Output of this strange loop related to matrices - linear-algebra

Let us consider the following pseudocode:
int n=n;
int A[][]
scanf(A[][],%d);
for i=1:n;i++
{
x=A[i][i]
for j=1:n;j++
{
if x<A[i][j]
a=x;
x=A[i][j];
A[i][i]=x;
A[i][j]=a;
return A[][]
I am fumbling on this pseudo code.the question, I think is just that the diagonal entries are compared and exchanged for the greatest entries. But, will the output depend on the entries of the matrix or will be independent of it is my main question. Specifically, is there any general formula for the output? Is it dependent on the type of matrix A I think it should some power of A. Any hints? Thanks beforehand.

You could just write your code on any language you love.
n = 3
A = [[1,2,3], [3,5,6], [7,8,9]]
for i in range(n):
x=A[i][i]
for j in range(n):
a = None
if x < A[i][j]:
a = x
x=A[i][j]
A[i][i]=x
A[i][j]=a
print (A)
Gives you:
[[3, 1, 2], [None, 6, 3], [None, 7, None]]
But, will the output depend on the entries of the matrix or will be
independent of it is my main question.
Ofc it depends. Your can see the initial data in the output. That means output depends on data.
Specifically, is there any general formula for the output?
I believe NO, but I cant mathematically prove. Just look at Nones appear in output. I hardly imagine such formula.
Is it dependent on the type of matrix A I think it should some power
of A.
What is 'type of matrix' ?

Related

Problem with extracting values from vector for for loop

I am trying to extract values from a vector to generate random numbers from a GEV distribution. I keep getting an error. This is my code
x=rand(Truncated(Poisson(2),0,10),10)
t=[]
for i in 1:10 append!(t, maximum(rand(GeneralizedExtremeValue(2,4,3, x[i])))
I am new to this program and I think I am not passing the variable x properly. Any help will be appreciated. Thanks
If I am correctly understanding what you are trying to do, you might want something more like
x = rand(Truncated(Poisson(2),0,10),10)
t = Float64[]
for i in 1:10
append!(t, max(rand(GeneralizedExtremeValue(2,4,3)), x[i]))
end
Among other things, you were missing a paren, and probably want max instead of maximum here.
Also, while it would technically work, t = [] creates an empty array of type Any, which tends to be very inefficient, so you can avoid that by just telling Julia what type you want that array to hold with e.g. t = Float64[].
Finally, since you already know t only needs to hold ten results, you can make this again more efficient by pre-allocating t
x = rand(Truncated(Poisson(2),0,10),10)
t = Array{Float64}(undef,10)
for i in 1:10
t[i] = max(rand(GeneralizedExtremeValue(2,4,3)), x[i])
end

Get the mapping from each element of input to the bin of the histogram in Julia

Matlab's [n,mapx] = histc(x, bin_edged) returns the counts of x in each bin as n and returns a map, which is the same length of x which is the bin index that each element of x was placed into.
I can do the same thing in Julia as follows:
Using StatsBase
x = rand(1000)
bin_e = 0:0.1:1
h = fit(Histogram, x, bin_e)
yx = map((z) -> findnext(z.<=h.edges[1],1),x) .- 1
Is this the "right way" to do this? It seem a bit kludgy.
Inspired by this python question you should be able to define a small function that delivers the desired mapping (modulo conventions):
binindices(edges, data) = searchsortedlast.(Ref(edges), data)
Note that the bin edges are sorted and we can use seachsortedlast to get the last bin edge smaller or equal than a datapoint. Broadcasting this over all of the data we obtain the mapping. Note that the Ref(edges) indicates that edges is a scalar under broadcasting (that means that the full array is considered in each call).
Although conceptionally identical to your solution, this approach is about 13x faster on my machine.
I filed an issue over at StatsBase.jl's github page suggesting to add this as a feature.
After looking through the code for Histogram.jl I found that they already included a function binindex. So this solution is probably the best:
x = 0:0.001:10
h1 = fit(Histogram,x,0:10,closed=left)
xmap1 = StatsBase.binindex.(Ref(h1), x)
h2 = fit(Histogram,x,0:10,closed=right)
xmap2 = StatsBase.binindex.(Ref(h2), x)
I stumbled across this question when I was trying to figure out how many occurrences of each value I had in a list of values. If each value is in its own bin (as for categorical data, or integer data with a small number of unique values), this is what one would be plotting in a histogram.
If that is what you want, then countmap() in StatBase package is just what you need.

