1,0,0],
29: 0(06), 0(08), 1(10), 1(12) -> [0,0,1,1],
30: 0(06), 0(09), 1(12), 0(15) -> [0,0,1,0],
31: 0(06), 1(10), 0(14), 1(18) -> [0,1,0,1],
32: 1(07), 0(08), 0(09), 1(10) -> [1,0,0,1],
33: 1(07), 0(09), 1(11), 1(13) -> [1,0,1,1],
34: 1(07), 1(10), 1(13), 0(16) -> [1,1,1,0],
35: 1(07), 1(11), 0(15), 1(19) -> [1,1,0,1],
36: 0(08), 0(09), 1(10), 1(11) -> [0,0,1,1],
37: 0(08), 1(10), 1(12), 0(14) -> [0,1,1,0],
38: 0(08), 1(11), 0(14), 1(17) -> [0,1,0,1],
39: 0(08), 1(12), 0(16), 0(20) -> [0,1,0,0],
40: 0(09), 1(10), 1(11), 1(12) -> [0,1,1,1],
41: 0(09), 1(11), 1(13), 0(15) -> [0,1,1,0],
42: 0(09), 1(12), 0(15), 1(18) -> [0,1,0,1],
43: 1(10), 1(11), 1(12), 1(13) -> [1,1,1,1],
44: 1(10), 1(12), 0(14), 0(16) -> [1,1,0,0],
45: 1(10), 1(13), 0(16), 1(19) -> [1,1,0,1],
46: 1(11), 1(12), 1(13), 0(14) -> [1,1,1,0],
47: 1(11), 1(13), 0(15), 1(17) -> [1,1,0,1],
48: 1(11), 0(14), 1(17), 0(20) -> [1,0,1,0],
49: 1(12), 1(13), 0(14), 0(15) -> [1,1,0,0],
50: 1(12), 0(14), 0(16), 1(18) -> [1,0,0,1],
51: 1(13), 0(14), 0(15), 0(16) -> [1,0,0,0],
52: 1(13), 0(15), 1(17), 1(19) -> [1,0,1,1],
53: 0(14), 0(15), 0(16), 1(17) -> [0,0,0,1],
54: 0(14), 0(16), 1(18), 0(20) -> [0,0,1,0],
55: 0(15), 0(16), 1(17), 1(18) -> [0,0,1,1],
56: 0(16), 1(17), 1(18), 1(19) -> [0,1,1,1],
57: 1(17), 1(18), 1(19), 0(20) -> [1,1,1,0],

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 采纳的回答

Andrei Bobrov
Andrei Bobrov 2013-8-21
编辑：Andrei Bobrov 2013-8-21

0 个投票

N = 20;
n = 4;
A = hankel(1:N-n+1,N-n+1:N);
k = 0:n-1;
idx = [];
for ii = 1:size(A,1)
p = A(ii,:);
while p(end,end) + k(end) <= N
p = [p;p(end,:)+k];
end
idx=[idx;p];
end
or
N = 20;
n = 4;
A = hankel(1:N-n+1,N-n+1:N);
k = 0:n-1;
c = ceil((N - A(:,end) + 1)/k(end));
i2 = cumsum(c);
i1 = i2 - c + 1;
idx = zeros(i2(end),n);
for jj = 1:N-n+1
idx(i1(jj):i2(jj),:) = bsxfun(@plus,A(jj,:),(0:c(jj)-1)'*k);
end
ADD
s = [1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0];
[j1,j2,j2] = unique(s(idx),'rows')
out = [j1, histc(j2,1:max(j2))/i2(end)]; % This row corrected

8 个评论
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FRANCISCO
FRANCISCO 2013-8-21
thank you very much, that command should now be used to calculate the number of times to repeat each subsequence? is to calculate the probability by dividing the number of occurrences of that subsequence by the total number of subsequences. But I'm not sure which command used to count the number of occurrences of each subsequence
Andrei Bobrov
Andrei Bobrov 2013-8-21
see ADD part in my answer
FRANCISCO
FRANCISCO 2013-8-21
I get the following error:
Error using horzcat CAT arguments dimensions are not consistent.
Do not find me subsequences. j2 be the number of occurrences of each subsequence?
FRANCISCO
FRANCISCO 2013-8-21
N = 20;
n = 4;
A = hankel(1:N-n+1,N-n+1:N);
k = 0:n-1;
idx = [];
for ii = 1:size(A,1) p = A(ii,:); while p(end,end) + k(end) <= N
p = [p;p(end,:)+k];
end
idx=[idx;p];
end
s = [1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0];
[j1,j2,j2] = unique(s(idx),'rows')
out = [j1, histc(j2,1:max(i2))];
This makes me:
Undefined function or variable 'i2'.
and interpret the outputs? are the number of occurrences of each subsequence?
many thanks
FRANCISCO
FRANCISCO 2013-8-21
sorry, I have not understood the code. This it does is calculate the number of times to repeat each subsequence?. It calculates the sub but if calculated occurrences each subsequence?. it?
Andrei Bobrov
Andrei Bobrov 2013-8-21
Again correct last row in my code.

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更多回答（2 个）

Roger Stafford
Roger Stafford 2013-8-20
编辑：Roger Stafford 2013-8-21

0 个投票

n = 20;
d = 4;
c = zeros(sum([1,floor((d:n-1)/(d-1))]),d); % Allocate space for c
j = 0;
for k = 1:n-d+1
r = 1;
while k+r*(d-1) <= n
j = j+1;
c(j,:) = k:r:k+r*(d-1);
r = r+1;
end
end
The c array will be a 57 x 4 matrix of subsequence indices taken from 1:20.
c =
1 2 3 4
1 3 5 7
1 4 7 10
.....
17 18 19 20
If you replace the line "c(j,:) = k:r:k+r*(d-1);" by
c(j,:) = s(k:r:k+r*(d-1));
where s is your string, this will generate the subsequence of binary strings you are (apparently) asking for.

3 个评论
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Roger Stafford
Roger Stafford 2013-8-21
I have modified the above code so as to allocate the proper size for the c array.
FRANCISCO
FRANCISCO 2013-8-21
thank you very much, that command should now be used to calculate the number of times to repeat each subsequence? is to calculate the probability by dividing the number of occurrences of that subsequence by the total number of subsequences. But I'm not sure which command used to count the number of occurrences of each subsequence
FRANCISCO
FRANCISCO 2013-10-19
One question, as I can do with structure for you automatically calculate subsequences of length 4-20? ie, d = 4:20 but applying for so I said why not have the same dimension:
if true
% code
for d=4:20
c(d)=zeros(sum([1,floor((d:n-1)/(d-1))]),d);
j=0;
for k=1:n-d+1
r=1;
while k+r*(d-1)<=n
j=j+1;
c(j,:)=s(k:r:k+r*(d-1));% s es la cadena binaria / me da las subsecuencias
r=r+1;
end
end
end
end

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Roger Stafford
Roger Stafford 2013-8-22
编辑：Roger Stafford 2013-8-22

0 个投票

Here is a slightly shorter version:
n = 20;
d = 4;
f2 = cumsum([0,floor((n-1:-1:d-1)/(d-1))]);
f1 = f2(1:end-1)+1;
f2 = f2(2:end);
c = repmat(0:d-1,f2(end),1);
for k = 1:length(f1)
c(f1(k),:) = c(f1(k),:) + k;
c(f1(k):f2(k),:) = cumsum(c(f1(k):f2(k),:),1);
end

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