Numpy Problems with Arrays of poly1d Objects - python

I'd like to first start this out with the fact that it is possible, in numpy, to create an array of poly1d objects:
random_poly = np.frompyfunc(lambda i, j: np.poly1d(np.random.randint(1, 4, 3)), 2, 1)
def random_poly_array(shape):
return np.fromfunction(random_poly, shape)
a1 = random_poly_array((3,3))
This works just fine, and we can even multiply matrices made from this form using np.dot:
a2 = random_poly_array((3,3))
a1_x_a2 = np.dot(a1, a2)
However, most other methods fail to work. For example, you can't take a list of certain poly1d objects and convert it into an array:
np.array([np.poly1d([1,2,3]), np.poly1d([1,2,3])])
As that will raise ValueError: cannot copy sequence with size 2 to array axis with dimension 3. To add to the confusion,
np.array([np.poly1d([1,2]), np.poly1d([1,2])])
will not raise an error, but instead create a 2x2 array of just 2's. Adding dtype=object has no affect, and numpy will still try to convert the poly1d objects to arrays.
The reason why this is problematic is that one cannot take an array of dimension d and convert it to an array of poly1d objects of dimension d-1. I would have expected
arr = np.arange(1, 10).reshape(3,3)
np.apply_along_axis(np.poly1d, 0, arr)
To return an array of poly1d objects, but instead it returns an unalterated array. Even worse, if arr=np.arange(9).reshape(3,3), it will throw an error, as the first poly1d object created will have a length of 2 instead of 3 due to the zero coefficient. Thus, my question is this: is there a feasible method to create poly1d arrays in numpy? If not, why not?


Using the concept of None forcing numpy to not broadcast an object into an array, something brought to my attention by Paul Panzer, I created a function which will transform the last axis into a poly1d object:
def array_to_poly(arr):
return np.apply_along_axis(lambda poly: [None, np.poly1d(poly)], -1, arr)[..., 1]
However, if we're okay with abusing more than one system in a single function, we can make it apply over arbitrary axes:
def array_to_poly(arr, axis=-1):
temp_arr = np.apply_along_axis(lambda poly: [None, np.poly1d(poly)], axis, arr)
n = temp_arr.ndim
s = [slice(None) if i != axis%n else 1 for i in range(n)]
return temp_arr[s]
Testing it with arr = np.arange(1, 25).reshape(2,3,4), we obtain:
In [ ]: array_to_poly(arr, 0)
Out[ ]:
array([[poly1d([ 1, 13]), poly1d([ 2, 14]), poly1d([ 3, 15]),
poly1d([ 4, 16])],
[poly1d([ 5, 17]), poly1d([ 6, 18]), poly1d([ 7, 19]),
poly1d([ 8, 20])],
[poly1d([ 9, 21]), poly1d([10, 22]), poly1d([11, 23]),
poly1d([12, 24])]], dtype=object)
In [ ]: array_to_poly(arr, 1)
Out[ ]:
array([[poly1d([1, 5, 9]), poly1d([ 2, 6, 10]), poly1d([ 3, 7, 11]),
poly1d([ 4, 8, 12])],
[poly1d([13, 17, 21]), poly1d([14, 18, 22]), poly1d([15, 19, 23]),
poly1d([16, 20, 24])]], dtype=object)
In [ ]: array_to_poly(arr, 2)
Out[ ]:
array([[poly1d([1, 2, 3, 4]), poly1d([5, 6, 7, 8]),
poly1d([ 9, 10, 11, 12])],
[poly1d([13, 14, 15, 16]), poly1d([17, 18, 19, 20]),
poly1d([21, 22, 23, 24])]], dtype=object)
as expected.

Related

Why do these two numpy.divide operations give such different results?

