I'm building a decision tree algorithm. The sorting is very expensive in this algorithm because for every split I need to sort each column. So at the beginning - even before tree construction I'm presorting variables - I'm creating a matrix so for each column in the matrix I save its ranking. Then when I want to sort the variable in some split I don't actually sort it but use the presorted ranking array. The problem is that I don't know how to do it in a space efficient manner.
A naive solution of this is below. This is only for 1 variabe (v) and 1 split (split_ind).
import numpy as np
v = np.array([60,70,50,10,20,0,90,80,30,40])
sortperm = v.argsort() #1 sortperm = array([5, 3, 4, 8, 9, 2, 0, 1, 7, 6])
rankperm = sortperm.argsort() #2 rankperm = array([6, 7, 5, 1, 2, 0, 9, 8, 3, 4])
split_ind = np.array([3,6,4,8,9]) # this is my split (random)
# split v and sortperm
v_split = v[split_ind] # v_split = array([10, 90, 20, 30, 40])
rankperm_split = rankperm[split_ind] # rankperm_split = array([1, 9, 2, 3, 4])
vsorted_dummy = np.ones(10)*-1 #3 allocate "empty" array[N]
vsorted_dummy[rankperm_split] = v_split
vsorted = vsorted_dummy[vsorted_dummy!=-1] # vsorted = array([ 10., 20., 30., 40., 90.])
Basically I have 2 questions:
Is double sorting necessary to create ranking array? (#1 and #2)
In the line #3 I'm allocating array[N]. This is very inefficent in terms of space because even if split size n << N I have to allocate whole array. The problem here is how to calculate rankperm_split. In the example original rankperm_split = [1,9,2,3,4] while it should be really [1,5,2,3,4]. This problem can be reformulated so that I want to create a "dense" integer array that has maximum gap of 1 and it keeps the ranking of the array intact.
UPDATE
I think that second point is the key here. This problem can be redefined as
A[N] - array of size N
B[N] - array of size N
I want to transform array A to array B so that:
Ranking of the elements stays the same (for each pair i,j if A[i] < A[j] then B[i] < B[j]
Array B has only elements from 1 to N where each element is unique.
A few examples of this transformation:
[3,4,5] => [1,2,3]
[30,40,50] => [1,2,3]
[30,50,40] => [1,3,2]
[3,4,50] => [1,2,3]
A naive implementation (with sorting) can be defined like this (in Python)
def remap(a):
a_ = sorted(a)
b = [a_.index(e)+1 for e in a]
return b
Related
Consider array a, holding some of the permutations of 1,2,3,4. (my actual arrays may be larger)
import numpy as np
n = 3
a = np.array([[1, 2, 3, 4],
[1, 2, 4, 3],
[1, 3, 4, 2],
[1, 4, 3, 2],
[2, 3, 4, 1],
[2, 4, 3, 1],
[3, 1, 4, 2],
[4, 1, 2, 3]]))
I want to identify sets of n rows (in this example, n=3) where each row position holds unique values.
In this example, the output would be:
out = [[0, 4, 7],
[2, 5, 7],
[3, 4, 7]]
The 1st row of out indicates that a[0], a[4], and a[7] have unique values in each row position
When n = 2, there are 11 row pairs that match the criteria: [[0,4], [0,6], [0,7], [1,5] ...etc
When n = 4, there are 0 rows that match the criteria.
I'm new enough to python that I can't find a good way to approach this situation.
Solving this problem efficiently is far from not easy. Indeed, the brute-force solution consisting is using n nested loop is very inefficient: its complexity is O(c r! / (r-n)!) where r is the number of rows of a and c is the number of columns of a (note that ! is the factorial). Since r is a number of permutation which already grow experientially with the number of unique items in a, this means the complexity of this solution is really bad.
A more efficient solution (but still not great) is to pick a row, filter the other rows that can match with it (ie. there is no items at the same position that are equal), and then recursively do the same thing n times (the picked row are only the one that are filtered). The several sets of row indices can be appended in a list during the recursion. It is hard to evaluate the complexity of this solution, but it is far much faster in practice since most rows hopefully does not match together and the filtered rows tends to decrease exponentially too. That being said, the complexity is certainly still exponential since the size of the output appears to grow exponentially too and the output needs to be written.
