Most efficient way to select elements in 2d array in Ruby - arrays

Let's say I have 2d array (or matrix) like,
matrix = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
And, I've to select all elements in eight directions of a certain element matrix[i][j], so I know,
up and down elements will have same j
left and right elements will have same i
back-inclined diagonal (\) elements will have same diff i-j
front-inclined diagonal (/) elements will have same sum i+j
How to select those elements, easily and efficiently? Any direct method?
For example, if my element is 4, then my selection should yield, [1, 7, 5, 6, 2, 8]. If it is, 5, then selection should be all except 5.
Edit: I've coded any solution yet, as I thought it'll be very poor, but here is my idea. Try it online!
matrix = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
res = []
i = 0
j = 1
# let [i,j] be index of my element
matrix.each_with_index{|row,r| row.each_with_index{|col,c|
res << col if c == j || r == i || r+c == i+j || r-c == i-j
}}
p res

m is the matrix you have taken from a user that is a symmetric or non-symmetric 2-dimensional array.
# inputs
m = [ [1, 2, 3], [4, 5, 6], [7, 8, 9] ]
p, q = 1, 0 # co-ordinate of number 4
# code for output
i, j, result = m.size, m[0].size, []
i.times { |r| result.push(m[r][q]) if r != p }
j.times { |c| result.push(m[p][c]) if c != q }
i.times do |r|
j.times do |c|
result.push(m[r][c]) if [p,q] != [r,c] && (p-r).abs == (q-c).abs
end
end
> result
=> [1, 7, 5, 6, 2, 8]
Note: if the order of output is not needed to be preserved, we can manage with only 2 loops

