I'm having trouble replicating the mod function in SQL sever.
In excel, mod (-3, 7) = 4. But in SQL, -3 % 7 = -3
Am I using % wrong, or does SQL do mod differently?
This will give a result between 0 and n - 1 for both positive and negative values of x:
((x % n) + n) % n
Well, modular arithmetic is done on equivalence classes of integers, so neither Excel nor any RDBMS is "doing % wrong". If you want a representative between 0 and 6, though, you can always do
select (-3 % 7) + 7;
Related
Is there any way to filter rows of a table where a numeric column contains a digit using maths?
I mean, currently, I'm solving that using:
where cast(t.numeric_column as varchar(255)) like "%2%"
However, I would like to know if could be possible to filter apply numeric operations...
Any ideas?
You could use division plus the modulus, if you knew the range of possible numbers. For example, assuming all expected numbers were positive and less than 100,000, you could use:
SELECT *
FROM yourTable
WHERE numeric_column % 10 = 2 OR
(numeric_column / 10) % 10 = 2 OR
(numeric_column / 100) % 10 = 2 OR
(numeric_column / 1000) % 10 = 2 OR
(numeric_column / 10000) % 10 = 2;
Although the above is ugly and unwieldy, it might actually outperform your approach which requires a costly conversion to string.
I have this code I need to explain and the Mod operator has confused me, MSDN has a page on it and I cannot understand it clearly. I have included a section of code below if you could refer to it in your answer, thanks.
number1 = (input1/ 10) - 0.5
number2 = input2 Mod 10
Result = number1 + number2
Mod in VB.NET is the Modulo operation. It returns the remainder when one number is divided by another.
For example, if you divided 4 by 2, your mod result would be 0 (no remainder). If you divided 5 by 2, your mod result would be 1.
Explanation of the Modulo operation
I am super confused what the percentage sign does in Objective C. Can someone explain to me in language that an average idiot like myself can understand?! Thanks.
% is the modulo operator, so for example 10 % 3 would result in 1.
If you have some numbers a and b, a % b gives you just the remainder of a divided by b.
So in the example 10 % 3, 10 divided by 3 is 3 with remainder 1, so the answer is 1.
If there is no remainder to a divided by b, the answer is zero, so for example, 4 % 2 = 0.
Here's a relevant SO question about modular arithmetic.
Same as what it does in C, it's "modulo" (also known as integer remainder).
% is the modulo operator. It returns the remainder of <number> / <number>. For example:
5 % 2
means 5 / 2, which equals 2 with a remainder of 1, so, 1 is the value that is returned. Here's some more examples:
3 % 3 == 0 //remainder of 3/3 is 0
6 % 3 == 0 //remainder of 6/3 is 0
5 % 3 == 2 //remainder of 5/3 is 2
15 % 4 == 3 //remainder of 15/4 is 3
99 % 30 == 9 //remainder of 99/30 is 9
The definition of modulo is:
mod·u·lo
(in number theory) with respect to or using a modulus of a specified number. Two numbers are congruent modulo a given number if they give the same remainder when divided by that number.
In Mathematics, The Percentage Sign %, Called Modulo (Or Sometimes The Remainder Operator) Is A Operator Which Will Find The Remainder Of Two Numbers x And y. Mathematically Speaking, If x/y = {(z, r) : y * z + r = x}, Where All x, y, and z Are All Integers, Then
x % y = {r: ∃z: x/y = (z, r)}. So, For Example, 10 % 3 = 1.
Some Theorems And Properties About Modulo
If x < y, Then x % y = x
x % 1 = 0
x % x = 0
If n < x, Then (x + n) % x = n
x Is Even If And Only If x % 2 = 0
x Is Odd If And Only If x % 2 = 1
And Much More!
Now, One Might Ask: How Do We Find x % y? Well, Here's A Fairly Simple Way:
Do Long Division. I Could Explain How To Do It, But Instead, Here's A Link To A Page Which Explains Long Division: https://www.mathsisfun.com/numbers/long-division-index.html
Stop At Fractions. Once We Reach The Part Where We Would Normally Write The Answer As A Fraction, We Should Stop. So, For Example, 101/2 Would Be 50.5, But, As We Said, We Would Stop At The Fractions, So Our Answer Ends Up Being 50.
Output What's Left As The Answer. Here's An Example: 103/3. First, Do Long Division. 103 - 90 = 13. 13 - 12 = 1. Now, As We Said, We Stop At The Fractions. So Instead Of Continuing The Process And Getting The Answer 34.3333333..., We Get 34. And Finally, We Output The Remainder, In This Case, 1.
NOTE: Some Mathematicians Write x mod y Instead Of x % y, But Most Programming Languages Only Understand %.
I'm trying to make a generic equation which converts a value. Here are some examples.
9,873,912 -> 9,900,000
125,930 -> 126,000
2,345 -> 2,400
280 -> 300
28 -> 30
In general, x -> n
Basically, I'm making a graph and I want to make values look nicer. If it's a 6 digit number or higher, there should be at least 3 zeros. If it's a 4 digit number or less, there should be at least 2 digit numbers, except if it's a 2 digit number, 1 zero is fine.
