I've read this SO question on how to rotate a two-dimensional array many times, and I was curious as to how you could expand this situation to work with a section of a two-dimensional array. I've thought about it for a while, and I can't seem to come up with a good equation. Essentially what I'm wanting to do is something like this:
1 2 3 4 5 6 13 7 1 4 5 6
7 8 9 10 11 12 14 8 2 10 11 12
13 14 15 16 17 18 15 9 3 16 17 18
19 20 21 22 23 24 ----> 19 20 21 22 23 24
25 26 27 28 29 30 25 26 27 28 29 30
31 32 33 34 35 36 31 32 33 34 35 36
I'm writing this in Ruby, currently, but I don't particularly care what language the solution to this kind of problem would be. I'm just curious as to how you'd go about solving this kind of problem.
Edit: To add some more specifics, the basic parameters for a function that could do this would be something that looked like this
def rotate(array, times=1, x=0, y=0, len=nil)
...
end
Check this question : How do you rotate a two dimensional array?
there is a simple way :
a 2-dim array, use
transpose.map &:reverse
Something like that?
def rotate a,len,ii=0,jj=0,t=1
t.times do
a = a.map.with_index do |line,i|
line.map.with_index do |e,j|
(ii...(ii+len))===i && (jj...(jj+len))===j ?
a[ii+jj+len-j-1][jj+i-ii] : e
end
end
end
a
end
t = (1..6).map{|i|(1..6).map{|j|j+6*i-6}}
t.each { |i| p i }
rotate(t,3,2,1).each { |i| p i }
rotate(t,6).each { |i| p i }
rotate(t,3,2,1,4).each { |i| p i }
rotate(t,6,0,0,4).each { |i| p i }
rotate(t,2,2,2,3).each { |i| p i }
Related
I want to select a control group from one data frame based of matching the age from a second data frame. As an example I have subject.df
subject.df
id age
1 1 55
2 2 62
3 3 73
4 4 54
5 5 66
I'd like to subset control.df based off of matching the age directly on a 1 to 1 matching from the subject.df dataframe.
control.df
id age
6 6 66
7 7 71
8 8 80
9 9 51
10 10 55
11 11 56
12 12 77
13 13 62
14 14 64
15 15 73
16 16 67
17 17 54
18 18 75
19 19 77
20 20 78
21 21 53
22 22 64
23 23 83
24 24 61
25 25 77
I'm fairly new to R. In the past I've used Matlab and in this instance would use a for loop to iterate over the control.df dataframe, but I've been told that R doesn't always like for loops and that it can be computationally difficult in R.
In the end I'll be doing this on a much larger data set where the subject group is around 250 and the control group is more than 40K so I know that 1:1 matching is possible.
Is there a way to sum pairwise in Octave?
If for example, I have a 10-row by 4 column. I want a new 10 row by 2 column, where each column is the sum of the pairs.
ex.
[ 1 2 3 4
2 3 4 5
...
]
=> [ 3 7
5 9
...
]
I know how to accomplish this using for loops and accumarray etc, but I'm just not sure if there's a way to do it that is completely vectorized.
Here are a few more options.
Given:
a = reshape(1:40, 10, 4)
a =
1 11 21 31
2 12 22 32
3 13 23 33
4 14 24 34
5 15 25 35
6 16 26 36
7 17 27 37
8 18 28 38
9 19 29 39
10 20 30 40
Keep it simple
b = [sum(a(:,1:2),2) sum(a(:,3:4),2)]
b =
12 52
14 54
16 56
18 58
20 60
22 62
24 64
26 66
28 68
30 70
Squeeze a little
b = squeeze(sum(reshape(a, [], 2, 2), 2))
b =
12 52
14 54
16 56
18 58
20 60
22 62
24 64
26 66
28 68
30 70
Or, my personal favorite...
Mathemagic
b = a * [1 1 0 0; 0 0 1 1].'
b =
12 52
14 54
16 56
18 58
20 60
22 62
24 64
26 66
28 68
30 70
Perhaps someone comes with a better idea:
a = [1 2 3 4; 2 3 4 5]
b = reshape (sum (reshape (a.', 2, [])), [], rows(a)).'
gives
b =
3 7
5 9
I'm looking for something in Julia like a comprehension but for a matrix instead of a vector. If i have some single-variable function f(x) and I want an array that is filled with f(i) for i in 1..10, I can do this:
[f(i) for i = 1:10]
If I have some two-variable function g(i,j) and I want a matrix from i=[1,10]; j=[1,10] filled with the function I can do this:
M = zeros (10,10)
for i in 1:10
for j in 1:10
M[i,j] = g(i,j)
end
end
Is there some shortcut that allows me to express that in a shorter way and without wasting time allocating all that zeros?
