matrix: move n-th row by n position efficiently - python

I have a numpy 2d array and I need to transform it in a way that the first row remains the same, the second row moves by one position to right, (it can wrap around or just have zero padded to the front). Third row shifts 3 positions to the right, etc.
I can do this through a "for loop" but that is not very efficient. I am guessing there should be a filtering matrix that multipled by the original one will have the same effect, or maybe a numpy trick that will help me doing this? Thanks!
I have looked into numpy.roll() but I don't think it can work on each row separately.
import numpy as np
p = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]])
'''
p = [ 1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16]
desired output:
p'= [ 1 2 3 4
0 5 6 7
0 0 9 10
0 0 0 13]
'''

We can extract sliding windows into a zeros padded version of the input to have a memory efficient approach and hence performant too. To get those windows, we can leverage np.lib.stride_tricks.as_strided based scikit-image's view_as_windows. More info on use of as_strided based view_as_windows.
Hence, the solution would be -
from skimage.util.shape import view_as_windows
def slide_by_one(p):
m,n = p.shape
z = np.zeros((m,m-1),dtype=p.dtype)
a = np.concatenate((z,p),axis=1)
w = view_as_windows(a,(1,p.shape[1]))[...,0,:]
r = np.arange(m)
return w[r,r[::-1]]
Sample run -
In [60]: p # generic sample of size mxn
Out[60]:
array([[ 1, 5, 9, 13, 17],
[ 2, 6, 10, 14, 18],
[ 3, 7, 11, 15, 19],
[ 4, 8, 12, 16, 20]])
In [61]: slide_by_one(p)
Out[61]:
array([[ 1, 5, 9, 13, 17],
[ 0, 2, 6, 10, 14],
[ 0, 0, 3, 7, 11],
[ 0, 0, 0, 4, 8]])
We can leverage the regular rampy pattern to have a more efficient approach with a more raw usage of np.lib.stride_tricks.as_strided, like so -
def slide_by_one_v2(p):
m,n = p.shape
z = np.zeros((m,m-1),dtype=p.dtype)
a = np.concatenate((z,p),axis=1)
s0,s1 = a.strides
return np.lib.stride_tricks.as_strided(a[:,m-1:],shape=(m,n),strides=(s0-s1,s1))
Another one with some masking -
def slide_by_one_v3(p):
m,n = p.shape
z = np.zeros((len(p),1),dtype=p.dtype)
a = np.concatenate((p,z),axis=1)
return np.triu(a[:,::-1],1)[:,::-1].flat[:-m].reshape(m,-1)

Here is a simple method based on zero-padding and reshaping. It is fast because it avoids advanced indexing and other overheads.
def pp(p):
m,n = p.shape
aux = np.zeros((m,n+m-1),p.dtype)
np.copyto(aux[:,:n],p)
return aux.ravel()[:-m].reshape(m,n+m-2)[:,:n].copy()

Related

Numpy where() using a condition that changes with the items position in the array

