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Get as many data processing

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FRANCISCO
FRANCISCO 2013-10-25
评论： FRANCISCO 2013-10-29
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

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回答（1 个）

Walter Roberson
Walter Roberson 2013-10-25
If you are going to do that, consider using accumarray() to do the additions.
If B is the array of bits, such as
B = [0 0 0 0; 1 0 0 0; 0 1 0 0; 1 0 0 0]
then
counts = accumarray( B(:,1) * 8 + B(:,2) * 4 + B(:,3) * 2 + B(:,4) * 1 + 1, 1 );
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FRANCISCO
FRANCISCO 2013-10-29
I tried several ways but it is impossible. Maybe I should use c #
FRANCISCO
FRANCISCO 2013-10-29
Walter, you know c #?. I have the code in c # but I would like to build it in matlab but nose if possible

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