I am trying to store vectors. When I run the program in the loop I see all the values, but when referred outside the loop only the last vector is evaluated and stored (the one that ends with prime number 953, see below). Any calculations done with the PVX vector are done only with the last entry. I want PVX to do calculations with all the results not just the last entry. How can I store these results to do calculations with?
This is the code:
PV=[2 3 5 7 11 13 17 19 23 29];
for numba=2:n
if mod(numba,PV)~=0;
xp=numba;
PVX=[2 3 5 7 11 13 17 19 23 29 xp]
end
end
The first few results looks like this:
PVX: Prime Vectors (Result)
PVX =
2 3 5 7 11 13 17 19 23 29 31
PVX =
2 3 5 7 11 13 17 19 23 29 37
PVX =
2 3 5 7 11 13 17 19 23 29 41
PVX =
2 3 5 7 11 13 17 19 23 29 43
PVX = ...........................................................
PVX =
2 3 5 7 11 13 17 19 23 29 953
If you want to store all PVX values, use a different row for each:
PV = [2 3 5 7 11 13 17 19 23 29];
PVX = [];
for numba=2:n
if mod(numba,PV)~=0;
xp = numba;
PVX = [PVX; 2 3 5 7 11 13 17 19 23 29 xp];
end
end
Of course if would be better to initiallize the PVX matrix to the appropriate size, but the number of rows is hard to predict.
Alternatively, build the PVX without loops:
xp = setdiff(primes(n), primes(29)).'; %'// all primes > 29 and <= n
PVX = [ repmat([2 3 5 7 11 13 17 19 23 29], numel(xp), 1) xp ];
As an example, for n=100, either of the above approaches gives
PVX =
2 3 5 7 11 13 17 19 23 29 31
2 3 5 7 11 13 17 19 23 29 37
2 3 5 7 11 13 17 19 23 29 41
2 3 5 7 11 13 17 19 23 29 43
2 3 5 7 11 13 17 19 23 29 47
2 3 5 7 11 13 17 19 23 29 53
2 3 5 7 11 13 17 19 23 29 59
2 3 5 7 11 13 17 19 23 29 61
2 3 5 7 11 13 17 19 23 29 67
2 3 5 7 11 13 17 19 23 29 71
2 3 5 7 11 13 17 19 23 29 73
2 3 5 7 11 13 17 19 23 29 79
2 3 5 7 11 13 17 19 23 29 83
2 3 5 7 11 13 17 19 23 29 89
2 3 5 7 11 13 17 19 23 29 97
I'm assuming you were going for this:
PVX=[2 3 5 7 11 13 17 19 23 29];
for numba=2:n
if mod(numba,PVX)~=0;
xp=numba;
PVX(end+1) = xp;
%// Or alternatively PVX = [PVX, xp];
end
end
but if you could get an estimate of how large PVX will be in the end, you should pre-allocate the array first for a significant speed up.
So, looks like you need all prime till n
As Dan said use this :
PVX=[2 3 5 7 11 13 17 19 23 29 ];
for numba=2:n
if mod(numba,PVX)~=0
xp=numba;
PVX=[ PVX xp];
end
end
Or why not simply use primes function ?
