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Assuming I have a vector of say four dimensions in which every variable lays in a special interval. Thus we got:
Vector k = (x1,x2,x3,x4) with x1 = (-2,2), x2 = (0,2), x3 = (-4,1), x4 = (-1,1)
I am only interested in the points constraint by the intervals.
So to say v1 = (0,1,2,0) is important where v2 = (-5,-5,5,5) is not.
In additon to that the point i+1 should be relatively close to point i among my journey. Therefore I dont want to jump around in space.
Is there a proper way of walking through those interesting points?
For example in 2D space with x1,x2 = (-2,2) like so:
Note: The frequenz of the red line could be higher
There are many ways to create a space-filling curve while preserving closeness. See the Wikipedia article for a few examples (some have associated algorithms for generating them): https://en.wikipedia.org/wiki/Space-filling_curve
Regardless, let's work with your zig-zag pattern for 2D and work on extending it to 3D and 4D. To extend it into 3D, we just add another zig to the zig-zag. Take a look at the (rough) diagram below:
Essentially, we repeat the pattern that we had in 2D but we now have multiple layers that represent the third dimension. The extra zig that we need to add is the switch between bottom-to-top and top-to-bottom every layer. This is pretty simple to abstract:
In 2D, we have x and y axes.
We move across the x domain switching between positive and negative
directions most frequently.
We move across the y domain once.
In 3D, we have x, y, and z axes.
We move across the x domain switching between positive and negative directions most frequently.
We move across the y domain switching between positive and negative directions second most frequently.
We move across the z domain once.
It should be clear how this generalizes to higher dimensions. Now, I'll present some (Python 3) code that implements the zig-zag pattern for 4D. Let's represent the position in 4D space as (x, y, z, w) and the ranges in each dimension as (x0, x1), (y0, y1), (z0, z1), (w0, w1). These are our inputs. Then, we also define xdir, ydir, and zdir to keep track of the direction of the zig-zag.
x, y, z, w = x0, y0, z0, w0
xdir, ydir, zdir = +1, +1, +1
for iw in range(w1 - w0):
for iz in range(z1 - z0):
for iy in range(y1 - y0):
for ix in range(x1 - x0):
print(x, y, z, w)
x = x + xdir
xdir = -xdir
print(x, y, z, w)
y = y + ydir
ydir = -ydir
print(x, y, z, w)
z = z + zdir
zdir = -zdir
print(x, y, z, w)
w = w + 1
This algorithm has the guarantee that no two points printed out after each other have a distance greater than 1.
Using recursion, you can clean this up to make a very nice generalizable method. I hope this helps; let me know if you have any questions.
With the work of #Matthew Miller I implemented this generalization for any given multidimenisonal space:
'''assuming that we take three points out of our intervals [0,2] for a,b,c
which every one of them is corresponding to one dimension i.e. a 3D-space'''
a = [0,1,2]
b = [0,1,2]
c = [0,1,2]
vec_in = []
vec_in.append(a)
vec_in.append(b)
vec_in.append(c)
result = []
hold = []
dir = [False] * len(vec_in)
def create_points(vec , index, temp, desc):
if (desc):
loop_x = len(vec[index])-1
loop_y = -1
loop_z = -1
else:
loop_x = 0
loop_y = len(vec[index])
loop_z = 1
for i in range(loop_x,loop_y,loop_z):
temp.append(vec[index][i])
if (index < (len(vec) - 1)):
create_points(vec, index + 1, temp, dir[index])
else:
u = []
for k in temp:
u.append(k)
result.append(u)
temp.pop()
if (dir[index] == False):
dir[index] = True
else:
dir[index] = False
if len(temp) != 0:
temp.pop()
#render
create_points(vec_in, 0, hold, dir[0])
for x in (result):
print(x)
The result is a journey which covers every possible postion in a continous way:
[0, 0, 0]
[0, 0, 1]
[0, 0, 2]
[0, 1, 2]
[0, 1, 1]
[0, 1, 0]
[0, 2, 0]
[0, 2, 1]
[0, 2, 2]
[1, 2, 2]
[1, 2, 1]
[1, 2, 0]
[1, 1, 0]
[1, 1, 1]
[1, 1, 2]
[1, 0, 2]
[1, 0, 1]
[1, 0, 0]
[2, 0, 0]
[2, 0, 1]
[2, 0, 2]
[2, 1, 2]
[2, 1, 1]
[2, 1, 0]
[2, 2, 0]
[2, 2, 1]
[2, 2, 2]
I have two arrays that I want to iterate over at the same time.
I'm using this:
julia> xs = [1,2,3];
julia> ys = [4,5,6];
julia> for i in 1:length(xs)
x = xs[i]
y = ys[i]
#show x, y
end
(x, y) = (1, 4)
(x, y) = (2, 5)
(x, y) = (3, 6)
Is there a better way to iterate over multiple arrays in Julia?
