PureScript contains a method in the Integer library fromNumber.
Here is an example of how it might be used:
myInteger = fromMaybe 0 (fromNumber myNumber)
However the docs provide this puzzling explanation:
Creates an Int from a Number value. The number must already be an integer and fall within the valid range of values for the Int type otherwise Nothing is returned.
Basically, your number must already be an Integer to convert it to an integer.
Assuming your number is not already an integer, a reasonable use case, how would you convert it to a number?
If it's not already an integer, there is no one true way of converting it to an integer. You could round to the nearest integer, round up, round down, do banker's rounding, or some sort of crazy conversion scheme of your own.
The Data.Int module offers several functions for different conversion strategies, such as floor, ceil, and round.
Related
I am trying to convert a double value to int and use it in the number of agents per arrival. I have a distribution but want it to round to the nearest integer. I looked up the anylogic math functions but the only one that makes sense is rint but that still returns a double.
you can do this:
(int) rint(yourVariable)
Another way is to use (int)floor(myDouble)) and (int)ceil(myDouble)) .
With these, you can decide if you want to round up or down (if the default rounding rules aren't to your liking)
In PostgreSQL, I would like to store signed values -999.9 - 9999.9.
Can I use numeric(5.1) for this?
Or what type should I use?
You can certainly use the arbitrary precision type numeric with a precision of 5 and a scale of 1, just like #Simon commented, but without the syntax error. Use a comma(,) instead of the dot (.) in the type modifier:
SELECT numeric(5,1) '-999.9' AS nr_lower
, numeric(5,1) '9999.9' AS nr_upper;
nr_lower | nr_upper
----------+----------
-999.9 | 9999.9
The minus sign and the dot in the string literal do not count against the allowed maximum of significant digits (precision).
If you don't need to restrict the length, just use numeric.
If you need to enforce minimum and maximum, add a check constraint:
CHECK (nr_column BETWEEN -999.9 AND 9999.9)
numeric stores your number exactly. If you don't need the absolute precision and tiny rounding errors are no problem, you might also use one of the floating point types double precision (float8) or real (float4).
Or, since you only allow a single fractional decimal digit, you can multiply by 10 and use integer, which would be the most efficient storage: 4 bytes, no rounding errors and fastest processing. Just use and document the number properly.
Details for numeric types in the manual.
Here is my code:
println(Double(2/5))
When I run this, it prints out
0.0
How can I fix this? I want it to come out to 0.4. It there some issue with the rounding?
The problem is that you're not converting to a Double until after you've done integer division between two integers. Let's take a look at order of operations. We start at the inside and move outward.
Perform integer division between the integer 2 and the integer 5, which results in the integer 0.
Create a double from the integer 0, which creates the double 0.0.
Call description on the double 0.0, which returns the string "0.0"
Call println on the string "0.0"
We can fix this by calling the Double constructor on each side of the division before we divide them.
println((Double(2)/Double(5)))
Now the order of operations is:
Convert the integer 2 to the floating point 2.0
Convert the integer 5 to the floating point 5.0
Perform floating point division between these floating point numbers, resulting in 0.4
Call description on the floating point number 0.4, which returns the string "0.4".
Call println on the string "0.4".
Note that it's not strictly necessary to convert both sides of the division to Double.
And as long as we're dealing with literals, we can just write println(2.0/5.0).
We could also get away with writing println((2 * 1.0)/5) which should now interpret all of our literals as floating point (as we've multiplied it by a floating point).
As long as either side of a math operating is a floating point type, the integer literal will be interpreted as a floating point type by Swift, but in my opinion, it's far better to explicitly convert our types so that we're excruciatingly clear on exactly what we want to happen. So let's get all of our numbers into the same type and be explicitly clear what we actually want.
If we're dealing with literals, we can add .0 to them to force them as floating point numbers:
println(2.0/5.0)
If we're doing with variables, we can use a constructor:
let myTwoInt: Int = 2
let myFiveInt: Int = 5
println((Double(myTwoInt)/Double(myFiveInt))
I think your issue is that you are dividing two integers which normally will return an integer.
I had a similar issue in java, adding a .0 to one or the other integers or converting either to a double by using the double function should fix it.
It's a feature of typed languages that creates a result of the same type as the values being divided.
Digits is correct about the cause; instead of the approach you're taking, try this:
print(2.0 / 5.0)
Given that 8-byte doubles can represent all 4-byte ints precisely, I'm wondering whether dividing a double A storing an int, by a double B storing an int (such that the integer B divides A) will always give the exact double corresponding to the integer that is their quotient? So, if B and C are integers, and B*C fits within a 32-bit int, then is it guaranteed that
int B,C = whatever s.t. B*C does not overflow 32-bit int
double(B*C)/double(C) == double((B*C)/C) ?
Does the IEEE754 standard guarantee this?
In my testing, it seems to work for all examples I've tried. In Python:
>>> (321312321.0*3434343.0)/321312321.0 == 3434343.0
True
The reason for asking is that Matlab makes it hard to work with ints, so I often just use the default doubles for integer calculations. And when I know that the integers are exactly divisible, and if I know that the answer to the present question is yes, then I could avoid doing casts to ints, idivide(..) etc., which is less readable.
Luis Mendo's comment does answer this question, but to specifically address the use in Matlab there are some handy utilities described here. You can use eps(numberOfInterest) to find the distance to the next largest double-precision floating point number. For example:
eps(1) = 2^(-52)
eps(2^52) = 1
This practically guarantees that mathematical operations with integers held in a double will be precise provided they don't overflow 2^52, which is quite a bit larger than what is held in a 32-bit int type.
I want to use xor for my double numbers in matlab,but bitxor is only working for int numbers. Is there a function that could convert double to int in Matlab?
The functions You are looking for might be: int8(number), int16(number), uint32(number) Any of them will convert Double to an Integer, but You must pick the best one for the result You want to achieve. Remember that You cannot cast from Double to Integer without rounding the number.
If I understood You correcly, You could create a function that would simply remove the "comma" from the Double number by multiplying your starting value by 2^n and then casting it to Integer using any of the functions mentioned earlier, performing whatever you want and then returning comma to its original position by dividing the number by 2^n
Multiplying the starting value by 2^n is a hack that will decrease the rounding error.
The perfect value for n would be the number of digits after the comma if this number is relatively small.
Please also specify, why are You trying to do this? This doesn't seem to be the optimal solution.
You can just cast to an integer:
a = 1.003
int8(a)
ans =
1
That gives you an 8 bit signed integer, you can also get other size i.e. int16 or else unsigned i.e. uint8 depending on what you want to do