PVX = primes( n ) ;
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I have a huge data like the following
NDDDDTSVCLGTRQCSWFAGCTNRTWNSSA 0
VCLGTRQCSWFAGCTNRTWNSSAVPLIGLP 0
LTWSGNDTCLYSCQNQTKGLLYQLFRNLFC 0
CQNQTKGLLYQLFRNLFCSYGLTEAHGKWR 0
ITNDKGHDGHRTPTWWLTGSNLTLSVNNSG 0
GHRTPTWWLTGSNLTLSVNNSGLFFLCGNG 0
FLCGNGVYKGFPPKWSGRCGLGYLVPSLTR 0
KGFPPKWSGRCGLGYLVPSLTRYLTLNASQ 0
QSVCMECQGHGERISPKDRCKSCNGRKIVR 1
I want to use the following key to replace the letter with numbers
A 1
R 2
N 3
D 4
B 5
C 6
E 7
Q 8
Z 9
G 10
H 11
I 12
L 13
K 14
M 15
F 16
P 17
S 18
T 19
W 20
Y 21
V 22
at first I want to remove all the numbers close to letter and then replace the letters , so lets look at the first like
NDDDDTSVCLGTRQCSWFAGCTNRTWNSSA
will have this
3 4 4 4 4 19 18 22 6 19 13 10 19 2 8 6 18 20 16 1 10 6 19 3 2 19 20 3 18 18 1
and for the rest of lines the same as many lines as I have
perl -e'
use autodie;
my %charmap = (
A => 1, R => 2, N => 3, D => 4, B => 5, C => 6, E => 7, Q => 8,
Z => 9, G => 10, H => 11, I => 12, L => 13, K => 14, M => 15, F => 16,
P => 17, S => 18, T => 19, W => 20, Y => 21, V => 22,
);
while (<>) {
s{(.)}{ ($charmap{$1} // $1) . " " }ge;
print;
}
' file
Or just
perl -pe'
BEGIN { #charmap{ split //, "ARNDBCEQZGHILKMFPSTWYV" } = 1..22 }
s{(.)}{ ($charmap{$1} // $1) . " " }ge;
' file
With any awk in any shell on any UNIX box:
$ cat tst.awk
BEGIN {
chars = "ARNDBCEQZGHILKMFPSTWYV"
for (i=1; i<=length(chars); i++) {
char = substr(chars,i,1)
map[char] = i
}
}
{
out = ""
chars = $1
for (i=1; i<=length(chars); i++) {
char = substr(chars,i,1)
out = (out == "" ? "" : out " ") (char in map ? map[char] : char)
}
print out
}
$ awk -f tst.awk file
3 4 4 4 4 19 18 22 6 13 10 19 2 8 6 18 20 16 1 10 6 19 3 2 19 20 3 18 18 1
22 6 13 10 19 2 8 6 18 20 16 1 10 6 19 3 2 19 20 3 18 18 1 22 17 13 12 10 13 17
13 19 20 18 10 3 4 19 6 13 21 18 6 8 3 8 19 14 10 13 13 21 8 13 16 2 3 13 16 6
6 8 3 8 19 14 10 13 13 21 8 13 16 2 3 13 16 6 18 21 10 13 19 7 1 11 10 14 20 2
12 19 3 4 14 10 11 4 10 11 2 19 17 19 20 20 13 19 10 18 3 13 19 13 18 22 3 3 18 10
10 11 2 19 17 19 20 20 13 19 10 18 3 13 19 13 18 22 3 3 18 10 13 16 16 13 6 10 3 10
16 13 6 10 3 10 22 21 14 10 16 17 17 14 20 18 10 2 6 10 13 10 21 13 22 17 18 13 19 2
14 10 16 17 17 14 20 18 10 2 6 10 13 10 21 13 22 17 18 13 19 2 21 13 19 13 3 1 18 8
8 18 22 6 15 7 6 8 10 11 10 7 2 12 18 17 14 4 2 6 14 18 6 3 10 2 14 12 22 2
Alternative Perl solution:
#!/usr/bin/perl
use strict;
use warnings;
my %key = (
A => 1, R => 2, N => 3, D => 4, B => 5,
C => 6, E => 7, Q => 8, Z => 9, G => 10,
H => 11, I => 12, L => 13, K => 14, M => 15,
