I have a set of indices that I want to convert them to encodings. In order to do so :
i = [2, 1, 3, 4]
s = sparse(i, 1:lenght(i), 1)
s = full(s);
This works fine as expected but when the array i = [2, 1, 3, 3]. The full function gives a 3 by 4 matrix instead of 4 by 4. Julia thinks that the last row is unnecessary and deletes it which ,for my case, is not.
Is it possible to create a square matrix by using sparse and full when there are repetitions inside the index array i?
B.R.
Just supply the dimensions you want as additional arguments, e.g.:
s = sparse(i, 1:length(i), 1, 4, 4)
The details are explained in help for sparse.
Related
I'm writing a simple interpreter that people can use (among other things) to sort a list via a comparison function (preferably a stable sort). Sorting algorithms that I'm familiar with all seem to require a variable number of calls to that comparison function and you don't know ahead of time which items will be compared to each other. That won't work because of the nature of what work is done in the interpreted language vs the runtime at what times.
The steps required by the interpreter are:
Step 1: Create a list of as to be compared to another list of bs, one a & b at a time. Something like sort([1, 2, 3]) producing:
a = [2, 3]
b = [1, 2]
(2 is compared to 1 and then 3 is compared to 2 in the above example, going index by index.)
Step 2: Create two new lists (before and after) with the same number of items as a and b to represent the result of the comparison function. The values are any null or non-null value. Something like:
before = [2, 3]
after = [null, null]
(2 should come before 1, representing 1 from b as null. The non-null values are preserved, but any non-null value could be in 2's place.)
I can impose a limitation that the values in before and after must be items from the lists a and b, but I'd prefer not to if I possibly can. I mention it because I'm unsure how I could know where the non-null value came from (a or b). But if the items compared from a and b are the same but only one is null at the end, I have the same problem.
Step 3: Use those two lists before and after to sort the original list. Something like:
sort([1, 2, 3], greaterThan) => [3, 2, 1]
// a = [2, 3]
// b = [1, 2]
// before = [2, 3]
// after = [null, null]
(If both values are non-null or both null, it should favor their original order relative to each other, or a "stable" sort.)
In such a trivial example, the items in a and b are sufficient to sort the list. The Javascript (the language the interpreter is written in) Array.sort method will compare them like this:
(2, 1)
(3, 2)
and be done. But if the order of the original list were [2, 3, 1] then it has to do:
(3, 2)
(1, 3)
(1, 2)
(1, 3)
(I don't know why or what algorithm they use).
In that example, I would have to provide lists a and b as [3, 1, 1, 1] and [2, 3, 2, 3] respectively.
How do I get a list for a and b that will work given any comparison function or order of the original list -- and then use the resulting before and after lists to sort that original list?
I have a matrix
A = [[ 1. 2. 3.]
[ 4. 5. 6.]]
and a vector
b = [ 5. 10. 15.]
I want to add each column of A (A[:,i]) to b[i], i.e.
[[ 6. 12. 18.]
[ 9. 15. 21.]]
an easy way to do it would be
A = tf.constant([[1., 2, 3],[1, 2, 3]])
b = tf.constant([[5, 10, 15.]])
e = tf.ones((2,1))
a + tf.matmul( e, b ) # outer product "repmat"
But it seems terribly wasteful to do this and have to construct an entire auxiliary matrix which we will eventually throw out. Is there a more idomatic way of doing this without writing my own op?
As mentioned you can do A + b:
import tensorflow as tf
tf.InteractiveSession()
A = tf.constant([[1., 2, 3], [4, 5, 6]])
b = tf.constant([[5, 10, 15.]])
(A + b).eval()
returns:
array([[ 6., 12., 18.],
[ 9., 15., 21.]], dtype=float32)
The reason this works is because of array broadcasting. The Numpy broadcasting page has great info and tensorflow broadcasting works the same way. Basically for each dimension (moving from trailing dimension to leading dimension) tensorflow/numpy attempts checks to see if the dimensions are compatible (either they have the same number of elements or one of them only has 1 element).
In your case A is of shape [2, 3] and b is of shape [1, 3]. The second dimensions match, but because b has a first dimension with only a single element that element of b is "broadcast" along the first dimension of A (two elements).
Sorry for the bad title, but I don't know how to call this.
I have K lists, N elements in each, for example:
[8, 5, 6]
[4, 3, 2]
[6, 5, 0]
and I want to find such a permutation of the lists' elements, so that the sum of elements in first column, second column etc are as close to each other as possible (so the distribution is "fair").
In my example that would be (probably):
[8, 5, 6]
[4, 2, 3] -- the lists contain the same values
[0, 6, 5] just in different order
sums: 12, 13, 14
Is there some more elegant way than finding all the permutations for each list, and brute-force finding the "ideal" combination of them?