I'm trying to build a grid world using numpy.
The grid is 4*4 and laid out in a square.
The first and last squares (i.e. 1 and 16) are terminal squares.
At each time step you can move one step in any direction either: up, down , left or right.
Once you enter one of the terminal squares no further moves are possible and the game terminates.
The first and last columns are the left and right edges of the square whilst the first and last rows represent the top and bottom edges.
If you are on an edge, for example the left one and attempt to move left, instead of moving left you stay in the square you started in. Similarly you remain in the same square if you try and cross any of the other edges.
Although the grid is a square I've implemented it as an array.
States_r calculates the position of the states after a move right. 1 and 16 stay where they are because they are terminal states (note the code uses zero based counting so 1 and 16 are 0 and 15 respectively in the code).
The rest of the squares are in increased by one. The code for states_r works however those squares on the right edge i.e. (4, 8, 12) should also stay where they are but states_r code doesn't do that.
State_l is my attempt to include the edge condition for the left edge of the square. The logic is the same the terminal states (1, 16) should not move nor should those squares on the left edge (5, 9, 13). I think the general logic is correct but it's producing an error.
states = np.arange(16)
states_r = states[np.where((states + 1 <= 15) & (states != 0), states + 1, states)]
states_l = states[np.where((max(1, (states // 4) * 4) <= states - 1) & (states != 15), states - 1, states)]
The first example states_r works, it handles the terminal state but does not handle the edge condition.
The second example is my attempt to include the edge condition, however it is giving me the following error:
"The truth value of an array with more than one element is ambiguous."
Can someone please explain how to fix my code?
Or alternatively suggest another solution,ideally I want the code to be fast (so I can scale it up) so I want to avoid for loops if possible?
If I understood correctly you want arrays which indicate for each state where the next state is, depending on the move (right, left, up, down).
If so, I guess your implementation of state_r is not quit right. I would suggest to switch to a 2D representation of your grid, because a lot of the things you describe are easier and more intuitive to handle if you have x and y directly (at least for me).
import numpy as np
n = 4
states = np.arange(n*n).reshape(n, n)
states_r, states_l, states_u, states_d = (states.copy(), states.copy(),
states.copy(), states.copy())
states_r[:, :n-1] = states[:, 1:]
states_l[:, 1:] = states[:, :n-1]
states_u[1:, :] = states[:n-1, :]
states_d[:n-1, :] = states[1:, :]
# up [[ 0, 1, 2, 3],
# left state right [ 0, 1, 2, 3],
# down [ 4, 5, 6, 7],
# [ 8, 9, 10, 11]]
#
# [[ 0, 0, 1, 2], [[ 0, 1, 2, 3], [[ 1, 2, 3, 3],
# [ 4, 4, 5, 6], [ 4, 5, 6, 7], [ 5, 6, 7, 7],
# [ 8, 8, 9, 10], [ 8, 9, 10, 11], [ 9, 10, 11, 11],
# [12, 12, 13, 14]] [12, 13, 14, 15]] [13, 14, 15, 15]]
#
# [[ 4, 5, 6, 7],
# [ 8, 9, 10, 11],
# [12, 13, 14, 15],
# [12, 13, 14, 15]]
If you want to exclude the terminal states, you can do something like this:
terminal_states = np.zeros((n, n), dtype=bool)
terminal_states[0, 0] = True
terminal_states[-1, -1] = True
states_r[terminal_states] = states[terminal_states]
states_l[terminal_states] = states[terminal_states]
states_u[terminal_states] = states[terminal_states]
states_d[terminal_states] = states[terminal_states]
If you prefer the 1D approach:
import numpy as np
n = 4
states = np.arange(n*n)
valid_s = np.ones(n*n, dtype=bool)
valid_s[0] = False
valid_s[-1] = False
states_r = np.where(np.logical_and(valid_s, states % n < n-1), states+1, states)
states_l = np.where(np.logical_and(valid_s, states % n > 0), states-1, states)
states_u = np.where(np.logical_and(valid_s, states > n-1), states-n, states)
states_d = np.where(np.logical_and(valid_s, states < n**2-n), states+n, states)
Another way of doing it without preallocating arrays:
states = np.arange(16).reshape(4,4)
states_l = np.hstack((states[:,0][:,None],states[:,:-1],))
states_r = np.hstack((states[:,1:],states[:,-1][:,None]))
states_d = np.vstack((states[1:,:],states[-1,:]))
states_u = np.vstack((states[0,:],states[:-1,:]))
To get them all in 1-D, you can always flatten()/ravel()/reshape(-1) the 2-D arrays.
[[ 0 1 2 3]
[ 0 1 2 3]
[ 4 5 6 7]
[ 8 9 10 11]]
[[ 0 0 1 2] [[ 0 1 2 3] [[ 1 2 3 3]
[ 4 4 5 6] [ 4 5 6 7] [ 5 6 7 7]
[ 8 8 9 10] [ 8 9 10 11] [ 9 10 11 11]
[12 12 13 14]] [12 13 14 15]] [13 14 15 15]]
[[ 4 5 6 7]
[ 8 9 10 11]
[12 13 14 15]
[12 13 14 15]]
And for corners you can do:
states_u[-1,-1] = 15
states_l[-1,-1] = 15

How to break a Numpy ndarray into blocks [duplicate]