fixing race condition in tensorflow run

I would like to maintain a variable on the GPU, and perform some operations on that variable in place. The following snippet is a minimalish example of this.
import numpy as np
import tensorflow as tf
with tf.Graph().as_default():
i = tf.placeholder(tf.int32, [4], name='i')
y = tf.placeholder(tf.float32, [4], name='y')
_x = tf.get_variable('x', [4], initializer=tf.random_normal_initializer())
x = _x + tf.reduce_sum(tf.mul(_x,y))
assign_op = tf.assign(_x, x).op
permute_op = tf.assign(_x, tf.gather(_x, i))
ii = np.array([1,2,3,0])
yy = np.random.randn(4)
s = tf.Session()
s.run(tf.initialize_all_variables())
xxx0 = s.run(_x)
s.run([permute_op, assign_op], feed_dict={i: ii, y: yy})
xxx1 = s.run(_x)
print('assigned then permuted', np.allclose((xxx0+np.dot(xxx0,yy))[ii], xxx1))
print('permuted then assigned', np.allclose((xxx0[ii]+np.dot(xxx0[ii], yy)), xxx1))
The problem is that this program is ambiguous, in terms of the ordering of the assign_op and permute_op operations. Hence, one or the other of the final two print statements will be true, but which one that is varies randomly across multiple runs of the program. I could break this into two steps, the first running the permute_op and the second running the assign_op, but it seems this will be less efficient.
Is there an efficient way of breaking the race condition, and making the results predictable?
The easiest way to order the two assignments is to use the result of the first assignment as the variable input to the second one. This creates a data dependency between the assignments, which gives them a deterministic order. For example:
assigned = tf.assign(_x, x)
permuted = tf.assign(assigned, tf.gather(assigned, i))
sess.run(permuted.op) # Runs both assignments.
Note that I reversed the order of the permutation and assignment operations from what you said in your question, because doing the permutation first and then updating still has a race. Even if this isn't the semantics you wanted, the principle should hopefully be clear.
An alternative approach is to use with tf.control_dependencies(ops): blocks, where ops is a list of operations (such as assignments) that must run before the operations in the with block. This is slightly trickier to use, because you have to be careful about reading the updated value of a variable. (Like a non-volatile variable in C, the read may be cached.) The typical idiom to force a read is to use tf.identity(var.ref()), so the example would look something like:
assign_op = tf.assign(_x, x).op
with tf.control_dependencies([assign_op]):
# Read updated value of `_x` after `assign_op`.
new_perm = tf.gather(tf.identity(_x.ref()), i)
permute_op = tf.assign(_x, new_perm).op
sess.run(permute_op) # Runs both assignments.

using argmax or something simpler in R

I am trying to set up a Gibbs sampler in R where I update my value at each step.
I have a function in R that I want to maximise for 2 values; my previous value and a new one.
So I know the maximum outcome from the function applied to both values. But then how do I select the best input without doing it manually? (I need to do a lot of iterations). Here is an idea of the code and the variables:
g0<-function(k){sample(0:1,k,replace=T)}
this is a k dimensional vector with entries 1 or 0 uniformly. Initial starting point for my chain. If i=1 then include the i'th variable in the design matrix.
X1 design matrix
Xg<-function(g){
Xg<-cbind(X1[,1]*g[1],X1[,2]*g[2],X1[,3]*g[3],X1[,4]*g[4],X1[,5]*g[5],X1[,6]*g[6],X1[,7]*g[7])
return(Xg[,which(!apply(Xg,2,FUN = function(x){all(x == 0)}))])
}
Xg0<-Xg(g0)
reduced design matrix for g0
c<-1:100000
mp<-function(g){
mp<-sum((1/(c*(c+1)^-((q+1)/2)))*
(t(Y)%*%Y-(c/(c+1))*t(Y)%*%Xg(g)%*%solve(t(Xg(g))%*%Xg(g))%*%t(Xg(g))%*%Y)^(-27/2))
return(mp)
}
this is my function.
Therefore if I have mp(g) and mp(g*), for 2 inputs g and g*, such that the max is mp(g*) how can I return g*?
Thanks for any help and if you have any queries just ask. sorry about the messy code as well; I have not used this site before.
Like this:
inputs <- list(g, g2)
outputs <- sapply(inputs, mp)
best.input <- inputs[which.max(outputs)]

what does accessing zero element in R do?

if I have a vector a<-c(3, 5, 7, 8)
and run a[1], not surprisingly I will get 3
but if I will run a[0] I basically get numeric(0)
What does this mean?
And what does this do?
How can I use it for normal reasons?
Others have answered what x[0] does, so I thought I'd expand on why it's useful: generating test cases. It's great for making sure that your functions work with unusual data structure variants that users sometimes produce accidentally.
For example, it makes it easy to generate 0 row and 0 column data frames:
mtcars[0, ]
mtcars[, 0]
These can arise when subsetting goes wrong:
mtcars[mtcars$cyl > 10, ]
But in your testing code it's useful to flag that you're doing it deliberately.
http://cran.r-project.org/doc/manuals/r-release/R-lang.html#Indexing-by-vectors
As you can see it says: A special case is the zero index, which has null effects: x[0] is an empty vector and otherwise including zeros among positive or negative indices has the same effect as if they were omitted.

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