I'm not asking for code, just give me a hint how to do it, if you know.
Thanks!
ps. the lists can be quite large, and more of them - think ~20x~20 max.
If you can accept an approximation, I would do it iteratively :
Sort matrix lines by descending weight (sum of line elements).
Edit : Sorting first by max element in line could be better.
Each time you are going to add a new line to your result matrix, put smaller elements into higher columns.
Order lines of your result matrix back to their initial state (if you have to).
It works with your example, but will obviously not be always perfect.
Here is an example (javascript)
Sorry for the bad description in the title.
Consider a 2-dimensional list such as this:
list = [
[1, 2],
[2, 3],
[3, 4]
]
If I were to extract all possible "vertical" combinations of this list, for a total of 2*2*2=8 combinations, they would be the following sequences:
1, 2, 3
2, 2, 3
1, 3, 3
2, 3, 3
1, 2, 4
2, 2, 4
1, 3, 4
2, 3, 4
Now, let's say I remove some of these sequences. Let's say I only want to keep sequences which have either the number 2 in position #1 OR number 4 in position #3. Then I would be left with these sequences:
2, 2, 3
2, 3, 3
1, 2, 4
2, 2, 4
1, 3, 4
2, 3, 4
The problem
I would like to re-combine these remaining sequences to the least possible amount of 2-dimensional lists needed to contain all sequences but no less or no more.
By doing so, the resulting 2-dimensional lists in this particular example would be:
list_1 = [
[2],
[2, 3],
[3, 4]
]
list_2 = [
[1],
[2, 3],
[4]
]
In this particular case, the resulting lists can be thought out. But how would I go about if there were thousands of sequences yielding hundereds of 2-dimensional lists? I have been trying to come up with a good algorithm for two weeks now, but I am getting nowhere near a satisfying result.
Divide et impera, or divide and conquer. If we have a logical expression, stating that the value at position x should be a or the value at position y should be b, then we have 3 cases:
a is the value at position x and b is the value at position y
a is the value at position x and b is not the value at position y
a is not the value at position x and b is the value at position y
So, first you generate all your scenarios, you know now that you have 3 scenarios.
Then, you effectively separate your cases and handle all of them in a sub-routine as they were your main tasks. The philosophy behind divide et imera is to reduce your complex problem into several similar, but less complex problems, until you reach triviality.
We have two sorted arrays of the same size n. Let's call the array a and b.
How to find the middle element in an sorted array merged by a and b?
Example:
n = 4
a = [1, 2, 3, 4]
b = [3, 4, 5, 6]
merged = [1, 2, 3, 3, 4, 4, 5, 6]
mid_element = merged[(0 + merged.length - 1) / 2] = merged[3] = 3
More complicated cases:
Case 1:
a = [1, 2, 3, 4]
b = [3, 4, 5, 6]
Case 2:
a = [1, 2, 3, 4, 8]
b = [3, 4, 5, 6, 7]
Case 3:
a = [1, 2, 3, 4, 8]
b = [0, 4, 5, 6, 7]
Case 4:
a = [1, 3, 5, 7]
b = [2, 4, 6, 8]
Time required: O(log n). Any ideas?
Look at the middle of both the arrays. Let's say one value is smaller and the other is bigger.
Discard the lower half of the array with the smaller value. Discard the upper half of the array with the higher value. Now we are left with half of what we started with.
Rinse and repeat until only one element is left in each array. Return the smaller of those two.
If the two middle values are the same, then pick arbitrarily.
Credits: Bill Li's blog
Quite interesting task. I'm not sure about O(logn), but solution O((logn)^2) is obvious for me.
If you know position of some element in first array then you can find how many elements are smaller in both arrays then this value (you know already how many smaller elements are in first array and you can find count of smaller elements in second array using binary search - so just sum up this two numbers). So if you know that number of smaller elements in both arrays is less than N, you should look in to the upper half in first array, otherwise you should move to the lower half. So you will get general binary search with internal binary search. Overall complexity will be O((logn)^2)
Note: if you will not find median in first array then start initial search in the second array. This will not have impact on complexity
So, having
n = 4 and a = [1, 2, 3, 4] and b = [3, 4, 5, 6]
You know the k-th position in result array in advance based on n, which is equal to n.
The result n-th element could be in first array or second.
Let's first assume that element is in first array then
do binary search taking middle element from [l,r], at the beginning l = 0, r = 3;
So taking middle element you know how many elements in the same array smaller, which is middle - 1.
Knowing that middle-1 element is less and knowing you need n-th element you may have [n - (middle-1)]th element from second array to be smaller, greater. If that's greater and previos element is smaller that it's what you need, if it's greater and previous is also greater we need to L = middle, if it's smaller r = middle.
Than do the same for the second array in case you did not find solution for first.
In total log(n) + log(n)