I would like to correct the values in hyperspectral readings from a cameara using the formula described over here;
the captured data is subtracted by dark reference and divided with
white reference subtracted dark reference.
In the original example, the task is rather simple, white and dark reference has the same shape as the main data so the formula is executed as:
corrected_nparr = np.divide(np.subtract(data_nparr, dark_nparr),
np.subtract(white_nparr, dark_nparr))
However the main data is much larger in my experience. Shapes in my case are as following;
$ white_nparr.shape, dark_nparr.shape, data_nparr.shape
((100, 640, 224), (100, 640, 224), (4300, 640, 224))
that's why I repeat the reference arrays.
white_nparr_rep = white_nparr.repeat(43, axis=0)
dark_nparr_rep = dark_nparr.repeat(43, axis=0)
return np.divide(np.subtract(data_nparr, dark_nparr_rep), np.subtract(white_nparr_rep, dark_nparr_rep))
And it works almost perfectly, as can be seen in the image at the left. But this approach requires enormous amount of memory, so I decided to traverse the large array and replace the original values with corrected ones on-the-go instead:
ref_scale = dark_nparr.shape[0]
data_scale = data_nparr.shape[0]
for i in range(int(data_scale / ref_scale)):
data_nparr[i*ref_scale:(i+1)*ref_scale] =
np.divide
(
np.subtract(data_nparr[i*ref_scale:(i+1)*ref_scale], dark_nparr),
np.subtract(white_nparr, dark_nparr)
)
But that traversal approach gives me the ugliest of results, as can be seen in the right. I'd appreciate any idea that would help me fix this.
Note: I apply 20-times co-adding (mean of 20 readings) to obtain the images below.
EDIT: dtype of each array is as following:
$ white_nparr.dtype, dark_nparr.dtype, data_nparr.dtype
(dtype('float32'), dtype('float32'), dtype('float32'))

Your two methods don't agree because in the first method you used
white_nparr_rep = white_nparr.repeat(43, axis=0)
but the second method corresponds to using
white_nparr_rep = np.tile(white_nparr, (43, 1, 1))
If the first method is correct, you'll have to adjust the second method to act accordingly. Perhaps
for i in range(int(data_scale / ref_scale)):
data_nparr[i*ref_scale:(i+1)*ref_scale] =
np.divide
(
np.subtract(data_nparr[i*ref_scale:(i+1)*ref_scale], dark_nparr[i]),
np.subtract(white_nparr[i], dark_nparr[i])
)
A simple example with 2-d arrays that shows the difference between repeat and tile:
In [146]: z
Out[146]:
array([[ 1, 2, 3, 4, 5],
[11, 12, 13, 14, 15]])
In [147]: np.repeat(z, 3, axis=0)
Out[147]:
array([[ 1, 2, 3, 4, 5],
[ 1, 2, 3, 4, 5],
[ 1, 2, 3, 4, 5],
[11, 12, 13, 14, 15],
[11, 12, 13, 14, 15],
[11, 12, 13, 14, 15]])
In [148]: np.tile(z, (3, 1))
Out[148]:
array([[ 1, 2, 3, 4, 5],
[11, 12, 13, 14, 15],
[ 1, 2, 3, 4, 5],
[11, 12, 13, 14, 15],
[ 1, 2, 3, 4, 5],
[11, 12, 13, 14, 15]])
Off topic postscript: I don't know why the author of the page that you linked to writes NumPy expressions as (for example):
corrected_nparr = np.divide(
np.subtract(data_nparr, dark_nparr),
np.subtract(white_nparr, dark_nparr))
NumPy allows you to write that as
corrected_nparr = (data_nparr - dark_nparr) / (white_nparr - dark_nparr)
whick looks much nicer to me.

getting multiple array after performing subtraction operation within array elements

import numpy as np
m = []
k = []
a = np.array([[1,2,3,4,5,6],[50,51,52,40,20,30],[60,71,82,90,45,35]])
for i in range(len(a)):
m.append(a[i, -1:])
for j in range(len(a[i])-1):
n = abs(m[i] - a[i,j])
k.append(n)
k.append(m[i])
print(k)
Expected Output in k:
[5,4,3,2,1,6],[20,21,22,10,10,30],[25,36,47,55,10,35]
which is also a numpy array.
But the output that I am getting is
[array([5]), array([4]), array([3]), array([2]), array([1]), array([6]), array([20]), array([21]), array([22]), array([10]), array([10]), array([30]), array([25]), array([36]), array([47]), array([55]), array([10]), array([35])]
How can I solve this situation?

You want to subtract the last column of each sub array from themselves. Why don't you use a vectorized approach? You can do all the subtractions at once by subtracting the last column from the rest of the items and then column_stack together with unchanged version of the last column. Also note that you need to change the dimension of the last column inorder to be subtractable from the 2D array. For that sake we can use broadcasting.
In [71]: np.column_stack((abs(a[:, :-1] - a[:, None, -1]), a[:,-1]))
Out[71]:
array([[ 5, 4, 3, 2, 1, 6],
[20, 21, 22, 10, 10, 30],
[25, 36, 47, 55, 10, 35]])

