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How to merge 2 numpy arrays?

I feel like there is some documentation I am missing, but I can't find anything on this specific example - everything is just about concatenating or stacking arrays.
I have array x and array y both of shape (2,3)
x = [[1,2,3],[4,5,6]]
y = [[7,8,9],[10,11,12]]
x = 1 2 3
4 5 6
y = 7 8 9
10 11 12
I want array z with shape (2,3,2) that looks like this
z = [[[1,7],[2,8],[3,9]],[[4,10],[5,11],[6,12]]]
z = [1,7] [2,8] [3,9]
[4 10] [5 11] [6 12]
basically joins the x and y element at each position.

Sounds like the function you're looking for is stack(), using it to stack along the 3rd dimension.
import numpy as np
x = np.asarray([[1,2,3],[4,5,6]])
y = np.asarray([[7,8,9],[10,11,12]])
z = np.stack((x, y), 2)

If you have two-dimensional matrices, you can use numpy.dstack():
z = np.dstack((x, y))

In [39]: z = np.concatenate((x[...,None], y[...,None]), axis=2)
In [40]: z
Out[40]:
array([[[ 1, 7],
[ 2, 8],
[ 3, 9]],
[[ 4, 10],
[ 5, 11],
[ 6, 12]]])

Create 2d Array in Python Using For Loop Results

I run some code within a for loop. I want to take the results from my loop and place them in a 2d array with 2 columns and as many rows as times I iterate the loop. Here's a simplified version of what I have:
for i in range(10):
'bunch of stuff'
centerx = xc
centery = yc
How can I save my values for centerx and centery in a 2d array with 2 columns and 10 rows? Any help is appreciated, thanks!

You can try this:
import numpy as np
listvals = []
for i in range(10):
listvals.append((xc, yc))
mat_vals = np.vstack(listvals)
This will output an ndarray like this:
[[ 2 0]
[ 3 1]
[ 4 2]
[ 5 3]
[ 6 4]
[ 7 5]
[ 8 6]
[ 9 7]
[10 8]
[11 9]]
Or this maybe is better:
import numpy as np
list_xc = []
list_yc = []
for i in np.arange(10):
list_xc.append(xc)
list_yc.append(yc)
mat_xc = np.asarray(list_xc)
mat_yc = np.asarray(list_yc)
mat_f = np.column_stack((mat_xc, mat_yc))

You can do this:
aList = []
for i in range(10):
aList.append([xc, yc])

Try this comprehension,
[ [i, i*10] for i in range(5) ]
which delivers
[[0, 0], [1, 10], [2, 20], [3, 30], [4, 40], [5, 50]]

Replace elements in numpy array using list of old and new values

I want to replace elements in a numpy array using a list of old values and new values. See below for a code example (replace_old is the requested method). The method must work for both int, float and string elements. How do I do that?
import numpy as np
dat = np.hstack((np.arange(1,9), np.arange(1,4)))
print dat # [1 2 3 4 5 6 7 8 1 2 3]
old_val = [2, 5]
new_val = [11, 57]
new_dat = replace_old(dat, old_val, new_val)
print new_dat # [1 11 3 4 57 6 7 8 1 11 3]

You can use np.place :
>>> np.place(dat,np.in1d(dat,old_val),new_val)
>>> dat
array([ 1, 11, 3, 4, 57, 6, 7, 8, 1, 11, 3])
For creating the mask array you can use np.in1d(arr1,arr2) which will give you :
a boolean array the same length as ar1 that is True where an element of ar1 is in ar2 and False otherwise
Edit:Note that the preceding recipe will replace old_values based on those order and as #ajcr mentioned it wont work for another arrays,so as a general way for now I suggest the following way using a loop (which I don't think that was the best way):
>>> dat2 = np.array([1, 2, 1, 2])
>>> old_val = [1, 2]
>>> new_val = [33, 66]
>>> z=np.array((old_val,new_val)).T
>>> for i,j in z:
... np.place(dat2,dat2==i,j)
...
>>> dat2
array([33, 66, 33, 66])
In this case you create a new array (z) which is contains the relevant pairs from old_val and new_val and then you can pass them to np.place and replace them .