F => 16, P => 17, S => 18, T => 19, W => 20,
Y => 21, V => 22,
);
while (<STDIN>) {
my($text) = /^(\w+)/;
print join(' ',
map { $key{$_} }
split(//, $text)
), "\n";
}
exit 0;
Output with your given text:
$ perl dummy.pl <dummy.txt
3 4 4 4 4 19 18 22 6 13 10 19 2 8 6 18 20 16 1 10 6 19 3 2 19 20 3 18 18 1
22 6 13 10 19 2 8 6 18 20 16 1 10 6 19 3 2 19 20 3 18 18 1 22 17 13 12 10 13 17
13 19 20 18 10 3 4 19 6 13 21 18 6 8 3 8 19 14 10 13 13 21 8 13 16 2 3 13 16 6
6 8 3 8 19 14 10 13 13 21 8 13 16 2 3 13 16 6 18 21 10 13 19 7 1 11 10 14 20 2
12 19 3 4 14 10 11 4 10 11 2 19 17 19 20 20 13 19 10 18 3 13 19 13 18 22 3 3 18 10
10 11 2 19 17 19 20 20 13 19 10 18 3 13 19 13 18 22 3 3 18 10 13 16 16 13 6 10 3 10
16 13 6 10 3 10 22 21 14 10 16 17 17 14 20 18 10 2 6 10 13 10 21 13 22 17 18 13 19 2
14 10 16 17 17 14 20 18 10 2 6 10 13 10 21 13 22 17 18 13 19 2 21 13 19 13 3 1 18 8
8 18 22 6 15 7 6 8 10 11 10 7 2 12 18 17 14 4 2 6 14 18 6 3 10 2 14 12 22 2
On second thought...
As OP wants to obfuscate clear text then the more appropriate solution IMHO should be something like this:
$ bash <dummy.txt -c "$(echo /Td6WFoAAATm1rRGBMCtAbgBIQEWAAAAAAAAACsG0SbgALcApV0AOBlKq3igoJRmX9TqJifIRDIcDLdDtNRSv+tJBsifrrsdnlllNt2qqnlz0/uBmSnlO0FTKjKH/HXplJm9LaV7kXiNp/ZWDsyVqoV8EPjIEHHkXXd6jKahyq7tcCA4NGTHp/pwmk8jith6j/dcX67QCKmL0UtZUz9BqVWefD41lbrTNazbD8IP6zMLmAVxJav51SSTHzsUqhUfqhVmLsUg8sJkgloAAAAAAOMYtQXt21WNAAHJAbgBAABTvtYRscRn+wIAAAAABFla | base64 -d | xzcat)"
3 4 4 4 4 19 18 22 6 13 10 19 2 8 6 18 20 16 1 10 6 19 3 2 19 20 3 18 18 1
22 6 13 10 19 2 8 6 18 20 16 1 10 6 19 3 2 19 20 3 18 18 1 22 17 13 12 10 13 17
13 19 20 18 10 3 4 19 6 13 21 18 6 8 3 8 19 14 10 13 13 21 8 13 16 2 3 13 16 6
6 8 3 8 19 14 10 13 13 21 8 13 16 2 3 13 16 6 18 21 10 13 19 7 1 11 10 14 20 2
12 19 3 4 14 10 11 4 10 11 2 19 17 19 20 20 13 19 10 18 3 13 19 13 18 22 3 3 18 10
10 11 2 19 17 19 20 20 13 19 10 18 3 13 19 13 18 22 3 3 18 10 13 16 16 13 6 10 3 10
16 13 6 10 3 10 22 21 14 10 16 17 17 14 20 18 10 2 6 10 13 10 21 13 22 17 18 13 19 2
14 10 16 17 17 14 20 18 10 2 6 10 13 10 21 13 22 17 18 13 19 2 21 13 19 13 3 1 18 8
8 18 22 6 15 7 6 8 10 11 10 7 2 12 18 17 14 4 2 6 14 18 6 3 10 2 14 12 22 2
another awk
$ awk 'NR==FNR {a[$1]=$2; next}
{n=length($1);
for(i=1;i<=n;i++)
printf "%s", a[substr($1,i,1)] (i==n?ORS:OFS)}' mapfile datafile
3 4 4 4 4 19 18 22 6 13 10 19 2 8 6 18 20 16 1 10 6 19 3 2 19 20 3 18 18 1
22 6 13 10 19 2 8 6 18 20 16 1 10 6 19 3 2 19 20 3 18 18 1 22 17 13 12 10 13 17
13 19 20 18 10 3 4 19 6 13 21 18 6 8 3 8 19 14 10 13 13 21 8 13 16 2 3 13 16 6