Efficient numpy array random views with dropped dimensions

For computer vision training purposes, random cropping is often used as a data augmentation technique. At each iteration, a batch of random crops is generated and fed to the network being trained. This needs to be efficient, as it is done at each training iteration.
If the data has too many dimensions, random dimension selection might also be needed. Random frames can be selected in a video for example. The data can even have 4 dimensions (3 in space + time), or more.
How can one write an efficient generator of random views of lower dimension?
A very naïve version for getting 2D views from 3D data, and only one by one, could be:
import numpy as np
import numpy.random as nr
def views():
# suppose `data` comes from elsewhere
# data.shape is (n1, n2, n3)
while True:
drop_dim = nr.randint(0, 3)
drop_dim_keep = nr.randint(0, shape[drop_dim])
selector = np.zeros(shape, dtype=bool)
if drop_dim == 0:
selector[drop_dim_keep, :, :] = 1
elif drop_dim == 1:
selector[:, drop_dim_keep, :] = 1
else:
selector[:, :, drop_dim_keep] = 1
yield np.squeeze(data[selector])
A more elegant solution probably exists, where at least:
there is no ugly if/else on the randomly chosen dimension
views can take a batch_size integer argument and generate several views at once without a loop
the dimension of input/output data is not specified (e.g. can do 3D -> 2D as well as 4D -> 2D)

I tweaked your function to clarify what it's doing:
def views():
# suppose `data` comes from elsewhere
# data.shape is (n1, n2, n3)
while True:
drop_dim = nr.randint(0, 3)
dropshape = list(shape[:])
dropshape[drop_dim] -= 1
drop_dim_keep = nr.randint(0, shape[drop_dim])
print(drop_dim, drop_dim_keep)
selector = np.ones(shape, dtype=bool)
if drop_dim == 0:
selector[drop_dim_keep, :, :] = 0
elif drop_dim == 1:
selector[:, drop_dim_keep, :] = 0
else:
selector[:, :, drop_dim_keep] = 0
yield data[selector].reshape(dropshape)
A small sample run:
In [534]: data = np.arange(24).reshape(shape)
In [535]: data
Out[535]:
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]]])
In [536]: v = views()
In [537]: next(v)
2 1
Out[537]:
array([[[ 0, 2, 3],
[ 4, 6, 7],
[ 8, 10, 11]],
[[12, 14, 15],
[16, 18, 19],
[20, 22, 23]]])
In [538]: next(v)
0 0
Out[538]:
array([[[12, 13, 14, 15],
[16, 17, 18, 19],
[20, 21, 22, 23]]])
So it's picking one of the dimensions, and for that dimension dropping one 'column'.
The main efficiency issue is whether it's returning a view or a copy. In this case it has to return a copy.
You are using a boolean mask to select the return, exactly the same as what np.delete does in this case.
In [544]: np.delete(data,1,2).shape
Out[544]: (2, 3, 3)
In [545]: np.delete(data,0,0).shape
Out[545]: (1, 3, 4)
So you could replace much of your interals with delete, letting it take care of generalizing the dimensions. Look at its code to see how it handles those details (It isn't short and sweet!).
def rand_delete():
# suppose `data` comes from elsewhere
# data.shape is (n1, n2, n3)
while True:
drop_dim = nr.randint(0, 3)
drop_dim_keep = nr.randint(0, shape[drop_dim])
print(drop_dim, drop_dim_keep)
yield np.delete(data, drop_dim_keep, drop_dim)
In [547]: v1=rand_delete()
In [548]: next(v1)
0 1
Out[548]:
array([[[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]]])
In [549]: next(v1)
2 0
Out[549]:
array([[[ 1, 2, 3],
[ 5, 6, 7],
[ 9, 10, 11]],
[[13, 14, 15],
[17, 18, 19],
[21, 22, 23]]])
Replace the delete with take:
def rand_take():
while True:
take_dim = nr.randint(0, 3)
take_keep = nr.randint(0, shape[take_dim])
print(take_dim, take_keep)
yield np.take(data, take_keep, axis=take_dim)
In [580]: t = rand_take()
In [581]: next(t)
0 0
Out[581]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
In [582]: next(t)
2 3
Out[582]:
array([[ 3, 7, 11],
[15, 19, 23]])
np.take returns a copy, but the equivalent slicing does not
In [601]: data.__array_interface__['data']
Out[601]: (182632568, False)
In [602]: np.take(data,0,1).__array_interface__['data']
Out[602]: (180099120, False)
In [603]: data[:,0,:].__array_interface__['data']
Out[603]: (182632568, False)
A slicing tuple can be generated with expressions like
In [604]: idx = [slice(None)]*data.ndim
In [605]: idx[1] = 0
In [606]: data[tuple(idx)]
Out[606]:
array([[ 0, 1, 2, 3],
[12, 13, 14, 15]])
Various numpy functions that take an axis parameter construct an indexing tuple like this. (For example one or more of the apply... functions.

