Say I have this matrix in memory and I want to calculate the 3D FFT
T =
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15
16 17 18 19
20 21 22 23
24 25 26 27
28 29 30 31
32 33 34 35
36 37 38 39
40 41 42 43
44 45 46 47
44 45 46 47
52 53 54 55
56 57 58 59
60 61 62 63
real(fft2(T))
ans =
2000 -32 -32 -32
-128 0 0 0
-112 0 0 0
-128 0 0 0
-144 0 0 0
-128 0 0 0
-112 0 0 0
-128 0 0 0
-144 0 0 0
-128 0 0 0
-112 0 0 0
-128 0 0 0
-144 0 0 0
-128 0 0 0
-112 0 0 0
-128 0 0 0
real(fftn(T))
ans =
2000 -32 -32 -32
-128 0 0 0
-112 0 0 0
-128 0 0 0
-144 0 0 0
-128 0 0 0
-112 0 0 0
-128 0 0 0
-144 0 0 0
-128 0 0 0
-112 0 0 0
-128 0 0 0
-144 0 0 0
-128 0 0 0
-112 0 0 0
-128 0 0 0
Why am I getting the same result? How 3D FFTs can be done in Matlab/Octave?
A 3D-FFT should be applied to a 3D-array. If you apply the 3D-FFT to a 2D-array you get the same result as a 2D-FFT, because there is no third dimension in the array.
Think about it this way: an N-dimensional FFT is just N 1-dimensional FFT's, one along each dimension. If there is no third dimension in the array, the FFT along that dimension does nothing.
Related
I have code as below to implement a sparse matrix:
close all
n=10;
p = 1/11;
term13 = -(1 / p^2);
term2 = (2 / p^2);
e=ones(n,1);
z=sparse(n,n);
for j=1:n
vval(j) = barrier(j*p);
z(j)=term2 + vval(j);
end
h = spdiags([term13*e z(j)*e term13*e], -1:1, n,n);
t=full(h)
and barrier is a function that has the value of 600 if 0.4<= j*p <=0.6, otherwise it is zero.
Naturally, I expect the matrix to be as bellow:
242 -121 0 0 0 0 0 0 0 0
-121 242 -121 0 0 0 0 0 0 0
0 -121 242 -121 0 0 0 0 0 0
0 0 -121 242 -121 0 0 0 0 0
0 0 0 -121 842 -121 0 0 0 0
0 0 0 0 -121 842 -121 0 0 0
0 0 0 0 0 -121 242 -121 0 0
0 0 0 0 0 0 -121 242 -121 0
0 0 0 0 0 0 0 -121 242 -121
0 0 0 0 0 0 0 0 -121 242
but surprisingly it has the form
242 -121 0 0 0 0 0 0 0 0
-121 242 -121 0 0 0 0 0 0 0
0 -121 242 -121 0 0 0 0 0 0
0 0 -121 242 -121 0 0 0 0 0
0 0 0 -121 242 -121 0 0 0 0
0 0 0 0 -121 242 -121 0 0 0
0 0 0 0 0 -121 242 -121 0 0
0 0 0 0 0 0 -121 242 -121 0
0 0 0 0 0 0 0 -121 242 -121
0 0 0 0 0 0 0 0 -121 242
To be exact, I expect h(5,5) = h(6,6) = 842 i.e. 242+600.
but it takes barrier =0.
Why is this value incorrect?
I'm not sure:
why the for loop is needed,
why you allocate a nxn sparse matrix z, and then use it with a single index z(j),
where vval is allocated, or used for that matter,
and why you create a sparse diagonal matrix, only to call full in the next line.
But regardless, the issue you asked about is caused by the line
h = spdiags([term13*e z(j)*e term13*e], -1:1, n,n);
which should be
h = spdiags([term13*e, z*e, term13*e], -1:1, n,n);
instead (note z instead of z(j)).
When I run the following command to parse haproxy logs, the output doesn't contain any headers, and I'm not able to understand the meanings of the numbers in each of the columns.
