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sql rounding up to a multiple of 5?
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Closed 11 months ago.
I am trying a number of combinations in order to get to "the next highest number starting with 0" from a particular integer.
I've tried ROUND, CEILING and FLOOR and I can not get any of them to work.
Examples.
17 = 20
11 = 20
204 = 210
1001 = 1010
107 = 110
So basically any value should then return the next 0 valued number AFTER it.
Any help would be greatly appreciated. Thanks
assuming all is int's then you can divide and multiply with 10 to do it
(#i+10)/10*10
Since ceiling only works on 10^0 level, you have to divide, then ceiling, then multiply:
select 10*ceiling(#value/10)
Related
In Kdb+, how do I use the "roll" function to make the random numbers generated fall within a range that doesn't start with 0? For example what if I wanted the range to be within 2-10 instead of 0-10?
What do I have to add to the code to make it fall into a range instead of the default 0-x? I have tried and looked for every method but can't seem to find one.
You could also just roll from 0-8 then add two. This doesn't require a list to be pre-generated
q)2+5?9
10 2 7 10 7
Assuming you want 2-10 inclusive
// quick and simple method
q)10?2+til 8
6 2 4 3 4 3 4 5 4 7
// or function (x)=num to be dealt, (y) start range, (z) end range
q)f:{x?y+til 1+z-y}
q)f[10;10;20]
12 17 10 11 19 12 11 18 18 11
If you supply a list in the right hand argument then you will get a random value from that list. To roll for a random range from 2-10 you can use til to generate the range:
q)2+til 9
2 3 4 5 6 7 8 9 10
q)1?2+til 9
,6
You can even supply a general list to randomly draw from:
q)3?(`abc;2 3f;10;20;30;"text")
2 3f
`abc
"text"
Simple math function for random number generator is:
(rand() mod (1+max- min)) + min
q) f:{x+rand[0] mod 1+y-x}
q) f[5;10]
q) 7
Update: I failed to notice that you wanted to generate couple of random numbers in the range. You could easily modify above function for that:
q) f:{x+(z?0) mod 1+y-x}
q) f[2;10;4]
q) 6 4 7 2
Converting from base-10 to base-8 is a little trickier,
but still straightforward. We basically have to reverse the process from above. Let's start with an example: 150 of base-10.
We first find the largest power of 8 that is smaller than our number. Here, this is 82 or 64 (83 is 512). We count how many groups of 64 we can take from 150. This is 2, so the first digit in our base-8 number is 2. We have now accounted for 128 out of 150, so we have 22 left over.
The largest power of 8 that is smaller than 22 is 81 (that is, 8). How many groups of 8 can we take from 22? Two groups again, and thus our second digit is 2.
Finally, we are left with 6, and can obviously take 6 groups of one from this, our final digit. We end up with 226 of base-8.
In fact, we can make this process a touch clearer with math. Here are the steps:
150/82 = 2 remainder 22
22/81 = 2 remainder 6
6/80 = 6
Our final answer is then all of our non-remainder digits, or 226.
my question is same way if we want 65 of base-10 to convert to base-8 what happens ?
any help is appreciated..thanks in advance
Still new to the programing game but I need a little help! I'm not exactly sure how to describe what I want to do but I'll give it my best shot. I have a set of numbers produced by an algorithm I've put together. e.g. :
....
10 10 10
11 11 11
12 1 2
13 3 4
14 12 13
15 6 7
16 5 15
17 8 9
....
Essentially what I want to do is assign these index numbers to groups. Lets say I start with the number 14 in the first column. It is going to belong to group 1, so I label it in a new column in row 14 "1" for group one. The second and the third column show other index numbers that are grouped with the index 14. So I use a code like:
FindLHS = find(matrix(:,1)==matrix(14,2));
and
FindRHS = find(matrix(:,1)==matrix(14,3));
so clearly this will produce the results of
FindLHS = 12
FindRHS = 13
I will then proceed to label both 12 and 13 as belonging to group "1" as I did for 14
now my problem is I want to do this same procedure for both 12 and 13 of finding and labelling the indexs for 12 and 13 being (1,2) and (3,4). Is there a way to repeat that code for both idx of 1,2,3 and 4? because the real dataset has over 5000 data points in it...
Do you understand what I mean?
Thanks
James
All you really want to do is find wherever matrix(:,1) contains one of the numbers you've already found, include the numbers in the second and third columns into your group list (presuming they aren't already there), and stop when that list stops growing, right? This may not be the most efficient way of doing it but it gives you the basic idea:
while ~(numel(oldnum)==numel(num))
oldnum = num;
idx = ismember(matrix(:,1),oldnum)
num = unique(matrix(idx,:))
end
Output:
num =
1
2
3
4
12
13
14
Now if your first column is literally just your numbers 1 through 5000 in order, you don't need to even find the index, you can just use your number list directly.
To do this for multiple groups you would just need an outer loop that stores the information for each group, then picks out the next unused number. I'm presuming that your individual groups are consistent so that no matter which of those numbers you pick you end up with the same result - e.g. starting at 2 or 14 gives you the same result (if not, it becomes more complex).
This question already has answers here:
Assembly Random Number within Range using Easy 68K (68000)
(2 answers)
Closed 8 years ago.
I am doing my assignment and get stuck at this problem. Please help me. I just can create a random number from 0 to 74 .
Create a random number 0-9 and then add 65
I have an array with size m = 11 and my hash function is Division method : h(k) = k mod m
I have an integer k = 10 and 10 mod 11 is -1 so where should I put this key in the array? I should put this key in the slot which its index is 10?
please help me thanks
EDITED : for getting my answer well for example I have integers like k = 10,22,31,4,15,28,17,88,59
the array would be like this?thanks
10 9 8 7 6 5 4 3 2 1 0 index
10 31 59 17 28 4 15 88 22 keys
As it's usually done, 10 mod 11 is 10, so yes, you'd normally use index 10.
Edit: To generalize: at least as it's normally defined, given two positive inputs, a modulo will always produce a positive result. As such, your questions about what to do with negative results don't really make sense with respect to the normal definition.
If you really do have the possibility of getting a negative result, my immediate reaction would be to switch to some language that will produce a reasonable result. If you can't do that, then you'd probably want to move the value into the correct range by adding m to the negative number until you get a number in the range [0..m) so it fits the normal definition of mod, then use that as your index.