Use zip along with tuple destructuring:
julia> xs = [1,2,3];
julia> ys = [4,5,6];
julia> for (x, y) in zip(xs, ys)
#show x, y
end
(x, y) = (1, 4)
(x, y) = (2, 5)
(x, y) = (3, 6)
zip will stop iteration at the shortest array:
julia> for (x, y) in zip([1,2], [0,0,0])
#show x, y
end
(x, y) = (1, 0)
(x, y) = (2, 0)
This pattern can be generalized to an arbitrary number of lists:
julia> for (x, y, z) in zip([1,2], [3,4], [5,6])
#show x, y, z
end
(x, y, z) = (1, 3, 5)
(x, y, z) = (2, 4, 6)
One possibility consists in using the eachindex function: if it is given multiple Array-like arguments, it will return a iterable set of indices suitable to iterate on all arguments at once.
This is useful in particular in the following situations:
when you need to use the index itself (for example because you don't only need to access the elements of the collections, but also set some of them), or
when you want to check that both arrays indeed have the same number of elements (this might or might not be a desired property depending on your use case).
Example 1: using the index itself to fill the first array with values coming from the second
julia> x = [1,2,3];
julia> y = [4,5,6];
julia> #inbounds for i in eachindex(x, y)
x[i] = 2*y[i]
end
julia> x
3-element Array{Int64,1}:
8
10
12
Example 2: check that the arrays have the same range
julia> x = [1,2];
julia> y = [4,5,6];
julia> #inbounds for i in eachindex(x, y)
x[i] = 2*y[i]
end
ERROR: DimensionMismatch("all inputs to eachindex must have the same indices, got [1, 2] and [1, 2, 3]")
Example 3: note that eachindex generalizes well for multi-dimensional arrays too.
julia> x = zeros(2, 3);
julia> y = ones(2, 3);
julia> #inbounds for i in eachindex(x, y)
x[i] = 2*y[i]
end
julia> x
2×3 Array{Float64,2}:
2.0 2.0 2.0
2.0 2.0 2.0
You can iterate over multiple collections using map and foreach. For example, with map:
julia> x, y = 1:3, 4:6;
julia> map(hypot, x, y)
3-element Array{Float64,1}:
4.123105625617661
5.385164807134504
6.708203932499369
For more complicated multi-line anonymous functions, you can use do-block syntax:
julia> xs, ys = 1:4, 10:10:40;
julia> map(xs, ys) do x, y
if isodd(x)
x + y
else
x * y
end
end
4-element Array{Int64,1}:
11
40
33
160
foreach is very similar to map, but is intended for use when a function is applied for its side effect, like printing or plotting, rather than its return value. An example with foreach:
julia> x, y = ["a", "b", "c"], 1:3;
julia> foreach(println ∘ ^, x, y)
a
bb
ccc
Note the use of the function composition operator in the foreach call.
Using Prolog:
Write a predicate dispnth to display the nth element of a list. You may assume that the input list always has n or more elements.
For Example:
?- dispnth([1, [2, 3], 4, 5], 2, X). should return X = [2, 3]
I have this so far:
dispnth([X|_], 0, X).
dispnth([_|Xs], N, X) :-
dispnth(N1, X, Xs),
N is N1 + 1.
First let's give the predicate a more descriptive name, say list_nth_element/3. Next you might like to consider an auxiliary predicate list_nth_element_/4 with an additional argument, that holds the current position. From your given example I assume that you start counting at 1, so that's going to be the start value for the fourth argument. Then the predicates might look something like this:
list_nth_element(L,N,E) :-
list_nth_element_(L,N,E,1).
list_nth_element_([X|Xs],N,X,N). % if the 2nd and 4th elements are equal X is the nth element
list_nth_element_([_X|Xs],N,E,P0) :- % if the 2nd and 4th arguments
dif(P0,N), % differ
P1 is P0+1, % increment current position
list_nth_element_(Xs,N,E,P1). % and recurse
So essentially the fourth argument is used as a position indicator that is being incremented until you reached the desired position. However, there is no need to have this additional argument in the actual predicates interface, so it is "hidden" in the auxiliary predicate's interface.
Querying this predicate yields your desired result:
?- list_nth_element([1, [2, 3], 4, 5], 2, X).
X = [2,3] ? ;
no
You can also ask things like Which element is at what position?
?- list_nth_element([1, [2, 3], 4, 5], N, X).
N = X = 1 ? ;
N = 2,
X = [2,3] ? ;
N = 3,
X = 4 ? ;
N = 4,
X = 5 ? ;
no
How can I include an outer product (of the previous feature vector and itself) as a layer in chainer, especially in a way that's compatible with batching?