Code
def extract(matrix, row, col)
last_row = matrix.size-1
last_col = matrix.first.size-1
diag_sum = row + col
diag_first_row = diag_sum - [diag_sum, last_col].min
diag_last_row = [last_row, diag_sum].min
ante_diag_diff = row - col
ante_diag_first_row = [ante_diag_diff, 0].max
ante_diag_last_row = ante_diag_diff +
[last_col, last_row - ante_diag_diff].min
arr = []
(0..last_col).each { |j| arr << matrix[row][j] unless j == col }
(0..last_row).each { |i| arr << matrix[i][col] unless i == row }
(diag_first_row..diag_last_row).each do |i|
arr << matrix[i][diag_sum - i] unless i == row
end
(ante_diag_first_row..ante_diag_last_row).each do |i|
arr << matrix[i][i - ante_diag_diff] unless i == row
end
arr
end
Example
matrix = Array.new(5) { Array.new(5) { rand(10..99) } }
#=> [[52, 29, 61, 35, 27],
# [68, 99, 67, 18, 67],
# [79, 10, 73, 15, 36],
# [49, 94, 28, 24, 53],
# [37, 26, 65, 65, 43]]
(0..4).each do |i|
(0..4).each do |j|
puts "#{i}, #{j}: #{extract(matrix,i,j)}"
end
end
i j
----------------------------------------------------------------------
0, 0: [29, 61, 35, 27, 68, 79, 49, 37, 99, 73, 24, 43]
0, 1: [52, 61, 35, 27, 99, 10, 94, 26, 68, 67, 15, 53]
0, 2: [52, 29, 35, 27, 67, 73, 28, 65, 99, 79, 18, 36]
0, 3: [52, 29, 61, 27, 18, 15, 24, 65, 67, 10, 49, 67]
0, 4: [52, 29, 61, 35, 67, 36, 53, 43, 18, 73, 94, 37]
1, 0: [99, 67, 18, 67, 52, 79, 49, 37, 29, 10, 28, 65]
1, 1: [68, 67, 18, 67, 29, 10, 94, 26, 61, 79, 52, 73, 24, 43]
1, 2: [68, 99, 18, 67, 61, 73, 28, 65, 35, 10, 49, 29, 15, 53]
1, 3: [68, 99, 67, 67, 35, 15, 24, 65, 27, 73, 94, 37, 61, 36]
1, 4: [68, 99, 67, 18, 27, 36, 53, 43, 15, 28, 26, 35]
2, 0: [10, 73, 15, 36, 52, 68, 49, 37, 61, 99, 94, 65]
2, 1: [79, 73, 15, 36, 29, 99, 94, 26, 35, 67, 49, 68, 28, 65]
2, 2: [79, 10, 15, 36, 61, 67, 28, 65, 27, 18, 94, 37, 52, 99, 24, 43]
2, 3: [79, 10, 73, 36, 35, 18, 24, 65, 67, 28, 26, 29, 67, 53]
2, 4: [79, 10, 73, 15, 27, 67, 53, 43, 24, 65, 61, 18]
3, 0: [94, 28, 24, 53, 52, 68, 79, 37, 35, 67, 10, 26]
3, 1: [49, 28, 24, 53, 29, 99, 10, 26, 27, 18, 73, 37, 79, 65]
3, 2: [49, 94, 24, 53, 61, 67, 73, 65, 67, 15, 26, 68, 10, 65]
3, 3: [49, 94, 28, 53, 35, 18, 15, 65, 36, 65, 52, 99, 73, 43]
3, 4: [49, 94, 28, 24, 27, 67, 36, 43, 65, 29, 67, 15]
4, 0: [26, 65, 65, 43, 52, 68, 79, 49, 27, 18, 73, 94]
4, 1: [37, 65, 65, 43, 29, 99, 10, 94, 67, 15, 28, 49]
4, 2: [37, 26, 65, 43, 61, 67, 73, 28, 36, 24, 79, 94]
4, 3: [37, 26, 65, 43, 35, 18, 15, 24, 53, 68, 10, 28]
4, 4: [37, 26, 65, 65, 27, 67, 36, 53, 52, 99, 73, 24]
Explanation
First observe that if the target element is in row i and column j, the elements [p,q] on the diagonal that passes through that point have the property that p+q == i+j. Similarly, the elements [p,q] on the ante-diagonal that passes through that point have the property that p-q == i-j
Suppose
row = 1
col = 2
then
last_row = matrix.size-1
#=> 4
last_col = matrix.first.size-1
#=> 4
diag_sum = row + col
#=> 3
diag_first_row = diag_sum - [diag_sum, last_col].min
#=> 0
diag_last_row = [last_row, diag_sum].min
#=> 3
ante_diag_diff = row - col
#=> -1
ante_diag_first_row = [ante_diag_diff, 0].max
#=> 0
ante_diag_last_row = ante_diag_diff +
[last_col, last_row - ante_diag_diff].min
#=> 3
The remaining calculations are straightforward.
Here's a variant of the above that uses less code and may be slightly less efficient.
def extract(matrix, row, col)
row_range = 0..matrix.size-1
col_range = 0..matrix.first.size-1
diag_sum = row + col
ante_diag_diff = row - col
arr = []
col_range.each { |j| arr << matrix[row][j] unless j == col }
row_range.each do |i|
next if i == row
arr << matrix[i][col]
j = diag_sum - i
arr << matrix[i][j] if col_range.cover?(j)
j = i - ante_diag_diff
arr << matrix[i][j] if col_range.cover?(j)
end
arr
end

Here is sample code, you can modify/simplify it further
class Matrix
attr_reader :arr, :ele
def initialize(arr, ele)
#arr = arr
#ele = ele
end
def find_elements
position = arr.flatten.index(ele)
len = arr.length
x = position % len
y = position / len
horizontal_elements(y) + vertical_elements(x, y, len) + remaining_elements(x, y, len)
end
def horizontal_elements y
arr[y] - [ele]
end
def vertical_elements x, y, len
rows = (0..(len-1)).to_a - [y]
rows.map{|row| arr[row][x] }
end
def remaining_elements(x, y, len)
rows = []
rows << y + 1 if y + 1 < len
rows << y - 1 if y - 1 >= 0
c_rows = []
c_rows << x + 1 if x + 1 < len
c_rows << x - 1 if x - 1 >= 0
rows.map{|row| c_rows.map{|c_row| arr[row][c_row] } }.flatten
end
end
matrix = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
ele = 4
(1..9).each do |i|
puts '*' * 20
puts "For #{i}"
puts Matrix.new(matrix, i).find_elements.inspect
puts '*' * 20
end
O/P will be something like following
Desktop $ ruby something.rb
********************
For 1
[2, 3, 4, 7, 5]
********************
********************
For 2
[1, 3, 5, 8, 6, 4]
********************
********************
For 3
[1, 2, 6, 9, 5]
********************
********************
For 4
[5, 6, 1, 7, 8, 2]
********************
********************
For 5
[4, 6, 2, 8, 9, 7, 3, 1]
********************
********************
For 6
[4, 5, 3, 9, 8, 2]
********************
********************
For 7
[8, 9, 1, 4, 5]
********************
********************
For 8
[7, 9, 2, 5, 6, 4]
********************
********************
For 9
[7, 8, 3, 6, 5]
********************

