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I want to check if a variable is already defined/exists in scala or not. Lets say a function called checkVar do this operation:
var x = 10
checkVar(x) -> returns boolean True
checkVar(y) -> returns boolean False
I am asking this question because I want to create a mechanism to define a variable if it doesn't exist.
Variables only exist at compile time so you can't dynamically create or delete variables at runtime. So both x and y must be defined at compile time or else the compiler will reject the code.
What you can do is use Option to indicate whether a variable has a value or not:
def checkVar(v: Option[Int]) = v.nonEmpty
var x = Some(10)
checkVar(x) // True
val y = None
checkVar(y) // False
x = None
checkVar(x) // False
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I have a few problem with my derivative because it shows me an error when I approach Denominator to zero .
func derivativeOf(fn: (Double) -> Double, atX x: Double) -> Double {
let h -> 0
return (fn(x + h) - fn(x))/h
}
i know my syntax sucks but currently it is common in calculus and mathematical Differential.
You can't express "number approaching zero" as let h -> 0 - this is just an invalid syntax in Swift.
There's also no specific operator for "number approaching zero". But depending on what you need, you could for example express "smallest possible positive number", using Double.leastNonzeroMagnitude:
let h = Double.leastNonzeroMagnitude
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I want to compute the square of mean-square for each element of l
l=[0.02817088, -0.74100320, -0.54062120, -0.24612808, 0.06945337, -0.58415690, -0.51238549,
-0.07862326, -0.42417337, -0.33482340, -0.21339753, -0.03890844, -0.59325371, 0.28154593,
-0.32133359,-0.13534792, 0.14060645, 0.32204972, 0.44438052, -0.21750973,-0.59107599,
-0.60809913]'
k= -0.2224834
sum(l-k)^2/22
I am not sure if sum(l-k)^2/22 is the sum of each (l[j]-k) for j=1,2,...,22?
ans = 2.4223e-14
I guess what you need might be
>> mean((l-k).^2)
ans = 0.10945
Data (You need ... for line continuation if you have data in different lines for l)
l=[0.02817088, -0.74100320, -0.54062120, -0.24612808, 0.06945337, -0.58415690, -0.51238549, ...
-0.07862326, -0.42417337, -0.33482340, -0.21339753, -0.03890844, -0.59325371, 0.28154593, ...
-0.32133359,-0.13534792, 0.14060645, 0.32204972, 0.44438052, -0.21750973,-0.59107599, ...
-0.60809913]'
k= -0.2224834
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For below loop and inner loop to express the performance in big o notation is :
O(N squared) as its performance is proportional to the square of the size of the input data set.
var counter = 0
var counterval = 0;
for ((key, value) <- m2.par){
for ((key2, value2) <- m2.par){
counter = counter + 1;
println(counter)
}
println(counterval)
}
Is this correct ?
Yes, if you consider the size of m2 to be the input size and that increasing counter and printing it are both O(1) (which is a very reasonable assumption).
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In the case of matrices the following command works fine for me every time I want to make sure that its contents are REAL NUMBERS, but always BIGGER THAN ZERO and FINITE. But, it doesn't work for datasets.
ispositive = ( ~isnumeric(batch_data) ...
| ~all(isfinite(batch_data(:))) ...
| ~isreal(batch_data) ...
| ~(any(batch_data(:) <= 0)) );
if (ispositive)
end
Any idea on how to modify it?
ispositive = ( ~isnumeric(batch_data) ...
| ~all(isfinite(batch_data(:))) ...
| ~isreal(batch_data) ...
| ~(any(batch_data(:) <= 0)) );
This does NOT do what you say.
According to this statement the following are positive:
batch_data = Inf
batch_data = -Inf
batch_data = 'ralph'
batch_data = 1j;
batch_data = -1j;
Related to what you wrote, this works:
positive = all(isnumeric(batch_data(:)) ...
&& all(isfinite(batch_data(:))) ...
&& isreal(batch_data) ... % isreal breaks convention of is* functions
&& all(batch_data(:) > 0)) ;