Is there a way to slice a 2d array in numpy into smaller 2d arrays?
Example
[[1,2,3,4], -> [[1,2] [3,4]
[5,6,7,8]] [5,6] [7,8]]
So I basically want to cut down a 2x4 array into 2 2x2 arrays. Looking for a generic solution to be used on images.
There was another question a couple of months ago which clued me in to the idea of using reshape and swapaxes. The h//nrows makes sense since this keeps the first block's rows together. It also makes sense that you'll need nrows and ncols to be part of the shape. -1 tells reshape to fill in whatever number is necessary to make the reshape valid. Armed with the form of the solution, I just tried things until I found the formula that works.
You should be able to break your array into "blocks" using some combination of reshape and swapaxes:
def blockshaped(arr, nrows, ncols):
"""
Return an array of shape (n, nrows, ncols) where
n * nrows * ncols = arr.size
If arr is a 2D array, the returned array should look like n subblocks with
each subblock preserving the "physical" layout of arr.
"""
h, w = arr.shape
assert h % nrows == 0, f"{h} rows is not evenly divisible by {nrows}"
assert w % ncols == 0, f"{w} cols is not evenly divisible by {ncols}"
return (arr.reshape(h//nrows, nrows, -1, ncols)
.swapaxes(1,2)
.reshape(-1, nrows, ncols))
turns c
np.random.seed(365)
c = np.arange(24).reshape((4, 6))
print(c)
[out]:
[[ 0 1 2 3 4 5]
[ 6 7 8 9 10 11]
[12 13 14 15 16 17]
[18 19 20 21 22 23]]
into
print(blockshaped(c, 2, 3))
[out]:
[[[ 0 1 2]
[ 6 7 8]]
[[ 3 4 5]
[ 9 10 11]]
[[12 13 14]
[18 19 20]]
[[15 16 17]
[21 22 23]]]
I've posted an inverse function, unblockshaped, here, and an N-dimensional generalization here. The generalization gives a little more insight into the reasoning behind this algorithm.
Note that there is also superbatfish's
blockwise_view. It arranges the
blocks in a different format (using more axes) but it has the advantage of (1)
always returning a view and (2) being capable of handling arrays of any
dimension.
It seems to me that this is a task for numpy.split or some variant.
e.g.
a = np.arange(30).reshape([5,6]) #a.shape = (5,6)
a1 = np.split(a,3,axis=1)
#'a1' is a list of 3 arrays of shape (5,2)
a2 = np.split(a, [2,4])
#'a2' is a list of three arrays of shape (2,5), (2,5), (1,5)
If you have a NxN image you can create, e.g., a list of 2 NxN/2 subimages, and then divide them along the other axis.
numpy.hsplit and numpy.vsplit are also available.
There are some other answers that seem well-suited for your specific case already, but your question piqued my interest in the possibility of a memory-efficient solution usable up to the maximum number of dimensions that numpy supports, and I ended up spending most of the afternoon coming up with possible method. (The method itself is relatively simple, it's just that I still haven't used most of the really fancy features that numpy supports so most of the time was spent researching to see what numpy had available and how much it could do so that I didn't have to do it.)
def blockgen(array, bpa):
"""Creates a generator that yields multidimensional blocks from the given
array(_like); bpa is an array_like consisting of the number of blocks per axis
(minimum of 1, must be a divisor of the corresponding axis size of array). As
the blocks are selected using normal numpy slicing, they will be views rather
than copies; this is good for very large multidimensional arrays that are being
blocked, and for very large blocks, but it also means that the result must be
copied if it is to be modified (unless modifying the original data as well is
intended)."""
bpa = np.asarray(bpa) # in case bpa wasn't already an ndarray
# parameter checking
if array.ndim != bpa.size: # bpa doesn't match array dimensionality
raise ValueError("Size of bpa must be equal to the array dimensionality.")
if (bpa.dtype != np.int # bpa must be all integers
or (bpa < 1).any() # all values in bpa must be >= 1
or (array.shape % bpa).any()): # % != 0 means not evenly divisible
raise ValueError("bpa ({0}) must consist of nonzero positive integers "
"that evenly divide the corresponding array axis "
"size".format(bpa))
# generate block edge indices
rgen = (np.r_[:array.shape[i]+1:array.shape[i]//blk_n]
for i, blk_n in enumerate(bpa))
# build slice sequences for each axis (unfortunately broadcasting
# can't be used to make the items easy to operate over
c = [[np.s_[i:j] for i, j in zip(r[:-1], r[1:])] for r in rgen]
# Now to get the blocks; this is slightly less efficient than it could be
# because numpy doesn't like jagged arrays and I didn't feel like writing
# a ufunc for it.
for idxs in np.ndindex(*bpa):
blockbounds = tuple(c[j][idxs[j]] for j in range(bpa.size))
yield array[blockbounds]
You question practically the same as this one. You can use the one-liner with np.ndindex() and reshape():
def cutter(a, r, c):
lenr = a.shape[0]/r
lenc = a.shape[1]/c
np.array([a[i*r:(i+1)*r,j*c:(j+1)*c] for (i,j) in np.ndindex(lenr,lenc)]).reshape(lenr,lenc,r,c)
To create the result you want:
a = np.arange(1,9).reshape(2,1)
#array([[1, 2, 3, 4],
# [5, 6, 7, 8]])
cutter( a, 1, 2 )
#array([[[[1, 2]],
# [[3, 4]]],
# [[[5, 6]],
# [[7, 8]]]])
Some minor enhancement to TheMeaningfulEngineer's answer that handles the case when the big 2d array cannot be perfectly sliced into equally sized subarrays
def blockfy(a, p, q):
'''
Divides array a into subarrays of size p-by-q
p: block row size
q: block column size
'''
m = a.shape[0] #image row size
n = a.shape[1] #image column size
# pad array with NaNs so it can be divided by p row-wise and by q column-wise
bpr = ((m-1)//p + 1) #blocks per row
bpc = ((n-1)//q + 1) #blocks per column
M = p * bpr
N = q * bpc
A = np.nan* np.ones([M,N])
A[:a.shape[0],:a.shape[1]] = a
block_list = []
previous_row = 0
for row_block in range(bpc):
previous_row = row_block * p
previous_column = 0
for column_block in range(bpr):
previous_column = column_block * q
block = A[previous_row:previous_row+p, previous_column:previous_column+q]