A recursive way
I'd like to propose a different approach, not tested if it is more or less efficient but can provide some control on order and more.
Lets start with this matrix, where elements tell their indexing:
mat = [
['00', '01', '02'],
['10', '11', '12'],
['20', '21', '22'],
['30', '31', '32']
]
And define an helper method just to get the shape of the matrix and to check which is well formed:
def shape(mat)
raise 'Out of shape' if mat.map(&:size).uniq.size > 1
return mat.size, mat[0].size
end
Implementation
Now, let's define a method that takes as input the starting point, the limits of the matrix (shape) and the direction.
def walk(i_0, j_0, i_max, j_max, direction, res=[])
d_i, d_j = direction
next_step = [i_0 + d_i, j_0 + d_j]
if [-1, i_max].include?(next_step[0]) || [-1, j_max].include?(next_step[1])
return res
end
res << next_step
i_0, j_0 = next_step
walk(i_0, j_0, i_max, j_max, direction, res)
end
The method returns the list of indexes starting from the origin and walking along the direction.
Usage examples
Example 1
i_0, j_0 = [2, 0]
i_max, j_max = shape(mat)
direction = [-1, 1]
walk(i_0, j_0, *shape(mat), direction)
#=> [[1, 1], [0, 2]]
Example 2
You can use it to extract the indexes given a list of directions.
It's possible to arrange the list of directions to give a scan order (clockwise, cclockwise) or just skip some directions, as you will.
directions = [[1, 0], [-1, 0], [0, 1], [0, -1], [1, 1], [1, -1], [-1, 1], [-1, -1]]
i_0, j_0 = [2, 0]
directions = [[1, 0], [-1, 0], [0, 1], [0, -1], [1, 1], [1, -1], [-1, 1], [-1, -1]]
directions.flat_map { |direction| walk(i_0, j_0, *shape(mat), direction) }
#=> [[3, 0], [1, 0], [0, 0], [2, 1], [2, 2], [3, 1], [1, 1], [0, 2]]
Example 3
(Which is an extension of example 2)
You can directly get the elements from the matrix:
res = directions.flat_map do |direction|
walk(i_0, j_0, *shape(mat), direction).map do |coords|
i, j = coords
mat[i][j]
end
end
res
#=> ["30", "10", "00", "21", "22", "31", "11", "02"]
It can be possible to use a longer step (eg. direction = [2, 2]) upgrading the walk method.

