help with the resolution of the problem
1 次查看(过去 30 天)
显示 更早的评论
good, I previously had a binary sequence and my purpose was the creation of substrings of various lengths, eg length 4: Sequence
*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)*
_Substrings_
*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],*
using the following code
if true
% code
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
[j1,j2,j2] = unique(s(idx),'rows')
out = [j1, histc(j2,1:max(j2))/i2(end)]; % This row corrected
end
and at the end get a count of the times to repeat each pattern and their relative frequency:
*0 0 0 0------ 161697-- 0,0606515378844711
0 0 0 1------ 163593-- 0,0613627156789197
0 0 1 0------ 164201-- 0,0615907726931733
0 0 1 1------ 166680-- 0,0625206301575394
0 1 0 0------ 164105-- 0,0615547636909227
0 1 0 1------ 166501-- 0,0624534883720930
0 1 1 0------ 167099-- 0,0626777944486122
0 1 1 1------ 168835-- 0,0633289572393098
1 0 0 0------ 164086-- 0,0615476369092273
1 0 0 1------ 166963-- 0,0626267816954239
1 0 1 0------ 166931-- 0,0626147786946737
1 0 1 1------ 169470-- 0,0635671417854464
1 1 0 0------ 166622-- 0,0624988747186797
1 1 0 1------ 169326-- 0,0635131282820705
1 1 1 0------ 169251-- 0,0634849962490623
1 1 1 1------ 170640-- 0,0640060015003751*
The problem that arises is that when I processed this way I only processes some 4000 data and need to process many more. I have 4GB of RAM and Matlab 2012. What I thought is this: Assign each patron an integer:
*0 0 0 0------ 1
0 0 0 1-------2
0 0 1 0-------3
0 0 1 1-------4
0 1 0 0-------5
0 1 0 1-------6
0 1 1 0-------7
0 1 1 1-------8
1 0 0 0-------9
1 0 0 1-------10
1 0 1 0-------11
1 0 1 1-------12
1 1 0 0-------13
1 1 0 1-------14
1 1 1 0-------15
1 1 1 1-------16*
and set as a counter to assign the number of times to repeat that integer. In this way perhaps get as many data processing. thank you very much Can anyone help me do this processing, since otherwise the large number of data means that no processing occurs
回答(0 个)
另请参阅
产品
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!