# remove nan columns and nan rows
nan_cols = np.all(np.isnan(block), axis=0)
block = block[:, ~nan_cols]
nan_rows = np.all(np.isnan(block), axis=1)
block = block[~nan_rows, :]
## append
if block.size:
block_list.append(block)
return block_list
Examples:
a = np.arange(25)
a = a.reshape((5,5))
out = blockfy(a, 2, 3)
a->
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]])
out[0] ->
array([[0., 1., 2.],
[5., 6., 7.]])
out[1]->
array([[3., 4.],
[8., 9.]])
out[-1]->
array([[23., 24.]])
For now it just works when the big 2d array can be perfectly sliced into equally sized subarrays.
The code bellow slices
a ->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]])
into this
block_array->
array([[[ 0, 1, 2],
[ 6, 7, 8]],
[[ 3, 4, 5],
[ 9, 10, 11]],
[[12, 13, 14],
[18, 19, 20]],
[[15, 16, 17],
[21, 22, 23]]])
p ang q determine the block size
Code
a = arange(24)
a = a.reshape((4,6))
m = a.shape[0] #image row size
n = a.shape[1] #image column size
p = 2 #block row size
q = 3 #block column size
block_array = []
previous_row = 0
for row_block in range(blocks_per_row):
previous_row = row_block * p
previous_column = 0
for column_block in range(blocks_per_column):
previous_column = column_block * q
block = a[previous_row:previous_row+p,previous_column:previous_column+q]
block_array.append(block)
block_array = array(block_array)
If you want a solution that also handles the cases when the matrix is
not equally divided, you can use this:
from operator import add
half_split = np.array_split(input, 2)
res = map(lambda x: np.array_split(x, 2, axis=1), half_split)
res = reduce(add, res)
Here is a solution based on unutbu's answer that handle case where matrix cannot be equally divided. In this case, it will resize the matrix before using some interpolation. You need OpenCV for this. Note that I had to swap ncols and nrows to make it works, didn't figured why.
import numpy as np
import cv2
import math
def blockshaped(arr, r_nbrs, c_nbrs, interp=cv2.INTER_LINEAR):
"""
arr a 2D array, typically an image
r_nbrs numbers of rows
r_cols numbers of cols
"""
arr_h, arr_w = arr.shape
size_w = int( math.floor(arr_w // c_nbrs) * c_nbrs )
size_h = int( math.floor(arr_h // r_nbrs) * r_nbrs )
if size_w != arr_w or size_h != arr_h:
arr = cv2.resize(arr, (size_w, size_h), interpolation=interp)
nrows = int(size_w // r_nbrs)
ncols = int(size_h // c_nbrs)
return (arr.reshape(r_nbrs, ncols, -1, nrows)
.swapaxes(1,2)
.reshape(-1, ncols, nrows))
a = np.random.randint(1, 9, size=(9,9))
out = [np.hsplit(x, 3) for x in np.vsplit(a,3)]
print(a)
print(out)
yields
[[7 6 2 4 4 2 5 2 3]
[2 3 7 6 8 8 2 6 2]
[4 1 3 1 3 8 1 3 7]
[6 1 1 5 7 2 1 5 8]
[8 8 7 6 6 1 8 8 4]
[6 1 8 2 1 4 5 1 8]
[7 3 4 2 5 6 1 2 7]
[4 6 7 5 8 2 8 2 8]
[6 6 5 5 6 1 2 6 4]]
[[array([[7, 6, 2],
[2, 3, 7],
[4, 1, 3]]), array([[4, 4, 2],
[6, 8, 8],
[1, 3, 8]]), array([[5, 2, 3],
[2, 6, 2],
[1, 3, 7]])], [array([[6, 1, 1],
[8, 8, 7],
[6, 1, 8]]), array([[5, 7, 2],
[6, 6, 1],
[2, 1, 4]]), array([[1, 5, 8],
[8, 8, 4],
[5, 1, 8]])], [array([[7, 3, 4],
[4, 6, 7],
[6, 6, 5]]), array([[2, 5, 6],
[5, 8, 2],
[5, 6, 1]]), array([[1, 2, 7],
[8, 2, 8],
[2, 6, 4]])]]
I publish my solution. Notice that this code doesn't' actually create copies of original array, so it works well with big data. Moreover, it doesn't crash if array cannot be divided evenly (but you can easly add condition for that by deleting ceil and checking if v_slices and h_slices are divided without rest).
import numpy as np
from math import ceil
a = np.arange(9).reshape(3, 3)
p, q = 2, 2
width, height = a.shape
v_slices = ceil(width / p)
h_slices = ceil(height / q)
for h in range(h_slices):
for v in range(v_slices):
block = a[h * p : h * p + p, v * q : v * q + q]
# do something with a block
This code changes (or, more precisely, gives you direct access to part of an array) this:
[[0 1 2]
[3 4 5]
[6 7 8]]
Into this:
[[0 1]
[3 4]]
[[2]
[5]]
[[6 7]]
[[8]]
If you need actual copies, Aenaon code is what you are looking for.
If you are sure that big array can be divided evenly, you can use numpy splitting tools.
to add to #Aenaon answer and his blockfy function, if you are working with COLOR IMAGES/ 3D ARRAY here is my pipeline to create crops of 224 x 224 for 3 channel input
def blockfy(a, p, q):
'''
Divides array a into subarrays of size p-by-q
p: block row size
q: block column size
'''
m = a.shape[0] #image row size
n = a.shape[1] #image column size
# pad array with NaNs so it can be divided by p row-wise and by q column-wise
bpr = ((m-1)//p + 1) #blocks per row
bpc = ((n-1)//q + 1) #blocks per column
M = p * bpr
N = q * bpc
A = np.nan* np.ones([M,N])
A[:a.shape[0],:a.shape[1]] = a
block_list = []
previous_row = 0
for row_block in range(bpc):
previous_row = row_block * p
previous_column = 0
for column_block in range(bpr):
previous_column = column_block * q
block = A[previous_row:previous_row+p, previous_column:previous_column+q]
# remove nan columns and nan rows
nan_cols = np.all(np.isnan(block), axis=0)
block = block[:, ~nan_cols]
nan_rows = np.all(np.isnan(block), axis=1)
block = block[~nan_rows, :]
## append
if block.size:
block_list.append(block)
return block_list
then extended above to
for file in os.listdir(path_to_crop): ### list files in your folder
img = io.imread(path_to_crop + file, as_gray=False) ### open image
r = blockfy(img[:,:,0],224,224) ### crop blocks of 224 x 224 for red channel
g = blockfy(img[:,:,1],224,224) ### crop blocks of 224 x 224 for green channel
b = blockfy(img[:,:,2],224,224) ### crop blocks of 224 x 224 for blue channel
for x in range(0,len(r)):
img = np.array((r[x],g[x],b[x])) ### combine each channel into one patch by patch
img = img.astype(np.uint8) ### cast back to proper integers
img_swap = img.swapaxes(0, 2) ### need to swap axes due to the way things were proceesed
img_swap_2 = img_swap.swapaxes(0, 1) ### do it again
Image.fromarray(img_swap_2).save(path_save_crop+str(x)+"bounding" + file,
format = 'jpeg',
subsampling=0,
quality=100) ### save patch with new name etc