Related

Sort odd numbers in array while keeping even numbers in place

I'm working on a coding challenge to sort the odd elements of an array while keeping all even elements in their initial position.
I can sort the odd elements in groups between even elements while leaving the evens alone, but I want them to be ordered throughout the array. Here is my code:
def sort_array source_array
return [] if source_array.empty?
a = source_array
n = a.length
loop do
swapped = false
(n - 1).times do |i|
if (a[i] > a[i + 1]) && a[i].odd? && a[i + 1].odd?
a[i], a[i + 1] = a[i + 1], a[i]
swapped = true
end
end
break if not swapped
end
a
end
source_array = [7, 5, 35, 8, 12, 17, 47, 47, 37, 64, 22, 55, 13]
p sort_array(source_array)
# => [5, 7, 35, 8, 12, 17, 37, 47, 47, 64, 22, 13, 55]
You could try this approach:
Create an array with all odds (leaving source_array intact). Sort
this array.
Consider source_array element by element. Is the
element even then leave it . If it is odd, then replace it with the
first element on the sorted odds list. Remove the first element of the
sorted list.
Here are a couple of ways to do that.
arr = [7, 5, 35, 8, 12, 17, 47, 47, 37, 64, 22, 55, 13]
Both begin with the following calculation
odd, even = arr.each_with_index.partition { |n,i| n.odd? }
#=> [[[7, 0], [5, 1], [35, 2], [17, 5], [47, 6], [47, 7], [37, 8], [55, 11], [13, 12]],
# [[8, 3], [12, 4], [64, 9], [22, 10]]]
odd
#=> [[7, 0], [5, 1], [35, 2], [17, 5], [47, 6], [47, 7], [37, 8], [55, 11], [13, 12]]
even
#=> [[8, 3], [12, 4], [64, 9], [22, 10]]
See Enumerable#each_with_index, Enumerable#partition and Integer#odd?.
#1
odd_val, odd_idx = odd.transpose
#=> [[7, 5, 35, 17, 47, 47, 37, 55, 13],
# [0, 1, 2, 5, 6, 7, 8, 11, 12]]
a = even.concat(odd_val.sort.zip(odd_idx))
#=> [[8, 3], [12, 4], [64, 9], [22, 10], [5, 0], [7, 1], [13, 2],
# [17, 5], [35, 6], [37, 7], [47, 8], [47, 11], [55, 12]]
h = a.map(&:reverse).to_h
#=> {3=>8, 4=>12, 9=>64, 10=>22, 0=>5, 1=>7, 2=>13, 5=>17, 6=>35, 7=>37,
# 8=>47, 11=>47, 12=>55}
h.values_at(*0..arr.size-1)
#=> [5, 7, 13, 8, 12, 17, 35, 37, 47, 64, 22, 47, 55]
Note:
odd_val.sort.zip(odd_idx)
#=> [5, 7, 13, 17, 35, 37, 47, 47, 55].zip([0, 1, 2, 5, 6, 7, 8, 11, 12])
#=> [[5, 0], [7, 1], [13, 2], [17, 5], [35, 6], [37, 7], [47, 8], [47, 11], [55, 12]]
See Array#sort, Array#zip and Hash#values_at.
#2
a = odd.map(&:first).sort
#=> [5, 7, 13, 17, 35, 37, 47, 47, 55]
even.each { |n,i| a.insert(i,n) }
a #=> [5, 7, 13, 8, 12, 17, 35, 37, 47, 64, 22, 47, 55]
See Array#insert.
source_array = [7, 5, 35, 8, 12, 17, 47, 47, 37, 64, 22, 55, 13]
odd_sorted = source_array.select(&:odd?).sort
source_array.map {|input| input.even? ? input : odd_sorted.shift }
# [5, 7, 13, 8, 12, 17, 35, 37, 47, 64, 22, 47, 55]

Swift using eratosthenes method to get prime numbers

I want to use eratosthenes method
to get prime numbers with swift. I create first function to return new array without those numbers that can be divided for specific multiplier, then create second function to create new array every time with new P multiplier. I wonder why its not work (look like it somehow pass old array, i dont know why). It should print new array of prime numbers at the end:
var simpleArr : [Int] = []
for i in 2...100 {
simpleArr.append(i)
}
func arrayEcludingDivingByP (p: Int, arrToCheck : [Int]) -> Array<Int>{
var tmp : [Int] = []
for (ob, index) in arrToCheck.enumerated() {
var isDividible : Bool = ob % p == 0 ? true : false
if (!isDividible){
tmp.append(ob)
}
}
return tmp
}
var p : Int = 2
func getSimpleNumbersArrayFromArray (p : Int, arrPassed : [Int]) -> Array <Int>{
var tmp : [Int] = []
var newArr = arrayEcludingDivingByP(p: p, arrToCheck: arrPassed)
if (newArr.isEmpty){
// No more p availible, just return tmp
} else {
let newP = p + 1
getSimpleNumbersArrayFromArray(p: newP, arrPassed: newArr)
tmp = newArr
print("tmp array? \(tmp)")
}
return tmp
}
getSimpleNumbersArrayFromArray(p: p, arrPassed: simpleArr)
In console it prints:
tmp array? [1]
tmp array? [1, 2]
tmp array? [1, 2, 3]
tmp array? [1, 2, 3, 4]
tmp array? [1, 2, 3, 4, 5]
tmp array? [1, 2, 3, 4, 5, 6]
tmp array? [1, 2, 3, 4, 5, 6, 7]
tmp array? [1, 2, 3, 4, 5, 6, 7, 8]
tmp array? [1, 2, 3, 4, 5, 6, 7, 9, 10, 11]
tmp array? [1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13]
tmp array? [1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17]
tmp array? [1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23]
tmp array? [1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 17, 18, 19, 21, 22, 23, 25, 26, 27, 29, 30, 31]
tmp array? [1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 28, 29, 31, 32, 34, 35, 37, 38, 40, 41, 43, 44, 46, 47]
tmp array? [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97]
But it suppose to print final iteration with tmp filled with prime numbers. What did i wrong?
You are using newArr to represent the values in the sieve that remain. Your tmp array should be the primes found so far. Also you are using a recursive call which is confusing. I suggest simply grabbing the first value out of newArr which is a prime, add that prime to your list of primes, and then call arrayEcludingDividingByP to filter out the non-primes divisible by that prime. Repeat until newArr isEmpty which will happen when the conditional binding statement while let newP = newArr.first fails and the loop ends:
func getPrimes(arrPassed: [Int]) -> [Int] {
var primes: [Int] = []
var newArr = arrPassed
while let newP = newArr.first {
primes.append(newP)
newArr = arrayEcludingDivingByP(p: newP, arrToCheck: newArr)
}
return primes
}
print(getPrimes(arrPassed: Array(2...100)))
This can be further optimized because once newP * newP is greater than the largest number in newArr (newArr.last!) you are done and you can simply append newArr to primes and set newArr to [].
Also, arrayEcludingDivingByP(p: newP, arrToCheck: newArr) can be replaced by newArr.filter { $0 % newP != 0 }