6 8 3 8 19 14 10 13 13 21 8 13 16 2 3 13 16 6 18 21 10 13 19 7 1 11 10 14 20 2
12 19 3 4 14 10 11 4 10 11 2 19 17 19 20 20 13 19 10 18 3 13 19 13 18 22 3 3 18 10
10 11 2 19 17 19 20 20 13 19 10 18 3 13 19 13 18 22 3 3 18 10 13 16 16 13 6 10 3 10
16 13 6 10 3 10 22 21 14 10 16 17 17 14 20 18 10 2 6 10 13 10 21 13 22 17 18 13 19 2
14 10 16 17 17 14 20 18 10 2 6 10 13 10 21 13 22 17 18 13 19 2 21 13 19 13 3 1 18 8
8 18 22 6 15 7 6 8 10 11 10 7 2 12 18 17 14 4 2 6 14 18 6 3 10 2 14 12 22 2
however, there is no provision of missing mappings that are not specified, i.e. if you have chars not listed on the mapfile they will be ignored.
If the goal is encryption I'll propose a different approach:
First let's generate a mapping (or encryption key)
$ key=$(printf "%s\n" {A..Z} | shuf | paste -sd' ' | tr -d ' ')
$ echo "$key"
CNYSGFRDKQTOXJVLEWBAHZPMUI
now you can encrypt/decrypt your file contents, simply
$ tr [A-Z] "$key" < datafile > file.encrypted
and to reverse
$ tr "$key" [A-Z] < file.encrypted > file.decrypted
obviously, you need to save the key.
In MATLAB, we can use im2col and col2im to transform from columns to blocks and back, for example
>> A = floor(30*rand(4,6))
A =
8 5 2 13 15 11
22 11 27 13 24 24
5 18 23 9 23 15
20 23 14 15 19 10
>> B = im2col(A,[2 2],'distinct')
B =
8 5 2 23 15 23
22 20 27 14 24 19
5 18 13 9 11 15
11 23 13 15 24 10
>> col2im(B,[2 2],[4,6],'distinct')
ans =
8 5 2 13 15 11
22 11 27 13 24 24
5 18 23 9 23 15
20 23 14 15 19 10
my question is that: after using im2col with sliding mode
>> B = im2col(A,[2 2],'sliding')
B =
8 22 5 5 11 18 2 27 23 13 13 9 15 24 23
22 5 20 11 18 23 27 23 14 13 9 15 24 23 19
5 11 18 2 27 23 13 13 9 15 24 23 11 24 15
11 18 23 27 23 14 13 9 15 24 23 19 24 15 10
I wish to get a 4-by-6 matrix C from B(without knowing A) that the value at each site equals the original value multiple the times of sampling.
In other word, C(1,1)=A(1,1), C(1,2)=A(1,2)*2, C(2,2) = A(2,2)*4
Though we can easily implement with a for-loop, but the efficiency is critically low. So how to vectorize the implementation?
If I'm understanding correctly, you're desired output is
C = [ 8 10 4 26 30 11
44 44 108 52 96 48
10 72 92 36 92 30
20 46 28 30 38 10 ]
which I got by computing C = A.*S where
S = [ 1 2 2 2 2 1
2 4 4 4 4 2
2 4 4 4 4 2
1 2 2 2 2 1 ]_
The entries in S represent how many sliding blocks each entry is a member of.
I believe your question boils down to how to construct the matrix S.
Solution:
S = min(min(1:M,M:-1:1),x)'*min(min(1:N,N:-1:1),y)
C = A.*S
where A is size M-by-N, and your sliding block is size x-by-y.