Is there any way to find top-left and right-bottom pixels of cropped image from a full image using Python? [duplicate]

I have a large NumPy.array field_array and a smaller array match_array, both consisting of int values. Using the following example, how can I check if any match_array-shaped segment of field_array contains values that exactly correspond to the ones in match_array?
import numpy
raw_field = ( 24, 25, 26, 27, 28, 29, 30, 31, 23, \
33, 34, 35, 36, 37, 38, 39, 40, 32, \
-39, -38, -37, -36, -35, -34, -33, -32, -40, \
-30, -29, -28, -27, -26, -25, -24, -23, -31, \
-21, -20, -19, -18, -17, -16, -15, -14, -22, \
-12, -11, -10, -9, -8, -7, -6, -5, -13, \
-3, -2, -1, 0, 1, 2, 3, 4, -4, \
6, 7, 8, 4, 5, 6, 7, 13, 5, \
15, 16, 17, 8, 9, 10, 11, 22, 14)
field_array = numpy.array(raw_field, int).reshape(9,9)
match_array = numpy.arange(12).reshape(3,4)
These examples ought to return True since the pattern described by match_array aligns over [6:9,3:7].

Approach #1
This approach derives from a solution to Implement Matlab's im2col 'sliding' in python that was designed to rearrange sliding blocks from a 2D array into columns. Thus, to solve our case here, those sliding blocks from field_array could be stacked as columns and compared against column vector version of match_array.
Here's a formal definition of the function for the rearrangement/stacking -
def im2col(A,BLKSZ):
# Parameters
M,N = A.shape
col_extent = N - BLKSZ[1] + 1
row_extent = M - BLKSZ[0] + 1
# Get Starting block indices
start_idx = np.arange(BLKSZ[0])[:,None]*N + np.arange(BLKSZ[1])
# Get offsetted indices across the height and width of input array
offset_idx = np.arange(row_extent)[:,None]*N + np.arange(col_extent)
# Get all actual indices & index into input array for final output
return np.take (A,start_idx.ravel()[:,None] + offset_idx.ravel())
To solve our case, here's the implementation based on im2col -
# Get sliding blocks of shape same as match_array from field_array into columns
# Then, compare them with a column vector version of match array.
col_match = im2col(field_array,match_array.shape) == match_array.ravel()[:,None]
# Shape of output array that has field_array compared against a sliding match_array
out_shape = np.asarray(field_array.shape) - np.asarray(match_array.shape) + 1
# Now, see if all elements in a column are ONES and reshape to out_shape.
# Finally, find the position of TRUE indices
R,C = np.where(col_match.all(0).reshape(out_shape))
The output for the given sample in the question would be -
In [151]: R,C
Out[151]: (array([6]), array([3]))
Approach #2
Given that opencv already has template matching function that does square of differences, you can employ that and look for zero differences, which would be your matching positions. So, if you have access to cv2 (opencv module), the implementation would look something like this -
import cv2
from cv2 import matchTemplate as cv2m
M = cv2m(field_array.astype('uint8'),match_array.astype('uint8'),cv2.TM_SQDIFF)
R,C = np.where(M==0)
giving us -
In [204]: R,C
Out[204]: (array([6]), array([3]))
Benchmarking
This section compares runtimes for all the approaches suggested to solve the question. The credit for the various methods listed in this section goes to their contributors.
Method definitions -
def seek_array(search_in, search_for, return_coords = False):
si_x, si_y = search_in.shape
sf_x, sf_y = search_for.shape
for y in xrange(si_y-sf_y+1):
for x in xrange(si_x-sf_x+1):
if numpy.array_equal(search_for, search_in[x:x+sf_x, y:y+sf_y]):
return (x,y) if return_coords else True
return None if return_coords else False
def skimage_based(field_array,match_array):
windows = view_as_windows(field_array, match_array.shape)
return (windows == match_array).all(axis=(2,3)).nonzero()
def im2col_based(field_array,match_array):
col_match = im2col(field_array,match_array.shape)==match_array.ravel()[:,None]
out_shape = np.asarray(field_array.shape) - np.asarray(match_array.shape) + 1
return np.where(col_match.all(0).reshape(out_shape))
def cv2_based(field_array,match_array):
M = cv2m(field_array.astype('uint8'),match_array.astype('uint8'),cv2.TM_SQDIFF)
return np.where(M==0)
Runtime tests -
Case # 1 (Sample data from question):
In [11]: field_array
Out[11]:
array([[ 24, 25, 26, 27, 28, 29, 30, 31, 23],
[ 33, 34, 35, 36, 37, 38, 39, 40, 32],
[-39, -38, -37, -36, -35, -34, -33, -32, -40],
[-30, -29, -28, -27, -26, -25, -24, -23, -31],
[-21, -20, -19, -18, -17, -16, -15, -14, -22],
[-12, -11, -10, -9, -8, -7, -6, -5, -13],
[ -3, -2, -1, 0, 1, 2, 3, 4, -4],
[ 6, 7, 8, 4, 5, 6, 7, 13, 5],
[ 15, 16, 17, 8, 9, 10, 11, 22, 14]])
In [12]: match_array
Out[12]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
In [13]: %timeit seek_array(field_array, match_array, return_coords = False)
1000 loops, best of 3: 465 µs per loop
In [14]: %timeit skimage_based(field_array,match_array)
10000 loops, best of 3: 97.9 µs per loop
In [15]: %timeit im2col_based(field_array,match_array)
10000 loops, best of 3: 74.3 µs per loop
In [16]: %timeit cv2_based(field_array,match_array)
10000 loops, best of 3: 30 µs per loop
Case #2 (Bigger random data):
In [17]: field_array = np.random.randint(0,4,(256,256))
In [18]: match_array = field_array[100:116,100:116].copy()
In [19]: %timeit seek_array(field_array, match_array, return_coords = False)
1 loops, best of 3: 400 ms per loop
In [20]: %timeit skimage_based(field_array,match_array)
10 loops, best of 3: 54.3 ms per loop
In [21]: %timeit im2col_based(field_array,match_array)
10 loops, best of 3: 125 ms per loop
In [22]: %timeit cv2_based(field_array,match_array)
100 loops, best of 3: 4.08 ms per loop

