I am using SQL Server and I have an int that is 4 to 5 characters long.
I have a report that cast the first 3 digits as the location and last 1 to 2 digits as a cause.
So this is how they look
5142 = 514 = paint line 2 = paint to thin:
50528 = 505 = machining 28 = oblong hole:
SELECT [Suspect]
,left(Suspect,3) as SuspectOP
,Right(Suspect,2) as SuspectID
This query will return
5142 = SuspectOP = 514 SuspectID = 42
50528 = SuspectOP = 505 SuspectID = 28
So what i want is to read everything after the first 3 digits of the int.
Some of the things I have tried are as follows:
Select Cast(Suspect as Varchar(5)),
Substring(Suspect,3,2)
And
Select Suspect % 514 as SuspectID
Which does work as long as the first 3 digits are always 514 which in my case aren't.
You could use a conditional operators based on the length like this:
SELECT
[Suspect]
, SuspectOP = LEFT(Suspect,3)
, SuspectID = CASE
WHEN LEN(Suspect) = 5 THEN RIGHT(Suspect,2)
ELSE RIGHT(Suspect, 1)
END
Mind you, it's not ideal, you should really keep the values separate if your use case is like the one mentioned.
Related
I want to print by moving numbers from 1 to 22 using range (1,23)
Can you help me?
number = 1
numf = driver.find_elements_by_xpath('/html/body/div[1]/div/div/div[5]/div[2]/main/section/ul/li[3]/div[2]/div/div[2]/ul/li[%d]/button' %number)
for i in numf:
print(i.text)```
You can use either of the following solutions:
Using %s:
number = 1
numf = driver.find_elements_by_xpath("/html/body/div[1]/div/div/div[5]/div[2]/main/section/ul/li[3]/div[2]/div/div[2]/ul/li[%s]/button"% str(number))
Using format():
number = 1
numf = driver.find_elements_by_xpath("/html/body/div[1]/div/div/div[5]/div[2]/main/section/ul/li[3]/div[2]/div/div[2]/ul/li[{}]/button".format(str(number)))
Does anyone know why, using SQLServer 2005
SELECT CONVERT(DECIMAL(30,15),146804871.212533)/CONVERT(DECIMAL (38,9),12499999.9999)
gives me 11.74438969709659,
but when I increase the decimal places on the denominator to 15, I get a less accurate answer:
SELECT CONVERT(DECIMAL(30,15),146804871.212533)/CONVERT(DECIMAL (38,15),12499999.9999)
give me 11.74438969
For multiplication we simply add the number of decimal places in each argument together (using pen and paper) to work out output dec places.
But division just blows your head apart. I'm off to lie down now.
In SQL terms though, it's exactly as expected.
--Precision = p1 - s1 + s2 + max(6, s1 + p2 + 1)
--Scale = max(6, s1 + p2 + 1)
--Scale = 15 + 38 + 1 = 54
--Precision = 30 - 15 + 9 + 54 = 72
--Max P = 38, P & S are linked, so (72,54) -> (38,20)
--So, we have 38,20 output (but we don use 20 d.p. for this sum) = 11.74438969709659
SELECT CONVERT(DECIMAL(30,15),146804871.212533)/CONVERT(DECIMAL (38,9),12499999.9999)
--Scale = 15 + 38 + 1 = 54
--Precision = 30 - 15 + 15 + 54 = 84
--Max P = 38, P & S are linked, so (84,54) -> (38,8)
--So, we have 38,8 output = 11.74438969
SELECT CONVERT(DECIMAL(30,15),146804871.212533)/CONVERT(DECIMAL (38,15),12499999.9999)
You can do the same math if follow this rule too, if you treat each number pair as
146804871.212533000000000 and 12499999.999900000
146804871.212533000000000 and 12499999.999900000000000
To put it shortly, use DECIMAL(25,13) and you'll be fine with all calculations - you'll get precision right as declared: 12 digits before decimal dot, and 13 decimal digits after.
Rule is: p+s must equal 38 and you will be on safe side!
Why is this?
Because of very bad implementation of arithmetic in SQL Server!
Until they fix it, follow that rule.
I've noticed that if you cast the dividing value to float, it gives you the correct answer, i.e.:
select 49/30 (result = 1)
would become:
select 49/cast(30 as float) (result = 1.63333333333333)
We were puzzling over the magic transition,
P & S are linked, so:
(72,54) -> (38,29)
(84,54) -> (38,8)
Assuming (38,29) is a typo and should be (38,20), the following is the math:
i. 72 - 38 = 34,
ii. 54 - 34 = 20
i. 84 - 38 = 46,
ii. 54 - 46 = 8
And this is the reasoning:
i. Output precision less max precision is the digits we're going to throw away.
ii. Then output scale less what we're going to throw away gives us... remaining digits in the output scale.