Slice 2d array into smaller 2d arrays

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

Slicing python matrix into quadrants

Suppose I have the following matrix in python:
[[1,2,3,4],
[5,6,7,8],
[9,10,11,12],
[13,14,15,16]]
I want to slice it into the following matrices (or quadrants/corners):
[[1,2], [5,6]]
[[3,4], [7,8]]
[[9,10], [13,14]]
[[11,12], [15,16]]
Is this supported with standard slicing operators in python or is it necessary to use an extended library like numpy?

If you are always working with a 4x4 matrix:
a = [[1 ,2 , 3, 4],
[5 ,6 , 7, 8],
[9 ,10,11,12],
[13,14,15,16]]
top_left = [a[0][:2], a[1][:2]]
top_right = [a[0][2:], a[1][2:]]
bot_left = [a[2][:2], a[3][:2]]
bot_right = [a[2][2:], a[3][2:]]
You could also do the same for an arbitrary size matrix:
h = len(a)
w = len(a[1])
top_left = [a[i][:w // 2] for i in range(h // 2)]
top_right = [a[i][w // 2:] for i in range(h // 2)]
bot_left = [a[i][:w // 2] for i in range(h // 2, h)]
bot_right = [a[i][w // 2:] for i in range(h // 2, h)]

The question is already answered, but I think this solution is more general.
It can also be used numpy.split and list comprehension in the following way:
import numpy as np
A = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]])
B = [M for SubA in np.split(A,2, axis = 0) for M in np.split(SubA,2, axis = 1)]
Getting:
>>>[array([[1, 2],[5, 6]]),
array([[3, 4],[7, 8]]),
array([[ 9, 10],[13, 14]]),
array([[11, 12],[15, 16]])]
Now if you want to have them assigned into different variables, just:
C1,C2,C3,C4 = B
Have a look to numpy.split doc.
Changing the parameter indices_or_sections you can get a higher number of splits.

>>> a = [[1,2,3,4], [5,6,7,8], [9,10,11,12], [13,14,15,16]]
>>> x = map(lambda x:x[:2], a)
>>> x
[[1, 2], [5, 6], [9, 10], [13, 14]]
>>> y = map(lambda x: x[2:], a)
>>> y
[[3, 4], [7, 8], [11, 12], [15, 16]]
>>> x[:2] + y[:2] + x[2:] + y[2:]
[[1, 2], [5, 6], [3, 4], [7, 8], [9, 10], [13, 14], [11, 12], [15, 16]]

Although the answers can provide the required solution. These are not applicable for the arrays in different sizes. If you have a NumPy array in size of (6x7), then these methods will not create a solution. I have prepared a solution for myself and want to share it here.
Note: In my solution, there will be overlaps due to the different axis sizes.
I have created this solution to divide an astronomical image into four quadrants. I, then, use these quadrants to calculate the mean and median in an annulus.
import numpy as np
def quadrant_split2d(array):
"""Example function for identifiying the elements of quadrants in an array.
array:
A 2D NumPy array.
Returns:
The quadrants. True for the members of the quadrants, False otherwise.
"""
Ysize = array.shape[0]
Xsize = array.shape[1]
y, x = np.indices((Ysize,Xsize))
if not (Xsize==Ysize)&(Xsize % 2 == 0): print ('There will be overlaps')
sector1=(x<Xsize/2)&(y<Ysize/2)
sector2=(x>Xsize/2-1)&(y<Ysize/2)
sector3=(x<Xsize/2)&(y>Ysize/2-1)
sector4=(x>Xsize/2-1)&(y>Ysize/2-1)
sectors=(sector1,sector2,sector3,sector4)
return sectors
You can test the function with the different type of arrays.
For example:
test = np.arange(42).reshape(6,7)
print ('Original array:\n', test)
sectors = quadrant_split2d(test)
print ('Sectors:')
for ss in sectors: print (test[ss])
This will give us the following sectors:
[ 0 1 2 3 7 8 9 10 14 15 16 17]
[ 3 4 5 6 10 11 12 13 17 18 19 20]
[21 22 23 24 28 29 30 31 35 36 37 38]
[24 25 26 27 31 32 33 34 38 39 40 41]