(Ignore the commas. They are just there to help read the examples). Anyways, I want to convert a value x to this new value n. What is an equation g(x) which spits out n?
It is for an objective-c program (iPhone app).
Divide, truncate and multiply.
10**x * int(n / 10**(x-d))
What is "x"? In your examples it's about int(log10(n))-1.
What is "d"? That's the number of significant digits. 2 or 3.
Ahhh rounding is a bit awkward in programming in general. What I would suggest is dividing by the power of ten, int cast and multiplying back. Not remarkably efficient but it will work. There may be a library that can do this in Objective-C but that I do not know.
if ( x is > 99999 ) {
x = ((int)x / 1000) * 1000;
}
else if ( x > 999 ) {
x = ((int) x / 100) * 100;
}
else if ( x > 9 ) {
x = ((int) x / 10) * 10;
}
Use standard C functions like round() or roundf()... try man round at a command line, there are several different options depending on the data type. You'll probably want to scale the values first by dividing by an appropriate number and then multiplying the result by the same number, something like:
int roundedValue = round(someNumber/scalingFactor) * scalingFactor;
So I thought that negative numbers, when mod'ed should be put into positive space... I cant get this to happen in objective-c
I expect this:
-1 % 3 = 2
0 % 3 = 0
1 % 3 = 1
2 % 3 = 2
But get this
-1 % 3 = -1
0 % 3 = 0
1 % 3 = 1
2 % 3 = 2
Why is this and is there a workaround?
result = n % 3;
if( result < 0 ) result += 3;
Don't perform extra mod operations as suggested in the other answers. They are very expensive and unnecessary.
In C and Objective-C, the division and modulus operators perform truncation towards zero. a / b is floor(a / b) if a / b > 0, otherwise it is ceiling(a / b) if a / b < 0. It is always the case that a == (a / b) * b + (a % b), unless of course b is 0. As a consequence, positive % positive == positive, positive % negative == positive, negative % positive == negative, and negative % negative == negative (you can work out the logic for all 4 cases, although it's a little tricky).
If n has a limited range, then you can get the result you want simply by adding a known constant multiple of 3 that is greater that the absolute value of the minimum.
For example, if n is limited to -1000..2000, then you can use the expression:
result = (n+1002) % 3;
Make sure the maximum plus your constant will not overflow when summed.
We have a problem of language:
math-er-says: i take this number plus that number mod other-number
code-er-hears: I add two numbers and then devide the result by other-number
code-er-says: what about negative numbers?
math-er-says: WHAT? fields mod other-number don't have a concept of negative numbers?
code-er-says: field what? ...
the math person in this conversations is talking about doing math in a circular number line. If you subtract off the bottom you wrap around to the top.
the code person is talking about an operator that calculates remainder.
In this case you want the mathematician's mod operator and have the remainder function at your disposal. you can convert the remainder operator into the mathematician's mod operator by checking to see if you fell of the bottom each time you do subtraction.
If this will be the behavior, and you know that it will be, then for m % n = r, just use r = n + r. If you're unsure of what will happen here, use then r = r % n.
Edit: To sum up, use r = ( n + ( m % n ) ) % n
I would have expected a positive number, as well, but I found this, from ISO/IEC 14882:2003 : Programming languages -- C++, 5.6.4 (found in the Wikipedia article on the modulus operation):
The binary % operator yields the remainder from the division of the first expression by the second. .... If both operands are nonnegative then the remainder is nonnegative; if not, the sign of the remainder is implementation-defined
JavaScript does this, too. I've been caught by it a couple times. Think of it as a reflection around zero rather than a continuation.
Why: because that is the way the mod operator is specified in the C-standard (Remember that Objective-C is an extension of C). It confuses most people I know (like me) because it is surprising and you have to remember it.
As to a workaround: I would use uncleo's.
UncleO's answer is probably more robust, but if you want to do it on a single line, and you're certain the negative value will not be more negative than a single iteration of the mod (for example if you're only ever subtracting at most the mod value at any time) you can simplify it to a single expression:
int result = (n + 3) % 3;
Since you're doing the mod anyway, adding 3 to the initial value has no effect unless n is negative (but not less than -3) in which case it causes result to be the expected positive modulus.
There are two choices for the remainder, and the sign depends on the language. ANSI C chooses the sign of the dividend. I would suspect this is why you see Objective-C doing so also. See the wikipedia entry as well.
Not only java script, almost all the languages shows the wrong answer'
what coneybeare said is correct, when we have mode'd we have to get remainder
Remainder is nothing but which remains after division and it should be a positive integer....
If you check the number line you can understand that
I also face the same issue in VB and and it made me to forcefully add extra check like
if the result is a negative we have to add the divisor to the result
Instead of a%b
Use: a-b*floor((float)a/(float)b)
You're expecting remainder and are using modulo. In math they are the same thing, in C they are different. GNU-C has Rem() and Mod(), objective-c only has mod() so you will have to use the code above to simulate rem function (which is the same as mod in the math world, but not in the programming world [for most languages at least])
Also note you could define an easy to use macro for this.
#define rem(a,b) ((int)(a-b*floor((float)a/(float)b)))
Then you could just use rem(-1,3) in your code and it should work fine.