split numpy multidimensional array into equal pieces [duplicate]

Is there a way to slice a 2d array in numpy into smaller 2d arrays?
Example
[[1,2,3,4], -> [[1,2] [3,4]
[5,6,7,8]] [5,6] [7,8]]
So I basically want to cut down a 2x4 array into 2 2x2 arrays. Looking for a generic solution to be used on images.
There was another question a couple of months ago which clued me in to the idea of using reshape and swapaxes. The h//nrows makes sense since this keeps the first block's rows together. It also makes sense that you'll need nrows and ncols to be part of the shape. -1 tells reshape to fill in whatever number is necessary to make the reshape valid. Armed with the form of the solution, I just tried things until I found the formula that works.
You should be able to break your array into "blocks" using some combination of reshape and swapaxes:
def blockshaped(arr, nrows, ncols):
"""
Return an array of shape (n, nrows, ncols) where
n * nrows * ncols = arr.size
If arr is a 2D array, the returned array should look like n subblocks with
each subblock preserving the "physical" layout of arr.
"""
h, w = arr.shape
assert h % nrows == 0, f"{h} rows is not evenly divisible by {nrows}"
assert w % ncols == 0, f"{w} cols is not evenly divisible by {ncols}"
return (arr.reshape(h//nrows, nrows, -1, ncols)
.swapaxes(1,2)
.reshape(-1, nrows, ncols))
turns c
np.random.seed(365)
c = np.arange(24).reshape((4, 6))
print(c)
[out]:
[[ 0 1 2 3 4 5]
[ 6 7 8 9 10 11]
[12 13 14 15 16 17]
[18 19 20 21 22 23]]
into
print(blockshaped(c, 2, 3))
[out]:
[[[ 0 1 2]
[ 6 7 8]]
[[ 3 4 5]
[ 9 10 11]]
[[12 13 14]
[18 19 20]]
[[15 16 17]
[21 22 23]]]
I've posted an inverse function, unblockshaped, here, and an N-dimensional generalization here. The generalization gives a little more insight into the reasoning behind this algorithm.
Note that there is also superbatfish's
blockwise_view. It arranges the
blocks in a different format (using more axes) but it has the advantage of (1)
always returning a view and (2) being capable of handling arrays of any
dimension.
It seems to me that this is a task for numpy.split or some variant.
e.g.
a = np.arange(30).reshape([5,6]) #a.shape = (5,6)
a1 = np.split(a,3,axis=1)
#'a1' is a list of 3 arrays of shape (5,2)
a2 = np.split(a, [2,4])
#'a2' is a list of three arrays of shape (2,5), (2,5), (1,5)
If you have a NxN image you can create, e.g., a list of 2 NxN/2 subimages, and then divide them along the other axis.
numpy.hsplit and numpy.vsplit are also available.
There are some other answers that seem well-suited for your specific case already, but your question piqued my interest in the possibility of a memory-efficient solution usable up to the maximum number of dimensions that numpy supports, and I ended up spending most of the afternoon coming up with possible method. (The method itself is relatively simple, it's just that I still haven't used most of the really fancy features that numpy supports so most of the time was spent researching to see what numpy had available and how much it could do so that I didn't have to do it.)
def blockgen(array, bpa):
"""Creates a generator that yields multidimensional blocks from the given
array(_like); bpa is an array_like consisting of the number of blocks per axis
(minimum of 1, must be a divisor of the corresponding axis size of array). As
the blocks are selected using normal numpy slicing, they will be views rather
than copies; this is good for very large multidimensional arrays that are being
blocked, and for very large blocks, but it also means that the result must be
copied if it is to be modified (unless modifying the original data as well is
intended)."""
bpa = np.asarray(bpa) # in case bpa wasn't already an ndarray
# parameter checking
if array.ndim != bpa.size: # bpa doesn't match array dimensionality
raise ValueError("Size of bpa must be equal to the array dimensionality.")
if (bpa.dtype != np.int # bpa must be all integers
or (bpa < 1).any() # all values in bpa must be >= 1
or (array.shape % bpa).any()): # % != 0 means not evenly divisible
raise ValueError("bpa ({0}) must consist of nonzero positive integers "
"that evenly divide the corresponding array axis "
"size".format(bpa))
# generate block edge indices
rgen = (np.r_[:array.shape[i]+1:array.shape[i]//blk_n]
for i, blk_n in enumerate(bpa))
# build slice sequences for each axis (unfortunately broadcasting
# can't be used to make the items easy to operate over
c = [[np.s_[i:j] for i, j in zip(r[:-1], r[1:])] for r in rgen]
# Now to get the blocks; this is slightly less efficient than it could be
# because numpy doesn't like jagged arrays and I didn't feel like writing
# a ufunc for it.
for idxs in np.ndindex(*bpa):
blockbounds = tuple(c[j][idxs[j]] for j in range(bpa.size))
yield array[blockbounds]
You question practically the same as this one. You can use the one-liner with np.ndindex() and reshape():
def cutter(a, r, c):
lenr = a.shape[0]/r
lenc = a.shape[1]/c
np.array([a[i*r:(i+1)*r,j*c:(j+1)*c] for (i,j) in np.ndindex(lenr,lenc)]).reshape(lenr,lenc,r,c)
To create the result you want:
a = np.arange(1,9).reshape(2,1)
#array([[1, 2, 3, 4],
# [5, 6, 7, 8]])
cutter( a, 1, 2 )
#array([[[[1, 2]],
# [[3, 4]]],
# [[[5, 6]],
# [[7, 8]]]])
Some minor enhancement to TheMeaningfulEngineer's answer that handles the case when the big 2d array cannot be perfectly sliced into equally sized subarrays
def blockfy(a, p, q):
'''
Divides array a into subarrays of size p-by-q
p: block row size
q: block column size
'''
m = a.shape[0] #image row size
n = a.shape[1] #image column size
# pad array with NaNs so it can be divided by p row-wise and by q column-wise
bpr = ((m-1)//p + 1) #blocks per row
bpc = ((n-1)//q + 1) #blocks per column
M = p * bpr
N = q * bpc
A = np.nan* np.ones([M,N])
A[:a.shape[0],:a.shape[1]] = a
block_list = []
previous_row = 0
for row_block in range(bpc):
previous_row = row_block * p
previous_column = 0
for column_block in range(bpr):
previous_column = column_block * q
block = A[previous_row:previous_row+p, previous_column:previous_column+q]