Hope this helps anyone else trying to make sense of this.
Convert the expression not the arguments.
select CONVERT(DECIMAL(38,36),146804871.212533 / 12499999.9999)
Using the following may help:
SELECT COL1 * 1.0 / COL2
Using VB.Net
When i add a the number with zeros means, it is showing exact result without zero's
For Example
Dim a, b, c as int32
a = 001
b = 5
c = a + b
a = 009
b = 13
c = a + b
Showing output as 6 instead of 006, 22 instead of 022
Expected output
006
022
How to do this.
Need vb.net code help
You need to store a number as a string if you want to store the exact number of zeros. Then addition won't work though.
If you just want to display the number with 3 digits, you can store it as an integer and format the result when you print it.
c.ToString("D3")
zero is nothing.. If you do a regular mathematical calculation of 001 + 5 the result is still 6. I would suggest you check out string padding.
Given a NxN matrix and a (row,column) position, what is a method to select a different position in a random (or pseudo-random) order, trying to avoid collisions as much as possible?
For example: consider a 5x5 matrix and start from (1,2)
0 0 0 0 0
0 0 X 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
I'm looking for a method like
(x,y) hash (x,y);
to jump to a different position in the matrix, avoiding collisions as much as possible
(do not care how to return two different values, it doesn't matter, just think of an array).
Of course, I can simply use
row = rand()%N;
column = rand()%N;
but it's not that good to avoid collisions.
I thought I could apply twice a simple hash method for both row and column and use the results as new coordinates, but I'm not sure this is a good solution.
Any ideas?
Can you determine the order of the walk before you start iterating? If your matrices are large, this approach isn't space-efficient, but it is straightforward and collision-free. I would do something like:
Generate an array of all of the coordinates. Remove the starting position from the list.
Shuffle the list (there's sample code for a Fisher-Yates shuffle here)
Use the shuffled list for your walk order.
Edit 2 & 3: A modular approach: Given s array elements, choose a prime p of form 2+3*n, p>s. For i=1 to p, use cells (iii)%p when that value is in range 1...s-1. (For row-length r, cell #c subscripts are c%r, c/r.)
Effectively, this method uses H(i) = (iii) mod p as a hash function. The reference shows that as i ranges from 1 to p, H(i) takes on each of the values from 0 to p-1, exactly one time each.
For example, with s=25 and p=29 or 47, this uses cells in following order:
p=29: 1 8 6 9 13 24 19 4 14 17 22 18 11 7 12 3 15 10 5 16 20 23 2 21 0
p=47: 1 8 17 14 24 13 15 18 7 4 10 2 6 21 3 22 9 12 11 23 5 19 16 20 0
according to bc code like
s=25;p=29;for(i=1;i<=p;++i){t=(i^3)%p; if(t<s){print " ",t}}
The text above shows the suggestion I made in Edit 2 of my answer. The text below shows my first answer.
Edit 0: (This is the suggestion to which Seamus's comment applied): A simple method to go through a vector in a "random appearing" way is to repeatedly add d (d>1) to an index. This will access all elements if d and s are coprime (where s=vector length). Note, my example below is in terms of a vector; you could do the same thing independently on the other axis of your matrix, with a different delta for it, except a problem mentioned below would occur. Note, "coprime" means that gcd(d,s)=1. If s is variable, you'd need gcd() code.
Example: Say s is 10. gcd(s,x) is 1 for x in {1,3,7,9} and is not 1 for x in {2,4,5,6,8,10}. Suppose we choose d=7, and start with i=0. i will take on values 0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, which modulo 10 is 0, 7, 4, 1, 8, 5, 2, 9, 6, 3, 0.
Edit 1 & 3: Unfortunately this will have a problem in the two-axis case; for example, if you use d=7 for x axis, and e=3 for y-axis, while the first 21 hits will be distinct, it will then continue repeating the same 21 hits. To address this, treat the whole matrix as a vector, use d with gcd(d,s)=1, and convert cell numbers to subscripts as above.
If you just want to iterate through the matrix, what is wrong with row++; if (row == N) {row = 0; column++}?
If you iterate through the row and the column independently, and each cycles back to the beginning after N steps, then the (row, column) pair will interate through only N of the N^2 cells of the matrix.
If you want to iterate through all of the cells of the matrix in pseudo-random order, you could look at questions here on random permutations.
This is a companion answer to address a question about my previous answer: How to find an appropriate prime p >= s (where s = the number of matrix elements) to use in the hash function H(i) = (i*i*i) mod p.