Python: Doing Calculations on array elements in a list

I have a list of arrays, in which each array represents a cell and the array elements are the coordinates x,y and z, the time point and the cell id. Here a sector of it:
cells=[ ...,
[ 264.847, 121.056, 30.868, 42. , 375. ],
[ 259.24 , 116.875, 29.973, 43. , 375. ],
[ 260.757, 118.574, 32.772, 44. , 375. ]]), array([[ 263.967, 154.089, 55.5 , 38. , 376. ],
[ 260.744, 152.924, 55.5 , 39. , 376. ],
[ 258.456, 151.373, 55.5 , 40. , 376. ],
...,
[ 259.086, 159.564, 48.521, 53. , 376. ],
[ 258.933, 159.796, 48.425, 54. , 376. ],
[ 259.621, 158.719, 51.606, 55. , 376. ]]), array([[ 291.647, 57.582, 28.178, 38. , 377. ],
[ 284.625, 59.221, 30.028, 39. , 377. ],
[ 282.915, 59.37 , 30.402, 40. , 377. ],
...,
[ 271.224, 58.534, 23.166, 42. , 377. ],
[ 270.048, 58.738, 21.749, 43. , 377. ],
[ 268.38 , 58.138, 20.606, 44. , 377. ]]), array([[ 87.83 , 222.144, 26.258, 39. , 378. ],
[ 99.779, 223.631, 24.98 , 40. , 378. ],
[ 104.107, 224.177, 23.728, 41. , 378. ],
...,
[ 127.778, 222.205, 23.123, 63. , 378. ],
[ 126.815, 222.347, 23.934, 64. , 378. ],
[ 127.824, 221.048, 25.508, 65. , 378. ]]),...]
minimumCellCoors = cells
maximumCellCoors = cells
centoEdge = radius+fcr_size
Now i want to change the coordinates x, y and z, so the 0.,1. and 2. element of the arrays in the list to get them in a specific grid. The user gives the spacing for x,y and z and then the operation could look like:
x_Coo=round(x_element/x)*x
y_Coo=round(y_element/y)*y
z_Coo=round(z_element/z)*z
So the real question here is, how could i do a operation on all of the elements in the array ( or in this case the first three elements in the array in the list)?
EDIT
If i use list comprehension to the list like:
[np.round((cellID[:,0]-(centoEdge+1))/x)*x for cellID in minimumCellCoors]
[np.round((cellID[:,1]-(centoEdge+1))/y)*y for cellID in minimumCellCoors]
[np.round((cellID[:,2]-(centoEdge+1))/z)*z for cellID in minimumCellCoors]
[np.round((cellID[:,0]+(centoEdge+1))/x)*x for cellID in maximumCellCoors]
[np.round((cellID[:,1]+(centoEdge+1))/x)*y for cellID in maximumCellCoors]
[np.round((cellID[:,2]+(centoEdge+1))/x)*z for cellID in maximumCellCoors]
How could i fusion the single lists of arrays to one array again?
Best regards!
First off you need to convert your list to a numpy array. It's more proper to create a numpy array instead of a list at first place. Then you can take advantage of numpy's vectorized operation support:
Here is an example:
In [45]: arr = np.arange(100).reshape(4, 5, 5)
In [46]: arr
Out[46]:
array([[[ 0, 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, 41, 42, 43, 44],
[45, 46, 47, 48, 49]],
[[50, 51, 52, 53, 54],
[55, 56, 57, 58, 59],
[60, 61, 62, 63, 64],
[65, 66, 67, 68, 69],
[70, 71, 72, 73, 74]],
[[75, 76, 77, 78, 79],
[80, 81, 82, 83, 84],
[85, 86, 87, 88, 89],
[90, 91, 92, 93, 94],
[95, 96, 97, 98, 99]]])
In [51]: arr[:,:,:3] = np.round(arr[:,:,:3]/5)*5
In [52]: arr
Out[52]:
array([[[ 0, 0, 0, 3, 4],
[ 5, 5, 5, 8, 9],
[10, 10, 10, 13, 14],
[15, 15, 15, 18, 19],
[20, 20, 20, 23, 24]],
[[25, 25, 25, 28, 29],
[30, 30, 30, 33, 34],
[35, 35, 35, 38, 39],
[40, 40, 40, 43, 44],
[45, 45, 45, 48, 49]],
[[50, 50, 50, 53, 54],
[55, 55, 55, 58, 59],
[60, 60, 60, 63, 64],
[65, 65, 65, 68, 69],
[70, 70, 70, 73, 74]],
[[75, 75, 75, 78, 79],
[80, 80, 80, 83, 84],
[85, 85, 85, 88, 89],
[90, 90, 90, 93, 94],
[95, 95, 95, 98, 99]]])
Note that you can also perform the operations with same length arrays as well as scalars:
For instance you could also do the following:
In [53]: arr[:,:,:3] = np.round(arr[:,:,:3]/5)*[4, 5, 6]