# remove nan columns and nan rows
nan_cols = np.all(np.isnan(block), axis=0)
block = block[:, ~nan_cols]
nan_rows = np.all(np.isnan(block), axis=1)
block = block[~nan_rows, :]
## append
if block.size:
block_list.append(block)
return block_list
Examples:
a = np.arange(25)
a = a.reshape((5,5))
out = blockfy(a, 2, 3)
a->
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]])
out[0] ->
array([[0., 1., 2.],
[5., 6., 7.]])
out[1]->
array([[3., 4.],
[8., 9.]])
out[-1]->
array([[23., 24.]])
For now it just works when the big 2d array can be perfectly sliced into equally sized subarrays.
The code bellow slices
a ->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]])
into this
block_array->
array([[[ 0, 1, 2],
[ 6, 7, 8]],
[[ 3, 4, 5],
[ 9, 10, 11]],
[[12, 13, 14],
[18, 19, 20]],
[[15, 16, 17],
[21, 22, 23]]])
p ang q determine the block size
Code
a = arange(24)
a = a.reshape((4,6))
m = a.shape[0] #image row size
n = a.shape[1] #image column size
p = 2 #block row size
q = 3 #block column size
block_array = []
previous_row = 0
for row_block in range(blocks_per_row):
previous_row = row_block * p
previous_column = 0
for column_block in range(blocks_per_column):
previous_column = column_block * q
block = a[previous_row:previous_row+p,previous_column:previous_column+q]
block_array.append(block)
block_array = array(block_array)
If you want a solution that also handles the cases when the matrix is
not equally divided, you can use this:
from operator import add
half_split = np.array_split(input, 2)
res = map(lambda x: np.array_split(x, 2, axis=1), half_split)
res = reduce(add, res)
Here is a solution based on unutbu's answer that handle case where matrix cannot be equally divided. In this case, it will resize the matrix before using some interpolation. You need OpenCV for this. Note that I had to swap ncols and nrows to make it works, didn't figured why.
import numpy as np
import cv2
import math
def blockshaped(arr, r_nbrs, c_nbrs, interp=cv2.INTER_LINEAR):
"""
arr a 2D array, typically an image
r_nbrs numbers of rows
r_cols numbers of cols
"""
arr_h, arr_w = arr.shape
size_w = int( math.floor(arr_w // c_nbrs) * c_nbrs )
size_h = int( math.floor(arr_h // r_nbrs) * r_nbrs )
if size_w != arr_w or size_h != arr_h:
arr = cv2.resize(arr, (size_w, size_h), interpolation=interp)
nrows = int(size_w // r_nbrs)
ncols = int(size_h // c_nbrs)
return (arr.reshape(r_nbrs, ncols, -1, nrows)
.swapaxes(1,2)
.reshape(-1, ncols, nrows))
a = np.random.randint(1, 9, size=(9,9))
out = [np.hsplit(x, 3) for x in np.vsplit(a,3)]
print(a)
print(out)
yields
[[7 6 2 4 4 2 5 2 3]
[2 3 7 6 8 8 2 6 2]
[4 1 3 1 3 8 1 3 7]
[6 1 1 5 7 2 1 5 8]
[8 8 7 6 6 1 8 8 4]
[6 1 8 2 1 4 5 1 8]
[7 3 4 2 5 6 1 2 7]
[4 6 7 5 8 2 8 2 8]
[6 6 5 5 6 1 2 6 4]]
[[array([[7, 6, 2],
[2, 3, 7],
[4, 1, 3]]), array([[4, 4, 2],
[6, 8, 8],
[1, 3, 8]]), array([[5, 2, 3],
[2, 6, 2],
[1, 3, 7]])], [array([[6, 1, 1],
[8, 8, 7],
[6, 1, 8]]), array([[5, 7, 2],
[6, 6, 1],
[2, 1, 4]]), array([[1, 5, 8],
[8, 8, 4],
[5, 1, 8]])], [array([[7, 3, 4],
[4, 6, 7],
[6, 6, 5]]), array([[2, 5, 6],
[5, 8, 2],
[5, 6, 1]]), array([[1, 2, 7],
[8, 2, 8],
[2, 6, 4]])]]
I publish my solution. Notice that this code doesn't' actually create copies of original array, so it works well with big data. Moreover, it doesn't crash if array cannot be divided evenly (but you can easly add condition for that by deleting ceil and checking if v_slices and h_slices are divided without rest).
import numpy as np
from math import ceil
a = np.arange(9).reshape(3, 3)
p, q = 2, 2
width, height = a.shape
v_slices = ceil(width / p)
h_slices = ceil(height / q)
for h in range(h_slices):
for v in range(v_slices):
block = a[h * p : h * p + p, v * q : v * q + q]
# do something with a block
This code changes (or, more precisely, gives you direct access to part of an array) this:
[[0 1 2]
[3 4 5]
[6 7 8]]
Into this:
[[0 1]
[3 4]]
[[2]
[5]]
[[6 7]]
[[8]]
If you need actual copies, Aenaon code is what you are looking for.
If you are sure that big array can be divided evenly, you can use numpy splitting tools.
to add to #Aenaon answer and his blockfy function, if you are working with COLOR IMAGES/ 3D ARRAY here is my pipeline to create crops of 224 x 224 for 3 channel input
def blockfy(a, p, q):
'''
Divides array a into subarrays of size p-by-q
p: block row size
q: block column size
'''
m = a.shape[0] #image row size
n = a.shape[1] #image column size
# pad array with NaNs so it can be divided by p row-wise and by q column-wise
bpr = ((m-1)//p + 1) #blocks per row
bpc = ((n-1)//q + 1) #blocks per column
M = p * bpr
N = q * bpc
A = np.nan* np.ones([M,N])
A[:a.shape[0],:a.shape[1]] = a
block_list = []
previous_row = 0
for row_block in range(bpc):
previous_row = row_block * p
previous_column = 0
for column_block in range(bpr):
previous_column = column_block * q
block = A[previous_row:previous_row+p, previous_column:previous_column+q]
# remove nan columns and nan rows
nan_cols = np.all(np.isnan(block), axis=0)
block = block[:, ~nan_cols]
nan_rows = np.all(np.isnan(block), axis=1)
block = block[~nan_rows, :]
## append
if block.size:
block_list.append(block)
return block_list
then extended above to
for file in os.listdir(path_to_crop): ### list files in your folder
img = io.imread(path_to_crop + file, as_gray=False) ### open image
r = blockfy(img[:,:,0],224,224) ### crop blocks of 224 x 224 for red channel
g = blockfy(img[:,:,1],224,224) ### crop blocks of 224 x 224 for green channel
b = blockfy(img[:,:,2],224,224) ### crop blocks of 224 x 224 for blue channel
for x in range(0,len(r)):
img = np.array((r[x],g[x],b[x])) ### combine each channel into one patch by patch
img = img.astype(np.uint8) ### cast back to proper integers
img_swap = img.swapaxes(0, 2) ### need to swap axes due to the way things were proceesed
img_swap_2 = img_swap.swapaxes(0, 1) ### do it again
Image.fromarray(img_swap_2).save(path_save_crop+str(x)+"bounding" + file,
format = 'jpeg',
subsampling=0,
quality=100) ### save patch with new name etc