Command halog -pct < haproxy.log > percentiles.txt
the output that I see is:
0.1 3493 18 0 0 0
0.2 6986 25 0 0 0
0.3 10479 30 0 0 0
0.4 13972 33 0 0 0
0.5 17465 37 0 0 0
0.6 20958 40 0 0 0
0.7 24451 43 0 0 0
0.8 27944 46 0 0 0
0.9 31438 48 0 0 0
1.0 34931 49 0 0 0
1.1 38424 50 0 0 0
1.2 41917 51 0 0 0
1.3 45410 52 0 0 0
1.4 48903 53 0 0 0
1.5 52396 55 0 0 0
1.6 55889 56 0 0 0
1.7 59383 57 0 0 0
1.8 62876 58 0 0 0
1.9 66369 60 0 0 0
2.0 69862 61 0 0 0
3.0 104793 74 0 0 0
4.0 139724 80 0 1 0
5.0 174656 89 0 1 0
6.0 209587 94 0 1 0
7.0 244518 100 0 1 0
8.0 279449 106 0 1 0
9.0 314380 112 0 1 0
10.0 349312 118 0 1 0
15.0 523968 144 0 1 0
20.0 698624 168 0 1 0
25.0 873280 180 0 2 0
30.0 1047936 190 0 2 0
35.0 1222592 200 0 3 0
40.0 1397248 210 0 3 0
45.0 1571904 220 0 4 0
50.0 1746560 230 0 6 0
55.0 1921216 241 0 7 0
60.0 2095872 258 0 9 0
65.0 2270528 279 0 10 0
70.0 2445184 309 0 16 0
75.0 2619840 354 1 18 0
80.0 2794496 425 1 20 0
85.0 2969152 545 1 22 0
90.0 3143808 761 1 39 1
91.0 3178740 821 1 80 1
92.0 3213671 921 1 217 1
93.0 3248602 1026 1 457 1
94.0 3283533 1190 1 683 1
95.0 3318464 1408 1 889 1
96.0 3353396 1721 1 1107 1
97.0 3388327 2181 1 1328 1
98.0 3423258 2902 1 1555 1
98.1 3426751 3000 1 1580 1
98.2 3430244 3094 1 1607 1
98.3 3433737 3196 1 1635 1
98.4 3437231 3301 1 1666 1
98.5 3440724 3420 1 1697 1
98.6 3444217 3550 1 1731 1
98.7 3447710 3690 1 1770 1
98.8 3451203 3848 1 1815 1
98.9 3454696 4030 1 1864 1
99.0 3458189 4249 1 1923 2
99.1 3461682 4490 1 1993 2
99.2 3465176 4766 2 2089 2
99.3 3468669 5085 2 2195 2
99.4 3472162 5441 3 2317 97
99.5 3475655 5899 5 2440 365
99.6 3479148 6517 11 2567 817
99.7 3482641 7403 14 2719 1555
99.8 3486134 8785 16 2992 2779
99.9 3489627 11650 997 3421 4931
100.0 3493121 85004 4008 20914 71716
The first column looks to be the percentile, (like P50, P90, P99, etc) but the what are the values in the 2nd, 3rd, 4th, 5th and 6th columns? Also, are they total values (halog reports total times when provided with other options), or average values or maximum values?
<percentile> <request count> <Request Time*> <Connect Time**> <Response Time***> <Data Time****>
* Referred to as TR in the documentation.
** Referred to as Tc in the documentation.
*** Referred to as Tr in the documentation.
**** Referred to as Td in the documentation.
The source provides some good pointers.
If I have a cell containing 2 matrices:
A = {[10 0 0 10 0 20 15;
0 10 10 15 0 0 20;
10 0 0 12 11 0 0;
10 0 0 0 19 0 14;
18 13 0 0 0 15 0;
0 10 0 0 0 10 15]
[ 0 0 10 0 20 15;
10 10 15 0 0 20;
0 0 12 11 0 0;
0 0 0 19 0 14;
13 0 0 0 15 0;
11 0 13 0 15 0]}
B = [500 550 600 650 700 550]
I want to multiply each matrix with this vector in a way that 1st row of A is multiplied by 1st element of B, 2nd row of A with 2nd element of B, ..., and 6th row of A with 6th element of B. how to define this since A is a cell?
You can use cellfun to operate on each element in A separately, and bsxfun to multiply each row in the matrices in A by one element in B:
celltimes = #(A) bsxfun(#times,A,B.');
C = cellfun(celltimes,A,'UniformOutput',false)
the result:
>> celldisp(C)
C{1} =
Columns 1 through 4
5000 0 0 5000
0 5500 5500 8250
6000 0 0 7200
6500 0 0 0
12600 9100 0 0
0 5500 0 0
Columns 5 through 7
0 10000 7500
0 0 11000
6600 0 0
12350 0 9100
0 10500 0
0 5500 8250
C{2} =
Columns 1 through 4
0 0 5000 0
5500 5500 8250 0
0 0 7200 6600
0 0 0 12350
9100 0 0 0
6050 0 7150 0
Columns 5 through 6
10000 7500
0 11000
0 0
0 9100
10500 0
8250 0
As pointed by #Adriaan in the comments, in Matlab 2016b bsxfun is implicitly implemented, so you can just write:
C = cellfun(#(A) A.*B.',A,'UniformOutput',false);
I'm trying to change the following code so that the first matrix will become the second matrix:
function BellTri = matrix(n)
BellTri = zeros(n);
BellTri(1,1) = 1;
for i = 2:n
BellTri(i,1) = BellTri(i-1,i-1);
for j = 2:i
BellTri(i,j) = BellTri(i - 1,j-1) + BellTri(i,j-1);
end
end
BellTri
First matrix (when n = 7)
1 0 0 0 0 0 0
1 2 0 0 0 0 0
2 3 5 0 0 0 0
5 7 10 15 0 0 0
15 20 27 37 52 0 0
52 67 87 114 151 203 0
203 255 322 409 523 674 877
Second matrix
1 1 2 5 15 52 877
1 3 10 37 151 674 0
2 7 27 114 523 0 0
5 20 87 409 0 0 0
15 67 322 0 0 0 0
52 255 0 0 0 0 0
203 0 0 0 0 0 0
An option is to cyclically permute the columns using circshift.