F.matmul is also very handy.
Depending on the input shapes, you can combine it with F.expand_dims (of course F.reshape works, too) or use transa/transb arguments.
For details, refer to the official documentation of functions.
Code
import chainer.functions as F
import numpy as np
print("---")
x = np.array([[[1], [2], [3]], [[4], [5], [6]]], 'f')
y = np.array([[[1, 2, 3]], [[4, 5, 6]]], 'f')
print(x.shape)
print(y.shape)
z = F.matmul(x, y)
print(z)
print("---")
x = np.array([[[1], [2], [3]], [[4], [5], [6]]], 'f')
y = np.array([[[1], [2], [3]], [[4], [5], [6]]], 'f')
print(x.shape)
print(y.shape)
z = F.matmul(x, y, transb=True)
print(z)
print("---")
x = np.array([[1, 2, 3], [4, 5, 6]], 'f')
y = np.array([[1, 2, 3], [4, 5, 6]], 'f')
print(x.shape)
print(y.shape)
z = F.matmul(
F.expand_dims(x, -1),
F.expand_dims(y, -1),
transb=True)
print(z)
Output
---
(2, 3, 1)
(2, 1, 3)
variable([[[ 1. 2. 3.]
[ 2. 4. 6.]
[ 3. 6. 9.]]
[[ 16. 20. 24.]
[ 20. 25. 30.]
[ 24. 30. 36.]]])
---
(2, 3, 1)
(2, 3, 1)
variable([[[ 1. 2. 3.]
[ 2. 4. 6.]
[ 3. 6. 9.]]
[[ 16. 20. 24.]
[ 20. 25. 30.]
[ 24. 30. 36.]]])
---
(2, 3)
(2, 3)
variable([[[ 1. 2. 3.]
[ 2. 4. 6.]
[ 3. 6. 9.]]
[[ 16. 20. 24.]
[ 20. 25. 30.]
[ 24. 30. 36.]]])
You can use F.reshape and F.broadcast_to to explicitly handle array.
Assume you have 2-dim array h with shape (minibatch, feature).
If you want to calculate outer product of h and h, try below code.
Is this what you want to do?
import numpy as np
from chainer import functions as F
def outer_product(h):
s0, s1 = h.shape
h1 = F.reshape(h, (s0, s1, 1))
h1 = F.broadcast_to(h1, (s0, s1, s1))
h2 = F.reshape(h, (s0, 1, s1))
h2 = F.broadcast_to(h2, (s0, s1, s1))
h_outer = h1 * h2
return h_outer
# test code
h = np.arange(12).reshape(3, 4).astype(np.float32)
h_outer = outer_product(h)
print(h.shape)
print(h_outer.shape, h_outer.data)
I'm new to Prolog and I'm having trouble with the first part of my programming assignment:
Create a predicate split that that takes as input three parameters. The first and third parameters are lists and the second parameter is an element. You can think of the first parameter as being the input and the last two parameters being the output. The method computes all possible way of splitting a list into an element and the rest of the list. Here is an example run.
?- split([1,2,3,4,5],X,Y).
X = 1,
Y = [2, 3, 4, 5] ;
X = 2,
Y = [1, 3, 4, 5] ;
X = 3,
Y = [1, 2, 4, 5] ;
X = 4,
Y = [1, 2, 3, 5] ;
X = 5,
Y = [1, 2, 3, 4] ;
There are two rules in defining the predicate. The first rule simply gets the first element of the list and returns it as the second parameter and the rest of the list as the third parameter. The second rule generates the list by copying the first element of the list in the result (i.e., third parameter) and then recursively applying the method to the rest of the elements.
split([H|T], H, T).
split([H|T], X, [H|Y]) :-
split(T, X, Y).
There are two ways to take an element out of a list:
Take the head (the first element)
Set the head aside and take an element out of the tail (the rest of the list)
Notice that the predicate can run both ways; if the second and the third parameters are defined, it will yield all possible ways these two can be combined to form a list.
split(List,Elem,Rest) :- select(Elem,List,Rest).
| ?- select(X,[1,2,3],Y).
X = 1,
Y = [2,3] ? ;
X = 2,
Y = [1,3] ? ;
X = 3,
Y = [1,2] ? ;
no
and with split/3 ;
| ?- split([1,2,3,4],X,Y).
X = 1,
Y = [2,3,4] ? ;
X = 2,
Y = [1,3,4] ? ;
X = 3,
Y = [1,2,4] ? ;
X = 4,
Y = [1,2,3] ? ;
no
with Sicstus-prolog u need to export select from library/lists
:- use_module(library(lists)).