Here is the implementation:
def recursiveFindSets(a, i, n, availableRows, rowIndices, results):
if availableRows.size == 0:
return
for k in availableRows:
# Save the current choice
rowIndices[i] = k
# The next selected rows needs to be bigger than `k` so to prevent replicates
newAvailableRows = availableRows[availableRows > k]
# If there is no solutions with a[k], then choose another
if newAvailableRows.size == 0:
continue
# Find rows that contains different items of a[i]
goodMatches = np.all(a[newAvailableRows] != a[k], axis=1)
# Find the location relative to `a` and not `a[availableRows]`
newAvailableRows = newAvailableRows[goodMatches]
# If there is no solutions with a[k], then choose another
if newAvailableRows.size == 0:
continue
if i == n-2:
# Generate some solutions from `newAvailableRows`
for k2 in newAvailableRows:
rowIndices[i+1] = k2
results.append(rowIndices.copy())
elif i < n-2:
recursiveFindSets(a, i+1, n, newAvailableRows, rowIndices, results)
def findSets(a, n):
availableRows = np.arange(a.shape[0], dtype=int) # Filter
rowIndices = np.empty(n, dtype=np.int_) # Current set of row indices
results = [] # List of all the sets
recursiveFindSets(a, 0, n, availableRows, rowIndices, results)
if len(results) == 0:
return np.empty((0, n), dtype=int)
return np.vstack(results)
findSets(a, 3)
# Output:
# array([[0, 4, 7],
# [2, 5, 7],
# [3, 4, 7]])
You can reduce this problem to finding all cliques of size n in an undirected graph. Nodes in the graph are given by row indices of a. There is an edge between i and j if (a[i] != a[j]).all().
Here is one implementation based on networkx. A function enumerate_all_cliques(g) iterates over cliques in g in order of increasing size. We discard all cliques of size less than n, keep those of size n, and stop once the first clique of size greater than n is found or cliques run out.
from itertools import combinations
import networkx as nx
def f(arr, n):
nodes = np.arange(arr.shape[0])
g = nx.Graph()
g.add_nodes_from(nodes)
for i, j in combinations(nodes, 2):
if (arr[i] != arr[j]).all():
g.add_edge(i, j)
for c in nx.algorithms.clique.enumerate_all_cliques(g):
lc = len(c)
if lc == n:
yield c
elif lc > n:
break
print(list(f(a, 3)))
# [[0, 4, 7], [2, 5, 7], [3, 4, 7]]
Here is another approach: find all maximal cliques and yield all subsets of size n from each clique. This can lead to double-counting, hence set is used before the return statement.
def f(arr, n):
nodes = np.arange(arr.shape[0])
g = nx.Graph()
g.add_nodes_from(nodes)
for i, j in combinations(nodes, 2):
if (arr[i] != arr[j]).all():
g.add_edge(i, j)
cliques = set()
# iterate over maximal cliques
for c in nx.algorithms.clique.find_cliques(g):
# update the clique set subsets of c
cliques.update(map(frozenset, combinations(c, n)))
# return all cliques of size n without doublecounting
return [list(c) for c in cliques]
print(f(a, 3))
# [[2, 5, 7], [3, 4, 7], [0, 4, 7]]
The performance of either approach will vary depending on input values.
I use rand to reach random three element from a and add this m values to array but I want them to be unique. So, array can't be like this: array = [1,1,2]. How can I check when two elements are equal and how to prevent this other than sample method? I was thinking about this: Let's assume m=1 when times method runs the first time. If m =1 at the second time, I want to skip this value and reach a different one. Is there any code explanation for this ? Or maybe more different way?
a = [1, 2, 3, 4]
array = []
3.times do
m = a[rand(a.size)]
array << m
end
Use shuffle and slice 3 elements:
a = [1, 2, 3, 4]
shuffled = a.shuffle[0..2]
As I understand you wish to write a method similar to Array#sample, that returns a pseudo-random sample of a given size without replacement. I suggest the following, which I believe would be relatively efficient, particularly when the sample size is small or large relative to the size of array.
def sample(arr, sample_size)
n = arr.size
raise ArgumentError if n < sample_size
a = arr.dup
m = (sample_size < n/2) ? sample_size : n - sample_size
m.times do
i = rand(n)
n -= 1
a[i], a[n] = a[n], a[i]
end
n = arr.size
(sample_size < n/2) ? a[n-sample_size..] : a[0, sample_size]
end
a = [7, 5, 7, 1, 9, 6, 2, 0, 6, 7]
Notice that if sample_size >= arr.size/2 I sample arr.size - sample_size elements and return the unsampled elements.
