I have a numpy 2d array:
import numpy as np
A=np.array([[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16]])
print (A)
I would like to replace the diagonal elements with a = np.array([0,2,15,20]). That is the desired output should be:
A=[[0, 2, 3, 4],
[5, 2, 7, 8],
[9, 10, 15, 12],
[13, 14, 15, 20]]
I tried with the following code:
import numpy as np
A=np.array([[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16]])
a = np.array([0,2,15,20])
print(np.fill_diagonal(A, a))
But it resulted in None
As an alternative method if a be the main array and b the modified values array:
a_mod = a.ravel()
a_mod[::a.shape[0]+1] = b
result = a_mod.reshape(a.shape)
It can handle where the other diagonals of a matrix (instead the main diagonal) is of interest, by some modification. np.fill_diagonal works on the main diagonal.
try this
A[np.arange(4), np.arange(4)] = a
array([[ 0, 2, 3, 4],
[ 5, 2, 7, 8],
[ 9, 10, 15, 12],
[13, 14, 15, 20]])
Consider the following code fragment:
import numpy as np
mask = np.array([True, True, False, True, True, False])
val = np.array([9, 3])
arr = np.random.randint(1, 9, size = (5,len(mask)))
As expected, we get an array of random integers, 1 to 9, with 5 rows and 6 columns as below. The val array has not been used yet.
[[2, 7, 6, 9, 7, 5],
[7, 2, 9, 7, 8, 3],
[9, 1, 3, 5, 7, 3],
[5, 7, 4, 4, 5, 2],
[7, 7, 9, 6, 9, 8]]
Now I'll introduce val = [9, 3].
Where mask = True, I want the row element to be taken randomly from 1 to 9.
Where mask = False, I want the row element to be taken randomly from 1 to 3.
How can this be done efficiently? A sample output is shown below.
[[2, 7, 2, 9, 7, 1],
[7, 2, 1, 7, 8, 3],
[9, 1, 3, 5, 7, 3],
[5, 7, 1, 4, 5, 2],
[7, 7, 2, 6, 9, 1]]
One idea is to sample randomly between 0 to 1, then multiply with 9 or 3 depending on mask, and finally add 1 to move the sample.
rand = np.random.rand(5,len(mask))
is3 = (1-mask).astype(int)
# out is random from 0-8 or 0-2 depending on `is3`
out = (rand*val[is3]).astype(int)
# move out by `1`:
out = (out + 1)
Output:
array([[4, 9, 3, 6, 2, 1],
[1, 8, 2, 7, 1, 3],
[8, 2, 1, 2, 3, 2],
[4, 3, 2, 2, 3, 2],
[5, 8, 1, 5, 6, 1]])
Consider the numpy array below. I'd hoping to find a fast way to remove rows not having 4 distinct values.
import numpy as np
D = np.array([[2, 3, 6, 7],
[2, 4, 3, 4],
[4, 9, 0, 1],
[5, 5, 2, 5],
[7, 5, 4, 8],
[7, 5, 4, 7]])
In the small sample array show, the output should be:
D = np.array([[2, 3, 6, 7],
[4, 9, 0, 1],
[7, 5, 4, 8]])
Here's one way -
In [94]: s = np.sort(D,axis=1)
In [95]: D[(s[:,:-1] == s[:,1:]).sum(1) ==0]
Out[95]:
array([[2, 3, 6, 7],
[4, 9, 0, 1],
[7, 5, 4, 8]])
Alternatively -
In [107]: D[~(s[:,:-1] == s[:,1:]).any(1)]
Out[107]:
array([[2, 3, 6, 7],
[4, 9, 0, 1],
[7, 5, 4, 8]])
Or -
In [112]: D[(s[:,:-1] != s[:,1:]).all(1)]
Out[112]:
array([[2, 3, 6, 7],
[4, 9, 0, 1],
[7, 5, 4, 8]])
With pandas -
In [121]: import pandas as pd
In [122]: D[pd.DataFrame(D).nunique(1)==4]
Out[122]:
array([[2, 3, 6, 7],
[4, 9, 0, 1],
[7, 5, 4, 8]])
A working answer with np.unique
I found no way to use the axis keyword in np.unique to get rid of the list compression, perhaps someone can help?
D[np.array([np.max(np.unique(_,return_counts=True)[-1]) for _ in D])==1]
I have a 9x9 multidimensional array that represents a sudoku game. I need to break it into it's 9 3x3 many components. How would this be done? I have absolutely no idea where to begin, here.
game = [
[1, 3, 2, 5, 7, 9, 4, 6, 8],
[4, 9, 8, 2, 6, 1, 3, 7, 5],
[7, 5, 6, 3, 8, 4, 2, 1, 9],
[6, 4, 3, 1, 5, 8, 7, 9, 2],
[5, 2, 1, 7, 9, 3, 8, 4, 6],
[9, 8, 7, 4, 2, 6, 5, 3, 1],
[2, 1, 4, 9, 3, 5, 6, 8, 7],
[3, 6, 5, 8, 1, 7, 9, 2, 4],
[8, 7, 9, 6, 4, 2, 1, 5, 3]
]
Split into chunks, it becomes
chunk_1 = [
[1, 3, 2],
[4, 9, 8],
[7, 5, 6]
]
chunk_2 = [
[5, 7, 9],
[2, 6, 1],
[3, 8, 4]
]
...and so on
That was a fun exercise!
