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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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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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let players = ["Greg", "Jenn", "Steve", "Anthony", "Krista", "Marti", "Erin", "Brandon",].shuffled()
I want to loop over the array and have it print out all pairs after being shuffled... so if the above was the outcome after being shuffled... it would print out
Greg, Jenn
Steve, Anthony
Krista, Marti
Erin, Brandon
you could use this:
if !players.isEmpty {
let arrTpl = stride(from: 1, to: players.count, by: 2).map { (players[$0-1], players[$0]) }
print("\(arrTpl)")
}
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In my while statement, I cannot understand why my output is printed twice ?
I would like to print i only one time, where is my error ?
func fetch2(){
var i: Int = 0
while i <= (self.returned-1) {
let itemLookUp = "https://shopping.yahooapis.jp/ShoppingWebService/V1/json/itemLookup?appid=\(self.appId)&itemcode=\(self.arrayCodeProduct[i])&responsegroup=large"
print(i)
i = i+1
}
}
Here is the output that I obtain :
0
1
2
3
0
1
2
3
Thank you in advance.
It looks like fetch2() is called twice.
Add a print(#function) before you var i and check that fetch2() is not called several times.
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I have the vector below:
v={'T','AT','AS','C'};
I would like to see all the possible permutations for this vector. To do so I can use the command below:
p=perms(v)
But I want to go one step further as each of the elements has sub index of 1 to 4, for example, T1,T2,T3,T4 .....C1,C2,C3,C4. I would like to have all the possible permutations with its sub index as see such results
T1,AT1,AS1,C1
C3,AT3,AS3,t3
AS2,AT2,C2,T2
.
.
.
Could you please help me how to do that?
Thanks
You can do this by first using ndgrid to generate a set of indices for all your possible combinations:
v = {'T1', 'AT1', 'AS1', 'C1'; ...
'T2', 'AT2', 'AS2', 'C2'; ...
'T3', 'AT3', 'AS3', 'C3'; ...
'T4', 'AT4', 'AS4', 'C4'};
[ind1, ind2, ind3, ind4] = ndgrid(1:4);
c = [v(ind1(:), 1) v(ind2(:), 2) v(ind3(:), 3) v(ind4(:), 4)];
And c will be a 256-by-4 cell array, as expected (44 combinations). Now you can expand each row by it's total number of permutations using perms like so:
p = perms(1:4);
p = reshape(c(:, p.').', 4, []).';
And p will be a 6144-by-4 cell array, also as expected (24 permutations times 256 combinations).
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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).