What does the following code do:
rnorm(10, mean=2, sd=1:10)
The first number is from N(2,1)
The second number if from N(2,2)
The third number is from N(2,3)
etc...?
The first argument tells R how many random variates you want returned. In this case, it will give you back 10 values. Those values will be drawn from normal distributions with mean equal to 2. In addition, all 10 values will be drawn from distributions with different standard deviations, the first with SD=1, the second 2, ..., the 10th SD=10. Perhaps the thing to understand is that R, by its nature, is vectorized. That is, there is no such thing as a scalar, only a vector of length=1. (I recognize that that doesn't make a lot of sense within pure math, but it does in computer science.) As a result, arguments are often 'recycled' so that they will all match the length of the longest vector, i.e., you end up with a vector of 10 means, each equal to 2, to match your vector of 10 SDs. HTH.
Related
I am supposed to find the mean and standard deviation at each given sample size (N), using the "FOR LOOP". I started writing the code as below, I am required to save all the means into vector "p". How do I save all the means into one vector?
sample.sizes =c(3,10,50,100,500,1000)
mean.sds = numeric(0)
for ( N in sample.sizes ){
x <- rnorm(3,mean=0,sd=1)
mean.sds[i]
}
mean(x)
Actually you are doing many thing wrong?
If you are using variable N in for loop, you are not using it anywhere
for (N in 'some_vector') actually means N will take that value one by one. So N in sample sizes will first take, 3 then 10 then 50 and so on.
Now where does i come into picture?
You are calculating x for each iteration of N. In fact you are not using N anywhere in the loop?
first x will return 3 values. In the next line you intend to store these three values in just ith value of mean.sds where i is unknown and storing three values into one value, as it is, is not logically possible.
Do you want this?
sample.sizes =c(3,10,50,100,500,1000)
mean.sds = numeric(0)
for ( i in seq_along(sample.sizes )){
x <- rnorm(sample.sizes[i], mean=0, sd=1)
mean.sds[i] <- mean(x)
}
mean.sds
[1] 0.6085489531 -0.1547286299 0.0052106559 -0.0452804986 -0.0374094936 0.0005667246
I replaced N with seq_along(sample.sizes) which will give iterations equal to the number of that vector. Six in this example.
I passed each ith element to first argument of rnorm to generate these many random values.
Stored each random value into single vector. calculated its mean (one value only) and stored in ith value of your empty vector.
I have a vector that is being filled with random numbers within this range [0,1]. I want to somehow accept only the vectors, in which an element inside of it has a maximum deviation of 0,02 from its previous one and its next one.
For example I have the below vector [3,1]. This is acceptable, because the deviation of the 2nd element, between the first and the third element is not bigger than 0,02. Vector is not always consisted of 3 rows, it could be more.
**Vector**
0.32957
0.33097
0.33946
This is what i thought:
n=4
P=rand(1,n);
sort(P,"ascend");
for L=2:n
while P(L-1)-P(L)>0.02
P=rand(1,n);
endwhile
endfor
Vectorize this!
isvalid=~any(diff(sort(a))>0.02);
sort(a) : if its not sorted, sort
diff() : take the difference between adjacent elements
___ >0.02: Check if any of those differences is bigger than what you accept
~any(): if any is bigger, then return zero, "not valid".
From your code, it seems that there may be more to the question than what you ask, you seem to have the XY problem. You want to create a random vector that has the properties that you describe. You seem to be using uniform random numbers, so let me propose a way to generate your vector where your conditions are always true.
a(1)=rand(1); %or any other way to generate a first value.
length=100; %desired length.
a(2:length)=rand(length-1,1)*0.02; %generate random numbers never bigger than 0.02
a=cumsum(a); %cumulative sum
This ensures the vector is increasing in value, and never increasing more than 0.02
I have a double vector:
r = -50 + (50+50)*rand(10,1)
Now i want to ideally have all the numbers in the vector equal upto a tolerance of say 1e-4. I want to represent each r with a scalar say s(r) such that its value gives an idea of the quality of the vector. The vector is high quality if all elements in the vector are equal-like. I can easily run a for loop like
for i=1:10
for j=i+1:10
check equality upto the tolerance
end
end
But even then i cannot figure what computation to do inside the nested for loops to assign a scalar representing the quality . Is there a better way such that given any vector r length n, i can quickly calculate a scalar representing the quality of the vector.
Your double-loop algorithm is somewhat slow, of order O(n**2) where n is the number of dimensions of the vector. Here is a quick way to find the closeness of the vector elements, which can be done in order O(n), just one pass through the elements.
Find the maximum and the minimum of the vector elements. Just use two variables to store the maximum and minimum so far and run once through all the elements. The difference between the maximum and the minimum is called the range of the values, a commonly accepted measure of dispersion of the values. If the values are exactly equal, the range is zero which shows perfect quality. If the range is below 1e-4 then the vector is of acceptable quality. The bigger the range, the worse the equality.
The code is obvious for just about any given language, so I'll leave that to you. If the fact that the range only really considers the two extreme values of the vector bothers you, you could use other measures of variation such as the interquartile range, variance, or standard deviation. But the range seems to best fit what you request.
In this question Getting N random numbers that the sum is M, the object was to generate a set of random numbers that sums to a specific number N. After reading this question, I started playing around with the idea of generating sets of numbers that satisfy this condition
sum(A) == sum(B) && sum(B) == sum(A * B)
An example of this would be
A <- c(5, 5, -10, 6, 6, -12)
B <- c(5, -5, 0, 6, -6, 0)
In this case, the three sums equal zero. Obviously, those sets aren't random, but they satisfy the condition. Is there a way to generate 'random' sets of data that satisfy the above condition? (As opposed to using a little algorithm as in the above example.)
(Note: I tagged this as an R question, but the language really doesn't matter to me.)
You'd need to define the first vector in n-dimensional space, and the 2nd one will have N-2 degrees of freedom (i.e. random numbers) since the sum and one angle are already determined.
The 2nd vector would need to be transformed into N-dimensional space; There are infinitely many transforms that could work, so if you don't care about the probability distribution of the resulting vectors, just choose the one that's most intuitive to you.
There's a nice geometrical interpretation to the first constraint: it constrains the 2nd vector to a (hyper-)plane in N-dimensional space; the 2nd constraint doesn't have a simple geometric interpretation.
check out hyperspherical cooridnates.
You can generate one set completely randomly. And generate randomly all numbers in set B except for two numbers. Since you have two equations you should be able to solve for those two numbers.
Hello I am new to R and I can't find the way to do exactly what I want to. I have a vector of x numbers, and what i want to do is order it in increasing order, and then start making subtractions like this (let's say the vecto has 100 numbers for example):
[x(100)-x(99)]+[x(99)-x(98)]+[x(98)-x(97)]+[x(97)-x(96)]+...[x(2)-x(1)]
and then divide all that sum by the number of elements the vector has, in this case 100.
The only thing that I am able to do at the moment is order the vector with:
sort(nameOfTheVector)
Sorry for my bad English.
diff returns suitably lagged and iterated differences. In your case you want the default single lag. sum will return the sum any arguments passed to it, so....
sum(diff(sort(nameOfTheVector))) / length(nameOfTheVector)