Answer
game.each_slice(3).map{|stripe| stripe.transpose.each_slice(3).map{|chunk| chunk.transpose}}.flatten(1)
It would be cumbersome and not needed to define every chunk_1, chunk_2, ....
If you want chunk_2, you can use extract_chunks(game)[1]
It outputs [chunk_1, chunk_2, chunk_3, ..., chunk_9], so it's an Array of Arrays of Arrays :
1 3 2
4 9 8
7 5 6
5 7 9
2 6 1
3 8 4
4 6 8
3 7 5
2 1 9
6 4 3
5 2 1
...
You can define a method to check if this grid is valid (it is) :
def extract_chunks(game)
game.each_slice(3).map{|stripe| stripe.transpose.each_slice(3).map{|chunk| chunk.transpose}}.flatten(1)
end
class Array # NOTE: Use refinements if you don't want to patch Array
def has_nine_unique_elements?
self.flatten(1).uniq.size == 9
end
end
def valid?(game)
game.has_nine_unique_elements? &&
game.all?{|row| row.has_nine_unique_elements? } &&
game.all?{|column| column.has_nine_unique_elements? } &&
extract_chunks(game).all?{|chunk| chunk.has_nine_unique_elements? }
end
puts valid?(game) #=> true
Theory
The big grid can be sliced in 3 stripes, each containing 3 rows of 9 cells.
The first stripe will contain chunk_1, chunk_2 and chunk_3.
We need to cut the strip vertically into 3 chunks. To do so :
We transpose the strip,
Cut it horizontally with each_slice,
transpose back again.
We do the same for stripes #2 and #3.
To avoid returning an Array of Stripes of Chunks of Rows of Cells, we use flatten(1) to remove one level and return an Array of Chunks of Rows of Cells. :)
The method Matrix#minor is tailor-made for this:
require 'matrix'
def sub3x3(game, i, j)
Matrix[*game].minor(3*i, 3, 3*j, 3).to_a
end
chunk1 = sub3x3(game, 0, 0)
#=> [[1, 3, 2], [4, 9, 8], [7, 5, 6]]
chunk2 = sub3x3(game, 0, 1)
#=> [[5, 7, 9], [2, 6, 1], [3, 8, 4]]
chunk3 = sub3x3(game, 0, 2)
#=> [[4, 6, 8], [3, 7, 5], [2, 1, 9]]
chunk4 = sub3x3(game, 1, 0)
#=> [[6, 4, 3], [5, 2, 1], [9, 8, 7]]
...
chunk9 = sub3x3(game, 2, 2)
#=> [[6, 8, 7], [9, 2, 4], [1, 5, 3]]
Ruby has not concept of "rows" and "columns" of arrays. For convenience, therefore, I will refer to the 3x3 "subarray" of game, at offsets i and j (i = 0,1,2, j = 0,1,2), as the 3x3 submatrix of m = Matrix[*game] whose upper left value is at row offset 3*i and column offset 3*j of m, converted to an array.
This is relatively inefficient as a new matrix is created for the calculation of each "chunk". Considering the size of the array, this is not a problem, but rather than making that more efficient you really need to rethink the overall design. Creating nine local variables (rather than, say, an array of nine arrays) is not the way to go.
Here's a suggestion for checking the validity of game (that uses the method sub3x3 above) once all the open cells have been filled. Note that I've used the Wiki description of the game, in which the only valid entries are the digits 1-9, and I have assumed the code enforces that requirement when players enter values into cells.
def invalid_vector_index(game)
game.index { |vector| vector.uniq.size < 9 }
end
def sub3x3_invalid?(game, i, j)
sub3x3(game, i, j).flatten.uniq.size < 9
end
def valid?(game)
i = invalid_vector_index(game)
return [:ROW_ERR, i] if i
j = invalid_vector_index(game.transpose)
return [:COL_ERR, j] if j
m = Matrix[*game]
(0..2).each do |i|
(0..2).each do |j|
return [:SUB_ERR, i, j] if sub3x3_invalid?(game, i, j)
end
end
true
end
valid?(game)
#=> true
Notice this either returns true, meaning game is valid, or an array that both signifies that the solution is not valid and contains information that can be used to inform the player of the reason.