How to sum 2-dimensional arrays

I have a two-dimensional array with sub-arrays of equal size, for example:
array = [
[10, 12, 15 ,17], [16, 32, 65, 47], [45, 48, 41, 23],
[36, 25, 74, 98], [32, 19, 66, 88]
]
I would like to create a new array by summing the corresponding elements of every 4th sub-array, i.e. the elements that are "on top of each other" in the above example:
new_array = [
[10 + 36, 12 + 25, 15 + 74, 17 + 98],
[16 + 32, 32 + 19, 65 + 66, 47 + 88],
[45, 48, 4‌​1, 23]
]
These are just examples, the actual arrays can be larger.
Complete Matrix
You can use each_slice, transpose, map and transpose again to navigate your matrix.
The code first uses join('+') to show what is being calculated :
array= [[10,12,15,17],[16,32,65,47],[45,48,41,23],[36,25,74,98],[32,19,66,88],[1,2,3,4]]
array.each_slice(3).to_a.transpose.map{|r| r.transpose.map{|x| x.join('+')}}
# => [["10+36", "12+25", "15+74", "17+98"], ["16+32", "32+19", "65+66", "47+88"], ["45+1", "48+2", "41+3", "23+4"]]
array.each_slice(3).to_a.transpose.map{|r| r.transpose.map{|x| x.inject(:+)}}
# => [[46, 37, 89, 115], [48, 51, 131, 135], [46, 50, 44, 27]]
Warning!
You need to carefully select the each_slice parameter to suit your original array. transpose might raise an exception otherwise :
array = [[10,12,15,17],[19,32,65,47],[45,48,41,23],[36,25,74,98],[10,12,15,17],[16,98,65,47],[69,48,65,23],[66,25,74,98]]
array.each_slice(3).to_a.transpose.map{|r| r.transpose.map{|x| x.inject(:+)}}
#=> IndexError: element size differs (2 should be 3)
array.each_slice(4).to_a.transpose.map{|r| r.transpose.map{|x| x.inject(:+)}}
#=> [[20, 24, 30, 34], [35, 130, 130, 94], [114, 96, 106, 46], [102, 50, 148, 196]]
Incomplete Matrix
If the matrix size isn't a multiple of width :
array = [
[10, 12, 15 ,17], [16, 32, 65, 47], [45, 48, 41, 23],
[36, 25, 74, 98], [32, 19, 66, 88]
]
you could add subarrays full of 0s to get :
matrix = [
[10, 12, 15 ,17], [16, 32, 65, 47], [45, 48, 41, 23],
[36, 25, 74, 98], [32, 19, 66, 88], [ 0, 0, 0, 0]
]
Array#fill does the job :
def maxtrix_column_sums(array, width)
size = array.size
size2 = array.first.size
missing = (-size) % width
matrix = array.dup.fill(Array.new(size2, 0), size...size + missing)
matrix.each_slice(width).to_a.transpose.map { |r| r.transpose.map { |x| x.join('+') } }
end
p maxtrix_column_sums(array, 3)
#=> [["10+36", "12+25", "15+74", "17+98"], ["16+32", "32+19", "65+66", "47+88"], ["45+0", "48+0", "41+0", "23+0"]]
Here's a variation of Eric Duminil's answer using zip instead of transpose to account for an "odd" number of sub-arrays:
first, *rest = array.each_slice(3).to_a
first.zip(*rest).map { |r| r.compact.transpose.map { |x| x.inject(:+) } }
#=> [[46, 37, 89, 115], [48, 51, 131, 135], [45, 48, 41, 23]]
How it works:
each_slice separates the array into groups of 3:
array.each_slice(3).to_a
#=> [
# [[10, 12, 15, 17], [16, 32, 65, 47], [45, 48, 41, 23]],
# [[36, 25, 74, 98], [32, 19, 66, 88]]
# ]
first.zip(*rest) combines the first slice "column"-wise with the remaining slices, adding nil when a slice is missing:
first.zip(*rest)
#=> [
# [[10, 12, 15, 17], [36, 25, 74, 98]],
# [[16, 32, 65, 47], [32, 19, 66, 88]],
# [[45, 48, 41, 23], nil]
# ]
The map / compact / transpose part then restructures the sub-array while getting rid of nil values:
first.zip(*rest).map { |r| r.compact.transpose }
#=> [
# [[10, 36], [12, 25], [15, 74], [17, 98]],
# [[16, 32], [32, 19], [65, 66], [47, 88]],
# [[45], [48], [41], [23]]
# ]
And inject(:+) finally sums the inner elements.