Just use a multidimensional comprehension directly:
julia> g(x,y) = 2x+y
g (generic function with 1 method)
julia> [g(i,j) for i=1:10, j=1:10]
10x10 Array{Int64,2}:
3 4 5 6 7 8 9 10 11 12
5 6 7 8 9 10 11 12 13 14
7 8 9 10 11 12 13 14 15 16
9 10 11 12 13 14 15 16 17 18
11 12 13 14 15 16 17 18 19 20
13 14 15 16 17 18 19 20 21 22
15 16 17 18 19 20 21 22 23 24
17 18 19 20 21 22 23 24 25 26
19 20 21 22 23 24 25 26 27 28
21 22 23 24 25 26 27 28 29 30
This works for any number of dimensions, by adding variable ranges at the end.
Suppose we have a 2D (5x5) matrix:
test =
39 13 90 5 71
60 78 38 4 11
87 92 46 45 35
40 96 61 17 1
90 50 46 89 63
And a second 2D (5x2) matrix:
tidx =
1 3
2 4
2 3
2 4
4 5
And now we want to use tidx as an idex into test, so that we get the following output:
out =
39 90
78 4
92 46
96 17
89 63
One way to do this is with a for loop...
for i=1:size(test,1)
out(i,:) = test(i,tidx(i,:));
end
Question:
Is there a way to vectorize this so the same output is generated without a for loop?
Here is one way:
test(repmat([1:rows(test)]',1,columns(tidx)) + (tidx-1)*rows(test))
What you describe is an index problem. When you place a matrix all in one dimension, you get
test(:) =
39
60
87
40
90
13
78
92
96
50
90
38
46
61
46
5
4
45
17
89
71
11
35
1
63
This can be indexed using a single number. Here is how you figure out how to transform tidx into the correct format.
First, I use the above reference to figure out the index numbers which are:
outinx =
1 11
7 17
8 13
9 19
20 25
Then I start trying to figure out the pattern. This calculation gives a clue:
(tidx-1)*rows(test) =
0 10
5 15
5 10
5 15
15 20
This will move the index count to the correct column of test. Now I just need the correct row.
outinx-(tidx-1)*rows(test) =
1 1
2 2
3 3
4 4
5 5
This pattern is created by the for loop. I created that matrix with:
[1:rows(test)]' * ones(1,columns(tidx))
*EDIT: This does the same thing with a built in function.
repmat([1:rows(test)]',1,columns(tidx))
I then add the 2 together and use them as the index for test.
Imagine I've defined the following name in J:
m =: >: i. 2 4 5
This looks like the following:
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
26 27 28 29 30
31 32 33 34 35
36 37 38 39 40
I want to create a monadic verb of rank 1 that applies to each list in this list of lists. It will double (+:) or add 1 (>:) to each alternate item in the list. If we were to apply this verb to the first row, we'd get 2 3 6 5 10.
It's fairly easy to get a list of booleans which alternate with each item, e.g., 0 1 $~{:$ m gives us 0 1 0 1 0. I thought, aha! I'll use something like +:`>: #. followed by some expression, but I could never quite get it to work.
Any suggestions?
UPDATE
The following appears to work, but perhaps it can be refactored into something more elegant by a J pro.
poop =: monad define
(($ y) $ 0 1 $~{:$ y) ((]+:)`(]>:) #. [)"0 y
)
I would use the oblique verb, with rank 1 (/."1)- so it applies to successive elements of each list in turn.
You can pass a gerund into /. and it applies them in order, extending cyclically.
+:`>: /."1 m
2
3
6
5
10
12
8
16
10
20
22
13
26
15
30
32
18
36
20
40
42
23
46
25
50
52
28
56
30
60
62
33
66
35
70
72
38
76
40
80
I spent a long time and I looked at it, and I believe that I know why ,# works to recover the shape of the argument.
The shape of the arguments to the parenthesized phrase is the shape of the argument passed to it on the right, even though the rank is altered by the " conjugate (well, that is what trace called it, I thought it was an adverb). If , were monadic, it would be a ravel, and the result would be a vector or at least of a lower rank than the input, based on adverbs to ravel. That is what happens if you take the conjunction out - you get a vector.
So what I believe is happening is that the conjunction is making , act like a dyadic , which is called an append. The append alters what it is appending to what it is appending to. It is appending to nothing but that thing still has a shape, and so it ends up altering the intermediate vector back to the shape of the input.
Now I'm probably wrong. But $,"0#(+:>:/.)"1 >: i. 2 4 5 -> 2 4 5 1 1` which I thought sort of proved my case.
(,#(+:`>:/.)"1 a) works, but note that ((* 2 1 $~ $)#(+ 0 1 $~ $)"1 a) would also have worked (and is about 20 times faster, on large arrays, in my brief tests).