function [BellTri, Second] = matrix(n)
BellTri = zeros(n);
BellTri(1,1) = 1;
for i = 2:n
BellTri(i,1) = BellTri(i-1,i-1);
for j = 2:i
BellTri(i,j) = BellTri(i - 1,j-1) + BellTri(i,j-1);
end
end
Second = BellTri;
for i = 1:n
Second(:, i) = circshift(Second(:,i), 1-i);
end
for i = n-1:-1:2
Second(1, i) = Second(1, i-1);
end
end
Input: [BellTri, Second] = matrix(7)
Output:
BellTri =
1 0 0 0 0 0 0
1 2 0 0 0 0 0
2 3 5 0 0 0 0
5 7 10 15 0 0 0
15 20 27 37 52 0 0
52 67 87 114 151 203 0
203 255 322 409 523 674 877
Second =
1 1 2 5 15 52 877
1 3 10 37 151 674 0
2 7 27 114 523 0 0
5 20 87 409 0 0 0
15 67 322 0 0 0 0
52 255 0 0 0 0 0
203 0 0 0 0 0 0
One approach:
out = zeros(size(A));
out(logical(fliplr(triu(ones(size(A,1)))))) = A(logical(tril(ones(size(A,1)))));
Note: As Divakar pointed out, there should be a typo in the first row. This method gives the corrected one.
Results:
A = [1 0 0 0 0 0 0;
1 2 0 0 0 0 0;
2 3 5 0 0 0 0;
5 7 10 15 0 0 0;
15 20 27 37 52 0 0;
52 67 87 114 151 203 0;
203 255 322 409 523 674 877];
>> out
out =
1 2 5 15 52 203 877
1 3 10 37 151 674 0
2 7 27 114 523 0 0
5 20 87 409 0 0 0
15 67 322 0 0 0 0
52 255 0 0 0 0 0
203 0 0 0 0 0 0
I would like to create a three dimensional surface using this:
>> a=X
a =
Columns 1 through 8
0 50 100 150 200 250 300 350
Columns 9 through 16
400 450 500 550 600 650 700 750
Columns 17 through 21
800 850 900 950 1000
>> b=Y
b =
0
50
100
150
200
250
300
350
400
>> c=Z
c =
Columns 1 through 8
0 0 0 0 0 0 0 0
16 32 67 98 127 164 194 234
120 171 388 773 1086 1216 1770 2206
189 270 494 1978 2755 3134 5060 10469
133 166 183 348 647 937 1446 2304
192 162 154 113 161 189 266 482
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
Columns 9 through 16
0 0 0 0 0 0 0 0
366 604 529 504 346 226 228 179
4027 11186 10276 5349 2560 1322 996 799
27413 76387 37949 15591 5804 2654 1803 1069
9844 24152 14772 4613 1777 849 459 290
1288 2623 1538 582 280 148 90 56
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
Columns 17 through 21
0 0 0 0 0
108 94 79 0 0
646 476 612 0 0
884 858 722 0 0
266 215 139 0 0
48 48 31 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
>> surf(X,Y,Z)
At the same time I would like to define that Z values < = 1803 will be shown with red color on the surface graph, 1803 < Z < 2755 yellow and Z > = 2755 green. The limits of the colorbar can be the min and max values of Z (from 0 to 76387). How can I set the ranges of the colorbar in order to get this result?
This will do as you ask:
%# add red (row 1), yellow (row 2), green (row 2)
map = [1 0 0; 0 1 1; 0 1 0];
%# set the new map as the current map
colormap(map);
colors = zeros(size(c)); %# create colors array
colors(c <= 1803) = 1; %# red (1)
colors(c > 1803 & c < 2755) = 2; %# yellow (2)
colors(c >= 2755) = 3; %# green (3)
%# and pass it into surf
surf(a,b,c, colors)