NumPy random shuffle rows independently

I have the following array:
import numpy as np
a = np.array([[ 1, 2, 3],
[ 1, 2, 3],
[ 1, 2, 3]])
I understand that np.random.shuffle(a.T) will shuffle the array along the row, but what I need is for it to shuffe each row idependently. How can this be done in numpy? Speed is critical as there will be several million rows.
For this specific problem, each row will contain the same starting population.
import numpy as np
np.random.seed(2018)
def scramble(a, axis=-1):
"""
Return an array with the values of `a` independently shuffled along the
given axis
"""
b = a.swapaxes(axis, -1)
n = a.shape[axis]
idx = np.random.choice(n, n, replace=False)
b = b[..., idx]
return b.swapaxes(axis, -1)
a = a = np.arange(4*9).reshape(4, 9)
# 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]])
print(scramble(a, axis=1))
yields
[[ 3 8 7 0 4 5 1 2 6]
[12 17 16 9 13 14 10 11 15]
[21 26 25 18 22 23 19 20 24]
[30 35 34 27 31 32 28 29 33]]
while scrambling along the 0-axis:
print(scramble(a, axis=0))
yields
[[18 19 20 21 22 23 24 25 26]
[ 0 1 2 3 4 5 6 7 8]
[27 28 29 30 31 32 33 34 35]
[ 9 10 11 12 13 14 15 16 17]]
This works by first swapping the target axis with the last axis:
b = a.swapaxes(axis, -1)
This is a common trick used to standardize code which deals with one axis.
It reduces the general case to the specific case of dealing with the last axis.
Since in NumPy version 1.10 or higher swapaxes returns a view, there is no copying involved and so calling swapaxes is very quick.
Now we can generate a new index order for the last axis:
n = a.shape[axis]
idx = np.random.choice(n, n, replace=False)
Now we can shuffle b (independently along the last axis):
b = b[..., idx]
and then reverse the swapaxes to return an a-shaped result:
return b.swapaxes(axis, -1)
If you don't want a return value and want to operate on the array directly, you can specify the indices to shuffle.
>>> import numpy as np
>>>
>>>
>>> a = np.array([[1,2,3], [1,2,3], [1,2,3]])
>>>
>>> # Shuffle row `2` independently
>>> np.random.shuffle(a[2])
>>> a
array([[1, 2, 3],
[1, 2, 3],
[3, 2, 1]])
>>>
>>> # Shuffle column `0` independently
>>> np.random.shuffle(a[:,0])
>>> a
array([[3, 2, 3],
[1, 2, 3],
[1, 2, 1]])
If you want a return value as well, you can use numpy.random.permutation, in which case replace np.random.shuffle(a[n]) with a[n] = np.random.permutation(a[n]).
Warning, do not do a[n] = np.random.shuffle(a[n]). shuffle does not return anything, so the row/column you end up "shuffling" will be filled with nan instead.
Good answer above. But I will throw in a quick and dirty way:
a = np.array([[1,2,3], [1,2,3], [1,2,3]])
ignore_list_outpput = [np.random.shuffle(x) for x in a]
Then, a can be something like this
array([[2, 1, 3],
[4, 6, 5],
[9, 7, 8]])
Not very elegant but you can get this job done with just one short line.
Building on my comment to #Hun's answer, here's the fastest way to do this:
def shuffle_along(X):
"""Minimal in place independent-row shuffler."""
[np.random.shuffle(x) for x in X]
This works in-place and can only shuffle rows. If you need more options:
def shuffle_along(X, axis=0, inline=False):
"""More elaborate version of the above."""
if not inline:
X = X.copy()
if axis == 0:
[np.random.shuffle(x) for x in X]
if axis == 1:
[np.random.shuffle(x) for x in X.T]
if not inline:
return X
This, however, has the limitation of only working on 2d-arrays. For higher dimensional tensors, I would use:
def shuffle_along(X, axis=0, inline=True):
"""Shuffle along any axis of a tensor."""
if not inline:
X = X.copy()
np.apply_along_axis(np.random.shuffle, axis, X) # <-- I just changed this
if not inline:
return X
You can do it with numpy without any loop or extra function, and much more faster. E. g., we have an array of size (2, 6) and we want a sub array (2,2) with independent random index for each column.
import numpy as np
test = np.array([[1, 1],
[2, 2],
[0.5, 0.5],
[0.3, 0.3],
[4, 4],
[7, 7]])
id_rnd = np.random.randint(6, size=(2, 2)) # select random numbers, use choice and range if don want replacement.
new = np.take_along_axis(test, id_rnd, axis=0)
Out:
array([[2. , 2. ],
[0.5, 2. ]])
It works for any number of dimensions.
As of NumPy 1.20.0 released in January 2021 we have a permuted() method on the new Generator type (introduced with the new random API in NumPy 1.17.0, released in July 2019). This does exactly what you need:
import numpy as np
rng = np.random.default_rng()
a = np.array([
[1, 2, 3],
[1, 2, 3],
[1, 2, 3],
])
shuffled = rng.permuted(a, axis=1)
This gives you something like
>>> print(shuffled)
[[2 3 1]
[1 3 2]
[2 1 3]]
As you can see, the rows are permuted independently. This is in sharp contrast with both rng.permutation() and rng.shuffle().
If you want an in-place update you can pass the original array as the out keyword argument. And you can use the axis keyword argument to choose the direction along which to shuffle your array.