Explanation:
In the given example, M=4, N=6, x=2, and y=2.
Notice the solution S can be written as the outer product of two vectors:
S = [1;2;2;1] * [1,2,2,2,2,1]
We construct each of these two vectors using the values of M,N,x,y:
min(1:M,M:-1:1)' == min(1:4,4:-1:1)'
== min([1,2,3,4], [4,3,2,1])'
== [1,2,2,1]'
== [1;2;2;1]
In this case, the extra min(...,x) does nothing since all entries are already <=x.
min(1:N,N:-1:1) == min(1:6,6:-1:1)
== min([1,2,3,4,5,6],[6,5,4,3,2,1])
== [1,2,3,3,2,1]
This time the extra min(...,y) does matter.
min(min(1:N,N:-1:1),y) == min([1,2,3,3,2,1],y)
== min([1,2,3,3,2,1],2)
== [1,2,2,2,2,1]
I'm using MESH2D in Matlab in order to mesh ROI (Region Of Interest) from images. Now I would like to make binary masks from these triangular meshes. The outputs from [p,t] = mesh2d(node) are:
p = Nx2 array of nodal XY co-ordinates.
t = Mx3 array of triangles as indicies into P, defined with a counter-clockwise node ordering.
Example of an initial code (feel free to improve it!):
mask= logical([0 0 0 0 0; 0 1 1 0 0; 0 1 1 1 1; 0 1 1 0 0]) %let's say this is my ROI
figure, imagesc(mask)
lol=regionprops(mask,'all')
[p,t] = mesh2d(lol.ConvexHull); %it should mesh the ROI
How to make masks from this triangular mesh?
Thank you in advance!
This is p:
1,50000000000000 2
1,50000000000000 2,50000000000000
1,50000000000000 3
1,50000000000000 3,50000000000000
1,50000000000000 4
1,93703949778653 2,56171771423604
1,96936200278303 3,98632617574682
2 1,50000000000000
2 4,50000000000000
2,00975325040940 3,53647067507122
2,01137717786904 2,05700769275495
2,05400996239344 3,03376821385856
2,41193753423879 2,49774899749798
2,45957145752038 3,46313210038859
2,50000000000000 1,50000000000000
2,50000000000000 4,50000000000000
2,51246316199066 3,99053096338726
2,56500321259084 1,97186739050944
2,64423955240966 2,98576823004855
3 1,50000000000000
3 4,50000000000000
3,00248771086621 2,47385860181019
3,01650848812758 3,52665319517610
3,08981230082503 3,98949609178151
3,12731558449295 2,02370031640169
3,36937385842331 2,99811446160210
3,50000000000000 1,75000000000000
3,50000000000000 4,25000000000000
3,85193739480358 3,46578962137238
3,85353024582881 2,53499308989903
4 2
4 4
4,42246720814684 3,00037409439956
4,50000000000000 2,25000000000000
4,50000000000000 3,75000000000000
4,97304775909580 2,99999314296989
5 2,50000000000000
5 3,50000000000000
5,50000000000000 3
and t:
9 5 7
20 18 15
1 8 11
8 15 11
11 15 18
11 2 1
6 2 11
20 27 25
25 18 20
27 30 25
17 10 14
7 10 17
24 21 17
9 7 17
29 35 32
26 30 29
23 19 26
14 19 23
26 29 23
23 29 24
23 17 14
24 17 23
6 11 13
13 11 18
34 30 31
31 30 27
3 2 6
12 19 14
14 10 12
6 13 12
12 13 19
12 3 6
28 21 24
28 29 32
24 29 28
9 17 16
16 17 21
38 35 33
35 29 33
33 29 30
34 37 33
33 30 34
19 13 22
26 19 22
18 25 22
22 13 18
22 30 26
22 25 30
4 7 5
4 10 7
4 12 10
3 12 4
38 33 36
36 33 37
39 38 36
36 37 39
To get the mask for the ix-th triangle, use:
poly2mask(p(t(ix,:),1),p(t(ix,:),2),width,height)
t is used to index n to get the data for one triangle.
