HELP WITH MY PROBLEM
2 次查看(过去 30 天)
显示 更早的评论
I have doubts about how to do this. First explain what I have already done so you can understand what I do. I have a sequence, for example of 20 binary numbers:
0(1) 0(2) 0(3) 1(4) 1(5) 0(6) 0(7) 0(8) 0(9) 0(10) 1(11) 0(12) 0(13) 0(14) 0(15) 1(16)
0(17) 0(18) 1(19) 1(20).
After I created length subsequences 2,3 and 4 as follows:
Take the example of substrings of length 4. Shape creation is the same as for length 2 and 3:
1) 1(1) 0(2) 1(3) 1(4)
2) 1(1) 1(3) 0(5) 1(7)
3) 1(1) 1(4) 1(7) 1(10)
4) 1(1) 0(5) 0(9) 1(13)
5) 1(1) 0(6) 1(11) 0(16)
6) 1(1) 1(7) 1(13) 1(19)
7) 0(2) 1(3) 1(4) 0(5)
8) 0(2) 1(4) 0(6) 0(8)
9) 0(2) 0(5) 0(8) 1(11)
10) 0(2) 0(6) 1(10) 0(14)
11) 0(2) 1(7) 1(12) 1(17)
12) 0(2) 0(8) 0(14) 0(20)
13) 1(3) 1(4) 0(5) 0(6)
14) 1(3) 0(5) 1(7) 0(9)
15) 1(3) 0(6) 0(9) 1(12)
16) 1(3) 1(7) 1(11) 0(15)
17) 1(3) 0(8) 1(13) 1(18)
18) 1(4) 0(5) 0(6) 1(7)
19) 1(4) 0(6) 0(8) 1(10)
20) 1(4) 1(7) 1(10) 1(13)
21) 1(4) 0(8) 1(12) 0(16)
22) 1(4) 0(9) 0(14) 1(19)
23) 0(5) 0(6) 1(7) 0(8)
24) 0(5) 1(7) 0(9) 1(11)
25) 0(5) 0(8) 1(11) 0(14)
26) 0(5) 0(9) 1(13) 1(17)
27) 0(5) 1(10) 0(15) 0(20)
28) 0(6) 1(7) 0(8) 0(9)
29) 0(6) 0(8) 1(10) 1(12)
30) 0(6) 0(9) 1(12) 0(15)
31) 0(6) 1(10) 0(14) 1(18)
32) 1(7) 0(8) 0(9) 1(10)
33) 1(7) 0(9) 1(11) 1(13)
34) 1(7) 1(10) 1(13) 0(16)
35) 1(7) 1(11) 0(15) 1(19)
36) 0(8) 0(9) 1(10) 1(11)
37) 0(8) 1(10) 1(12) 0(14)
38) 0(8) 1(11) 0(14) 1(17)
39) 0(8) 1(12) 0(16) 0(20)
40) 0(9) 1(10) 1(11) 1(12)
41) 0(9) 1(11) 1(13) 0(15)
42) 0(9) 1(12) 0(15) 1(18)
43) 1(10) 1(11) 1(12) 1(13)
44) 1(10) 1(12) 0(14) 0(16)
45) 1(10) 1(13) 0(16) 1(19)
46) 1(11) 1(12) 1(13) 0(14)
47) 1(11) 1(13) 0(15) 1(17)
48) 1(11) 0(14) 1(17) 0(20)
49) 1(12) 1(13) 0(14) 0(15)
50) 1(12) 0(14) 0(16) 1(18)
51) 1(13) 0(14) 0(15) 0(16)
52) 1(13) 0(15) 1(17) 1(19)
53) 0(14) 0(15) 0(16) 1(17)
54) 0(14) 0(16) 1(18) 0(20)
55) 0(15) 0(16) 1(17) 1(18)
56) 0(16) 1(17) 1(18) 1(19)
57) 1(17) 1(18) 1(19) 0(20)
After these 57 patterns I calculate the relative frequency of all of them. Also get the relative frequencies of patterns substrings of length 2 and 3.
Okay, so far I have work already done with matlab. My doubts are from here
------------------------------------------------------------------------------------------------
Suppose you now want to know the probability that the number (21) of the above sequence is 0 ò 1, ie:
0(1) 0(2) 0(3) 1(4) 1(5) 0(6) 0(7) 0(8) 0(9) 0(10) 1(11) 0(12) 0(13) 0(14) 0(15) 1(16)
0(17) 0(18) 1(19) 1(20) X(21).
