Nope, I'll elaborate the result after the select. I need directly the SUM for each hour from output SQL select!
Maybe using grouping (and SUM) by partial, but I don't know how to implement yet!
---------- Post updated at 11:48 AM ---------- Previous update was at 11:40 AM ----------
Well, I found myself a solution, it was so simple!!! :-)
Code:
SELECT data, SUM(`tot`) as SUM, COUNT(tot) As HOUR FROM `data` GROUP BY MID(data, 8,2)
how would I write a command line that creates a new file named stuff.txt in the current working directory which contains the number of directories in the current working directory, followed by the number of empty files in the current working directory, followed by the name of the working directory? (3 Replies)
Hi
i data looks like this:
student 1
Subject1 45 55
Subject2 44 55
Subject3 33 44
//
student 2
Subject1 45 55
Subject2 44 55
Subject3 33 44
i would like to sum $2, $3 (marks) and divide each entry in $2 and $3 with their respective sums and print for each student as $4 and... (2 Replies)
Hi,
I have an array of strings. Each string has 4 comma separated values. I am binding this array to a SQL where I am required to do an INSERT after grouping.
The binding is done as :
$insertADWSth->bind_param_array(1,A_CONSTANT_STRING);... (1 Reply)
How do you do grouping in grep? Here's how I tried it at first:
egrep 'qualit(y|ies)' /usr/share/dict/words
-bash: syntax error near unexpected token `('
I'm using GNUgrep, and I found this on their site. grep regular expression syntax
So I tried this:
egrep 'qualit\(y\|ies\)'... (2 Replies)
Hi all,
I am using following command:
perl program.pl input.txt output.txt CUTOFF 3 > groups_3.txt
containing program.pl, two files (input.txt, output.txt) and getting output in groups_3.txt:
But, I wish to have 30 files corresponding to each CUTOFF ranging from 0 to 30 using the same... (1 Reply)
Hi Gurus,
I would like to do in a script something that is really easy in sql.
Basically, I have a file with a fomat like the following:
name1_test 501
name 1 33 510
test2 900
name1_test 300
So, I would like to group them as... (3 Replies)
awk 'FNR==NR {a; next} $NF in a' genes.txt refseq_exons.txt > output.txt
I can not figure out how to group the same name in $4 together.
Basically, all the SKI together in separate rows and all the TGFB2. Thank you :).
chr1 2160133 2161174 SKI
chr1 218518675 218520389 TGFB2... (1 Reply)
Hello,
I would like to group/sort a file of records by a particular field and then count how many records belong in that grouping.
For example say I have the following data:
1234|"ZZZ"|"Date"|"1"|"Y"|"ABC"|""|AA
ABCD|"ZZZ"|"Date"|"1"|"Y"|"ABC"|""|AA
EFGH|"ZZZ"|"Date"|"1"|"Y"|"ABC"|""|BB... (14 Replies)
Discussion started by: Nik44
14 Replies
LEARN ABOUT SUNOS
mlib_imagenormcrosscorrel
mlib_ImageNormCrossCorrel(3MLIB)mlib_ImageNormCrossCorrel(3MLIB)NAME
mlib_ImageNormCrossCorrel - normalized cross correlation
SYNOPSIS
cc [ flag... ] file... -lmlib [ library... ]
#include <mlib.h>
mlib_status mlib_ImageNormCrossCorrel(mlib_d64 *correl, const mlib_image *img1, const mlib_image *img2, const mlib_d64 *mean2, const
mlib_d64 *sdev2);
The mlib_ImageNormCrossCorrel() function computes the normalized cross-correlation coefficients between a pair of images, on a per-channel
basis.
It uses the following equations:
w-1 h-1
SUM SUM (d1[x][y][i] * d2[x][y][i])
x=0 y=0
correl[i] = -------------------------------------
s1[i] * s2[i]
d1[x][y][i] = img1[x][y][i] - m1[i]
d2[x][y][i] = img2[x][y][i] - m2[i]
1 w-1 h-1
m1[i] = ----- * SUM SUM img1[x][y][i]
w*h x=0 y=0
1 w-1 h-1
m2[i] = ----- * SUM SUM img2[x][y][i]
w*h x=0 y=0
w-1 h-1
s1[i] = sqrt{ SUM SUM (img1[x][y][i] - m1[i])**2 }
x=0 y=0
w-1 h-1
s2[i] = sqrt{ SUM SUM (img2[x][y][i] - m2[i])**2 }
x=0 y=0
where w and h are the width and height of the images, respectively; m1 and m2 are the mean arrays of the first and second images, respec-
tively; s1 and s2 are the un-normalized standard deviation arrays of the first and second images, respectively.
In usual cases, the normalized cross-correlation coefficient is in the range of [-1.0, 1.0]. In the case of (s1[i] == 0) or (s2[i] == 0),
where a constant image channel is involved, the normalized cross-correlation coefficient is defined as follows:
#define signof(x) ((x > 0) ? 1 : ((x < 0) ? -1 : 0))
if ((s1[i] == 0.) || (s2[i] == 0.)) {
if ((s1[i] == 0.) && (s2[i] == 0.)) {
if (signof(m1[i]) == signof(m2[i]) {
correl[i] = 1.0;
} else {
correl[i] = -1.0;
}
} else {
correl[i] = -1.0;
}
}
The two images must have the same type, the same size, and the same number of channels. They can have 1, 2, 3 or 4 channels. They can be
of type MLIB_BYTE, MLIB_SHORT, MLIB_USHORT or MLIB_INT.
If (mean2 == NULL) or (sdev2 == NULL), then m2 and s2 are calculated in this function according to the formulas shown above. Otherwise,
they are calculated as follows:
m2[i] = mean2[i];
s2[i] = sdev2[i] * sqrt(w*h);
where mean2 and sdev2 can be the output of mlib_ImageMean() and mlib_ImageStdDev(), respectively.
The function takes the following arguments:
correl Pointer to normalized cross correlation array on a channel basis. The array must be the size of channels in the images.
correl[i] contains the cross-correlation of channel i.
img1 Pointer to first image.
img2 Pointer to second image.
mean2 Pointer to the mean array of the second image.
sdev2 Pointer to the standard deviation array of the second image.
The function returns MLIB_SUCCESS if successful. Otherwise it returns MLIB_FAILURE.
See attributes(5) for descriptions of the following attributes:
+-----------------------------+-----------------------------+
| ATTRIBUTE TYPE | ATTRIBUTE VALUE |
+-----------------------------+-----------------------------+
|Interface Stability |Evolving |
+-----------------------------+-----------------------------+
|MT-Level |MT-Safe |
+-----------------------------+-----------------------------+
mlib_ImageAutoCorrel(3MLIB), mlib_ImageAutoCorrel_Fp(3MLIB), mlib_ImageCrossCorrel(3MLIB), mlib_ImageCrossCorrel_Fp(3MLIB), mlib_ImageNorm-
CrossCorrel_Fp(3MLIB), attributes(5)
23 May 2005 mlib_ImageNormCrossCorrel(3MLIB)
01-04-2012