How to bin a 2D array in numpy?

I'm new to numpy and I have a 2D array of objects that I need to bin into a smaller matrix and then get a count of the number of objects in each bin to make a heatmap. I followed the answer on this thread to create the bins and do the counts for a simple array but I'm not sure how to extend it to 2 dimensions. Here's what I have so far:
data_matrix = numpy.ndarray((500,500),dtype=float)
# fill array with values.
bins = numpy.linspace(0,50,50)
digitized = numpy.digitize(data_matrix, bins)
binned_data = numpy.ndarray((50,50))
for i in range(0,len(bins)):
for j in range(0,len(bins)):
k = len(data_matrix[digitized == i:digitized == j]) # <-not does not work
binned_data[i:j] = k
P.S. the [digitized == i] notation on an array will return an array of binary values. I cannot find documentation on this notation anywhere. A link would be appreciated.
You can reshape the array to a four dimensional array that reflects the desired block structure, and then sum along both axes within each block. Example:
>>> a = np.arange(24).reshape(4, 6)
>>> a
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]])
>>> a.reshape(2, 2, 2, 3).sum(3).sum(1)
array([[ 24, 42],
[ 96, 114]])
If a has the shape m, n, the reshape should have the form
a.reshape(m_bins, m // m_bins, n_bins, n // n_bins)
At first I was also going to suggest that you use np.histogram2d rather than reinventing the wheel, but then I realized that it would be overkill to use that and would need some hacking still.
If I understand correctly, you just want to sum over submatrices of your input. That's pretty easy to brute force: going over your output submatrix and summing up each subblock of your input:
import numpy as np
def submatsum(data,n,m):
# return a matrix of shape (n,m)
bs = data.shape[0]//n,data.shape[1]//m # blocksize averaged over
return np.reshape(np.array([np.sum(data[k1*bs[0]:(k1+1)*bs[0],k2*bs[1]:(k2+1)*bs[1]]) for k1 in range(n) for k2 in range(m)]),(n,m))
# set up dummy data
N,M = 4,6
data_matrix = np.reshape(np.arange(N*M),(N,M))
# set up size of 2x3-reduced matrix, assume congruity
n,m = N//2,M//3
reduced_matrix = submatsum(data_matrix,n,m)
# check output
print(data_matrix)
print(reduced_matrix)
This prints
print(data_matrix)
[[ 0 1 2 3 4 5]
[ 6 7 8 9 10 11]
[12 13 14 15 16 17]
[18 19 20 21 22 23]]
print(reduced_matrix)
[[ 24 42]
[ 96 114]]
which is indeed the result for summing up submatrices of shape (2,3).
Note that I'm using // for integer division to make sure it's python3-compatible, but in case of python2 you can just use / for division (due to the numbers involved being integers).
Another solution is to have a look at the binArray function on the comments here:
Binning a numpy array
To use your example :
data_matrix = numpy.ndarray((500,500),dtype=float)
binned_data = binArray(data_matrix, 0, 10, 10, np.sum)
binned_data = binArray(binned_data, 1, 10, 10, np.sum)
The result sum all square of size 10x10 in data_matrix (of size 500x500) to obtain a single value per square in binned_data (of size 50x50).
Hope this help !

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