We need to find a prime of form 3n+2, where n is any odd integer such that 3*n+2 >= s. Note that n odd gives 3n+2 = 3(2k+1)+2 = 6k+5 where k need not be odd. In the example code below, p = 5+6*(s/6); initializes p to be a number of form 6k+5, and p += 6; maintains p in this form.
The code below shows that half-a-dozen lines of code are enough for the calculation. Timings are shown after the code, which is reasonably fast: 12 us at s=half a million, 200 us at s=half a billion, where us denotes microseconds.
// timing how long to find primes of form 2+3*n by division
// jiw 20 Sep 2011
#include <stdlib.h>
#include <stdio.h>
#include <sys/time.h>
double ttime(double base) {
struct timeval tod;
gettimeofday(&tod, NULL);
return tod.tv_sec + tod.tv_usec/1e6 - base;
}
int main(int argc, char *argv[]) {
int d, s, p, par=0;
double t0=ttime(0);
++par; s=5000; if (argc > par) s = atoi(argv[par]);
p = 5+6*(s/6);
while (1) {
for (d=3; d*d<p; d+=2)
if (p%d==0) break;
if (d*d >= p) break;
p += 6;
}
printf ("p = %d after %.6f seconds\n", p, ttime(t0));
return 0;
}
Timing results on 2.5GHz Athlon 5200+:
qili ~/px > for i in 0 00 000 0000 00000 000000; do ./divide-timing 500$i; done
p = 5003 after 0.000008 seconds
p = 50021 after 0.000010 seconds
p = 500009 after 0.000012 seconds
p = 5000081 after 0.000031 seconds
p = 50000021 after 0.000072 seconds
p = 500000003 after 0.000200 seconds
qili ~/px > factor 5003 50021 500009 5000081 50000021 500000003
5003: 5003
50021: 50021
500009: 500009
5000081: 5000081
50000021: 50000021
500000003: 500000003
Update 1 Of course, timing is not determinate (ie, can vary substantially depending on the value of s, other processes on machine, etc); for example:
qili ~/px > time for i in 000 004 010 058 070 094 100 118 184; do ./divide-timing 500000$i; done
p = 500000003 after 0.000201 seconds
p = 500000009 after 0.000201 seconds
p = 500000057 after 0.000235 seconds
p = 500000069 after 0.000394 seconds
p = 500000093 after 0.000200 seconds
p = 500000099 after 0.000201 seconds
p = 500000117 after 0.000201 seconds
p = 500000183 after 0.000211 seconds
p = 500000201 after 0.000223 seconds
real 0m0.011s
user 0m0.002s
sys 0m0.004s
Consider using a double hash function to get a better distribution inside the matrix,
but given that you cannot avoid colisions, what I suggest is to use an array of sentinels
and mark the positions you visit, this way you are sure you get to visit a cell once.
I currently have two timelines (timeline1 and timeline2), with matching data (data1 and data2). Timelines almost, but not quite match (about 90% of common values).
I'm trying to find values from data1 and data2 that correspond to identical timestamps (ignoring all other values)
My first trivial implementation is as follows (and is obviously terribly slow, given that my timelines contain thousands of values). Any ideas on how to improve this? I'm sure there is a smart way of doing this while avoiding the for loop, or the find operation...
% We expect the common timeline to contain
% 0 1 4 5 9
timeline1 = [0 1 4 5 7 8 9 10];
timeline2 = [0 1 2 4 5 6 9];
% Some bogus data
data1 = timeline1*10;
data2 = timeline2*20;
reconstructedData1 = data1;
reconstructedData2 = zeros(size(data1));
currentSearchPosition = 1;
for t = 1:length(timeline1)
% We only look beyond the previous matching location, to speed up find
matchingIndex = find(timeline2(currentSearchPosition:end) == timeline1(t), 1);
if isempty(matchingIndex)
reconstructedData1(t) = nan;
reconstructedData2(t) = nan;
else
reconstructedData2(t) = data2(matchingIndex+currentSearchPosition-1);
currentSearchPosition = currentSearchPosition+matchingIndex;
end
end
% Remove values from data1 for which no match was found in data2
reconstructedData1(isnan(reconstructedData1)) = [];
reconstructedData2(isnan(reconstructedData2)) = [];
You can use Matlab's intersect function:
c = intersect(A, B)
Couldn't you just call INTERSECT?
commonTimeline = intersect(timeline1,timeline2);
commonTimeline =
0 1 4 5 9
You need to use the indexes returned from intersect.
[~ ia ib] = intersect(timeline1, timeline2);
recondata1 = data1(ia);
recondata2 = data2(ib);