I have a 21128x9 matrix in the following format:
x = ['Participant No.' 'yyyy' 'mm' 'dd' 'HH' 'MM' 'SS' 'question No.' 'response']
e.g.
x =
Columns 1 through 5
18 2011 10 26 15
18 2011 10 26 15
18 2011 10 26 15
18 2011 10 26 15
18 2011 10 26 15
19 2011 10 31 13
19 2011 10 31 13
19 2011 10 31 13
19 2011 10 31 13
19 2011 10 31 13
Columns 6 through 9
42 33 27 4
42 39 17 2
42 45 52 2
42 47 45 3
42 50 12 3
6 5 36 1
6 20 27 4
6 22 34 5
6 33 43 3
6 42 42 1
where columns 2-7 are date vectors.
The data are sorted by date/time.
I'd like to calculate the time taken to answer each question for each participant - i.e. the time elapsed between row 1 and 2, 2 and 3, 3 and 4, 4 and 5, and then 6 and 7, 7 and 8 etc. - to end up with a matrix, sorted by participant number, where I can then work out the mean time taken per question.
I've tried using the etime function, but to no avail.
EDIT: With regards to etime, just to see if it would work in practice, I tried to write:
etime(x(2,5:7),x(1,5:7))
to compare just columns 5-7 of rows 1 and 2, but i keep getting back:
??? Index exceeds matrix dimensions.
Error in ==> etime at 41
t = 86400*(datenummx(t1(:,1:3)) - datenummx(t0(:,1:3))) + ...
You were almost there! You needed to change the 5s to 2s, that's all:
etime(x(2,2:7),x(1,2:7))
Now to get them all lets make two matrices of the date vectors but one row out of synch with each other:
fisrt set up x:
x =[ 18 2011 10 26 15 42 33 27 4
18 2011 10 26 15 42 39 17 2
18 2011 10 26 15 42 45 52 2
18 2011 10 26 15 42 47 45 3
18 2011 10 26 15 42 50 12 3
19 2011 10 31 13 6 5 36 1
19 2011 10 31 13 6 20 27 4
19 2011 10 31 13 6 22 34 5
19 2011 10 31 13 6 33 43 3
19 2011 10 31 13 6 42 42 1]
now extract the times:
Tn = x(1:end-1, 2:7);
Tnplus1 = x(2:end, 2:7);
And no to get a vector of the difference in seconds between consecutive rows:
etime(Tnplus1, Tn)
Which results in:
ans =
6
6
2
3
422595
15
2
11
9
Also if you don't care about the year month day data just set them to zero i.e.
Tn(:, 1:3) = 0;
Tnplus1(:, 1:3) = 0;
etime(Tnplus1, Tn)
ans =
6
6
2
3
-9405
15
2
11
9
Here are some simple steps:
Calculate the difference between the two rows that you want to compare
Multiply with a vector that contains the number of seconds per unit
Small scale example:
% Hours Mins Secs:
difference = ([23 12 4] - [23 11 59]);
secvec = difference .* [3600 60 1];
secdiff = sum(secvec)
A simple MATLAB-problem:
coordinates=[1 6 ;9 20];
coordinates =
1 6
9 20
What i now want to have is:
idxList=[1 2 3 4 5 6 9 10 11 12 13 14 15 16 17 18 19 20];
idxList =
1 2 3 4 5 6 9 10 11 12 13 14 15 16 17 18 19 20
How i have to make that?
Here's one way:
>> cell2mat(cellfun(#(x) x(1):x(2), num2cell(coordinates, 2), 'UniformOutput', 0)')
ans =
1 2 3 4 5 6 9 10 11 12 13 14 15 16 17 18 19 20