Group an array of integers in a hash of ranges and counts

I have an array of integers like this:
[1, 1, 1, 2, 2, 4, 4, 5, 6, 11, 11, 12, 15, 22, 23, 23, 23, 31, 32, 32]
I am trying to convert this to a hash, grouping according to ranges at intervals of 10....
So, in this case it would be
{ [1..10] => 9, [11..20] => 4, [21..30] => 4, [31..40] => 3 }
I have tried a few things which haven't come close so it's a bit pointless putting them down here. I can convert the array to ranges
[1, 1, 1, 2, 2, 4, 4, 5, 6, 11, 11, 12, 15, 22, 23, 23, 23, 31, 32, 32].sort.uniq.inject([]) do |spans, n|
if spans.empty? || spans.last.last != n - 1
spans + [n..n]
else
spans[0..-2] + [spans.last.first..n]
end
end
But this is not what I am looking for. Any suggestions?
Hash[
your_array.group_by{|i| i / 10}.map{|k,v|
[(k*10+1..(k+1)*10), v.count]
}
]
#=> {1..10=>9, 11..20=>4, 21..30=>4, 31..40=>3}
arr.each_with_object(Hash.new(0)) do |e, hash|
i = e / 10
hash[i*10+1..i*10+10] += 1
end
#⇒ {
# 1..10 => 9,
# 11..20 => 4,
# 21..30 => 4,
# 31..40 => 3
# }
I've modified the example to have no numbers between 21 and 30, so that the hash should include the key-value pair 21..30=>0.
arr = [1, 1, 1, 2, 2, 4, 4, 5, 6, 11, 11, 12, 20, 32, 33, 33, 33, 41, 42, 42]
intervals = (1..arr.last-1).step(10).each_with_object({}) { |n,h| h[n..n+9] = 0 }
#=> {1..10=>0, 11..20=>0, 21..30=>0, 31..40=>0, 41..50=>0}
arr.each_with_object(intervals) do |n,intervals|
interval_end = 10*((n+9)/10)
intervals[interval_end-9..interval_end] += 1
end
#=> {1..10=>9, 11..20=>4, 21..30=>0, 31..40=>4, 41..50=>3}

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