My first question is how to create the following matrix:
0(18) 1(19) 1(20) X(21)-----------[0 1 1 X]
0(15) 0(17) 1(19) X(21)-----------[0 1 1 X]
0(12) 0(15) 0(18) X(21)-----------[0 0 0 X]
0(9) 0(13) 0(17) X(21)-------------[0 0 0 X]
0(6) 1(11) 1(16) X(21)-------------[0 1 1 X]
0(3) 0(9) 0(15) X(21)---------------[0 0 0 X]
Considering that if X = 1:
1-[0 1 1 1]
2-[0 1 1 1]
3-[0 0 0 1]
4-[0 0 0 1]
5-[0 1 1 1]
6-[0 0 0 1]
If X = 0:
7-[0 1 1 0]
8-[0 1 1 0]
9-[0 0 0 0]
10-[0 0 0 0]
11-[0 1 1 0]
12-[0 0 0 0]
With these patterns and taking into account the dependency between the previous numbers in the sequence, I have 12 patterns of which I have to study the probability that X = 0 ò X = 1.
As I calculated the relative frequencies of patterns of length 2, 3 and 4, do the following:
For example first pattern:
1-[0 1 1 1]
[0 1] ------------ relative frequency pattern of length 2
+
[0 1 1] ---------- relative frequency pattern of length 3
+
[0 1 1 1] ------- relative frequency pattern of length 4
= Frequency dependence assuming total pattern
Realize the same for the other 11 patterns and study the probability:
P (X = 1) = number of patterns with X = 1 with probability greater than X = 0 / total number of
patterns
P (X = 0) = number of patterns with X = 0 with probability greater than X = 1 / total number of patterns
So my question is how I can perform this process from the dotted line to the end, and I've tried a thousand ways but I don`t get the correct result.
thank you very much
5 个评论
Image Analyst
2013-9-24
Well for the first substring you have this: 1(1) 0(2) 1(3) 1(4). Where did this come from? From this "0(1) 0(2) 0(3) 1(4) 1(5) 0(6) 0(7) 0(8) 0(9) 0(10) 1(11) 0(12) 0(13) 0(14) 0(15) 1(16) 0(17) 0(18) 1(19) 1(20)" I take it your array is [0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 1 0 0 1 1]. So "1(1) 0(2) 1(3) 1(4)" = [1 0 1 1], right? And this does not occur anywhere in the array, right? All right, fine. But why does that string have (1) etc. in parentheses - what does it mean? Clearly it's not the indexes where they came from since element 1 is 0, not 1. And in "13) 1(3) 1(4) 0(5) 0(6)" you say that elements 3,4,5 and 6 are 1,1,0, & 0, yet in the long expression, 3,4,5 & 6 are 0,1,1,0, not 1,1,0,0. So at that point I threw up my hands and gave up.
回答(3 个)
Image Analyst
2013-9-24
This might be instructive:
% Define numerical matrix.
m=[0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 1 0 0 1 1]
% Turn it into a string.
strm = sprintf('%d', m)
% Define a pattern to look for.
patternToLookFor = '011'
% Find index(es) where that pattern begins.
indexes = strfind(strm, patternToLookFor)
In the command window:
m =
0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 1 0 0 1 1
strm =
00011000001000010011
patternToLookFor =
011
indexes =
3 18
Walter Roberson
2013-9-24
5 个评论
Walter Roberson
2013-9-25
n-grams work with symbols. They don't care whether the symbols are letters or bits.
Tally(1+bit1,1+bit2,1+bit3,1+bit4) = Tally(1+bit1,1+bit2,1+bit3,1+bit4) + 1;
and then
bit1 = bit2; bit2 = bit3; bit3 = bit4; bit4 = next bit
Now you can calculate conditional probabilities as Tally(1+A,1+B,1+C,1+D) / sum(Tally(1+A,1+B,1+C,:),4))
The "1+" adjust for the fact that indices start at 1 in MATLAB but bits are 0 or 1.
The probability of D after ABC is the count of ABCD divided by the count of (ABC0 + ABC1)
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Text Data Preparation 的更多信息
产品
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!