Now try
game[5], game[6] = game[6], game[5]
so
game
#=> [[1, 3, 2, 5, 7, 9, 4, 6, 8],
# [4, 9, 8, 2, 6, 1, 3, 7, 5],
# [7, 5, 6, 3, 8, 4, 2, 1, 9],
# [6, 4, 3, 1, 5, 8, 7, 9, 2],
# [5, 2, 1, 7, 9, 3, 8, 4, 6],
# [2, 1, 4, 9, 3, 5, 6, 8, 7],
# [9, 8, 7, 4, 2, 6, 5, 3, 1],
# [3, 6, 5, 8, 1, 7, 9, 2, 4],
# [8, 7, 9, 6, 4, 2, 1, 5, 3]]
valid?(game)
#=> [:SUB_ERR, 1, 0]
The rows and columns are obviously still valid, but this return value indicates that at least one 3x3 subarray is invalid and the array
[[6, 4, 3],
[5, 2, 1],
[2, 1, 4]]
was the first found to be invalid.
You could create a method that generates a single 3X3 chunk from a given index. since the sudoku board is of length 9, that will produce 9 3X3 chunks for you. see below.
#steps
#you'll loop through each index of the board
#to get the x value
#you divide the index by 3 and multiply by 3
#to get the y value
#you divide the index by 3, take remainder and multiply by 3
#for each x value, you can get 3 y values
#this will give you a single 3X3 box from one index so
def three_by3(index, sudoku)
#to get x value
x=(index/3)*3
#to get y value
y=(index%3)*3
(x...x+3).each_with_object([]) do |x,arr|
(y...y+3).each do |y|
arr<<sudoku[x][y]
end
end
end
sudoku = [ [1,2,3,4,5,6,7,8,9],
[2,3,4,5,6,7,8,9,1],
[3,4,5,6,7,8,9,1,2],
[1,2,3,4,5,6,7,8,9],
[2,3,4,5,6,7,8,9,1],
[3,4,5,6,7,8,9,1,2],
[1,2,3,4,5,6,7,8,9],
[2,3,4,5,6,7,8,9,1],
[3,4,5,6,7,8,9,1,2]]
p (0...sudoku.length).map {|i| three_by3(i,sudoku)}
#output:
#[[1, 2, 3, 2, 3, 4, 3, 4, 5],
# [4, 5, 6, 5, 6, 7, 6, 7, 8],
# [7, 8, 9, 8, 9, 1, 9, 1, 2],
# [1, 2, 3, 2, 3, 4, 3, 4, 5],
# [4, 5, 6, 5, 6, 7, 6, 7, 8],
# [7, 8, 9, 8, 9, 1, 9, 1, 2],
# [1, 2, 3, 2, 3, 4, 3, 4, 5],
# [4, 5, 6, 5, 6, 7, 6, 7, 8],
# [7, 8, 9, 8, 9, 1, 9, 1, 2]]
If I have the following data:
A = np.random.random((3, 4, 5))
# np.all(indices < A.shape) is true
indices = np.array([
[0, 0, 0],
[1, 2, 4],
...
[2, 3, 4]
])
How can I use each row of indices as a set of axis indices into A to give the following?
B = np.array([
A[0, 0, 0],
A[1, 2, 4],
...
A[2, 3, 4]
])
Here's a 2d example:
In [1]: A=np.arange(10,22).reshape(3,4)
In [2]: A
Out[2]:
array([[10, 11, 12, 13],
[14, 15, 16, 17],
[18, 19, 20, 21]])
In [3]: ind=np.array([[0,1],[1,3],[2,0],[0,2]])
In [4]: ind
Out[4]:
array([[0, 1],
[1, 3],
[2, 0],
[0, 2]])
In [5]: A[ind[:,0],ind[:,1]]
Out[5]: array([11, 17, 18, 12])
or for your variables,
A[indices[:,0], indices[:,1], indices[:,2]]
Or more generally:
In [8]: tuple(ind.T)
Out[8]: (array([0, 1, 2, 0]), array([1, 3, 0, 2]))
In [9]: A[tuple(ind.T)]
Out[9]: array([11, 17, 18, 12])
This is based on the idea that A[a,b] is the same as A[(a,b)]. And when a and b are matching lists or arrays, it selects values by pairing them up, roughly the same as
[A[i,j] for i,j in zip(a,b)]
For a product like indexing, the index arrays need to have more dimensions. ix_ is a handy way of generating such arrays:
In [53]: np.ix_(ind[:,0],ind[:,1])
Out[53]:
(array([[0],
[1],
[2],
[0]]), array([[1, 3, 0, 2]]))
In [54]: A[np.ix_(ind[:,0],ind[:,1])]
Out[54]:
array([[11, 13, 10, 12],
[15, 17, 14, 16],
[19, 21, 18, 20],
[11, 13, 10, 12]])
In [56]: A[ind[:,[0]],ind[:,1]]
Out[56]:
array([[11, 13, 10, 12],
[15, 17, 14, 16],
[19, 21, 18, 20],
[11, 13, 10, 12]])
You could use np.ravel_multi_index to generate the linear indices and then extract the selective elements from A with linear-indexing using np.take like so -
np.take(A,np.ravel_multi_index(indices.T,A.shape))