For an numpy 1d array such as:
In [1]: A = np.array([2,5,1,3,9,0,7,4,1,2,0,11])
In [2]: A
Out[2]: array([2,5,1,3,9,0,7,4,1,2,0,11])
I need to split the array by using the values as a sub-array length.
For the example array:
The first index has a value of 2, so I need the first split to occur at index 0 + 2, so it would result in ([2,5,1]).
Skip to index 3 (since indices 0-2 were gobbled up in step 1).
The value at index 3 = 3, so the second split would occur at index 3 + 3, and result in ([3,9,0,7]).
Skip to index 7
The value at index 7 = 4, so the third and final split would occur at index 7 + 4, and result in ([4,1,2,0,11])
I'm using this simple array as an example, because I think it will help in my actual use case, which is reading data from binary files (either as bytes or unsigned shorts). I'm guessing that numpy will be the fastest way to do it, but I could also use struct/bytearray/lists or whatever would be best.
I hope this makes sense. I had a hard time trying to figure out how best to word the question.
Here is an approach using standard python lists and a while loop:
def custom_partition(arr):
partitions = []
i = 0
while i < len(arr):
pariton_size = arr[i]
next_i = i + pariton_size + 1
partitions.append(arr[i:next_i])
i = next_i
return partitions
a = [2, 5, 1, 3, 9, 0, 7, 4, 1, 2, 0, 11]
b = custom_partition(a)
print(b)
Output:
[[2, 5, 1], [3, 9, 0, 7], [4, 1, 2, 0, 11]]
we are given an array of size n (n is even), we have to divide it into two equal-sized subarrays array1 and array2, sized n/2 each such that product of all the numbers of array1 equals to the product of all the numbers in array2.
Given array:
arr = [2, 4, 5, 12, 15, 18]
solution:
array2 = [4, 5, 18]
array1 = [2, 12, 15]
Explanation:
product of all elements in array1 is 360
product of all elements in array2 is 360.
This problem might be soved using dynamic programming. You need to get subset of size n/2 with product equal to p = sqrt(overall_product). Note that there is no solution when overall_product is not exact square.
Recursion might look like
solution(p, n/2, arr) = choose valid solution from
solution(p / arr[i], n/2-1, arr without arr[i])
return true for arguments (1,0,...)
use memoization or table to solve problem with merely large n values.
Assume that we are given two vectors:
Ａ＝（ａ₁，ａ₂，...，ａₘ）and Ｂ＝（ｂ₁，ｂ₂，...，ｂₘ）
and we need to do something for all the vectors between these two ones.
For example, for Ａ＝（１,１,０）and Ｂ＝（１,２,２）, all the vectors between Ａ and Ｂ are: ｛（１,１,１）,（１,１,２）,（１,２,０）,（１,２,１）｝.
An obvious way to generate such vectors is using ｍ loops (for loop), but probably it is not the best one. I would like to know if someone has some better idea.
Here's a fixed method. Returns a matrix where each row is one of the vectors of the result.
% Data
A = [0, 0, 1, 3, 5, 2]
B = [4, 8, 5, 7, 9, 6]
% Preallocate
b = cell(1,numel(A));
vec = cell(1,numel(A));
% Make a vector of values of each element of the result
for i = 1:numel(A)
vec{i} = A(i):B(i);
end
% Get all combinations using ndgrid
[b{:}] = ndgrid(vec{:});
b=cat(ndims(b{1})+1,b{:});
% Reshape the numel(A)+1 dimensional array into a 2D array
res = reshape(b,numel(b)/length(A),length(A));