There's no such search function built in to NumPy, but it is certainly possible to do in NumPy
As long as your arrays are not too massive*, you could use a rolling window approach:
from skimage.util import view_as_windows
windows = view_as_windows(field_array, match_array.shape)
The function view_as_windows is written purely in NumPy so if you don't have skimage you can always copy the code from here.
Then to see if the sub-array appears in the larger array, you can write:
>>> (windows == match_array).all(axis=(2,3)).any()
True
To find the indices of where the top-left corner of the sub-array matches, you can write:
>>> (windows == match_array).all(axis=(2,3)).nonzero()
(array([6]), array([3]))
This approach should also work for arrays of higher dimensions.
*although the array windows takes up no additional memory (only the strides and shape are changed to create a new view of the data), writing windows == match_array creates a boolean array of size (7, 6, 3, 4) which is 504 bytes of memory. If you're working with very large arrays, this approach might not be feasible.

One solution is to search the entire search_in array block-at-a-time (a 'block' being a search_for-shaped slice) until either a matching segment is found or the search_for array is exhausted. I can use it to get coordinates for the matching block, or just a bool result by sending True or False for the return_coords optional argument...
def seek_array(search_in, search_for, return_coords = False):
"""Searches for a contiguous instance of a 2d array `search_for` within a larger `search_in` 2d array.
If the optional argument return_coords is True, the xy coordinates of the zeroeth value of the first matching segment of search_in will be returned, or None if there is no matching segment.
If return_coords is False, a boolean will be returned.
* Both arrays must be sent as two-dimensional!"""
si_x, si_y = search_in.shape
sf_x, sf_y = search_for.shape
for y in xrange(si_y-sf_y+1):
for x in xrange(si_x-sf_x+1):
if numpy.array_equal(search_for, search_in[x:x+sf_x, y:y+sf_y]):
return (x,y) if return_coords else True # don't forget that coordinates are transposed when viewing NumPy arrays!
return None if return_coords else False
I wonder if NumPy doesn't already have a function that can do the same thing, though...

To add to the answers already posted, I'd like to add one that takes into account errors due to floating point precision in case that matrices come from, let's say, image processing for instance, where numbers are subject to floating point operations.
You can recurse the indexes of the larger matrix, searching for the smaller matrix. Then you can extract a submatrix of the larger matrix matching the size of the smaller matrix.
You have a match if the contents of both, the submatrix of 'large' and the 'small' matrix match.
The following example shows how to return the first indexes of the location in the large matrix found to match. It would be trivial to extend this function to return an array of locations found to match if that's the intent.
import numpy as np
def find_submatrix(a, b):
""" Searches the first instance at which 'b' is a submatrix of 'a', iterates
rows first. Returns the indexes of a at which 'b' was found, or None if
'b' is not contained within 'a'"""
a_rows=a.shape[0]
a_cols=a.shape[1]
b_rows=b.shape[0]
b_cols=b.shape[1]
row_diff = a_rows - b_rows
col_diff = a_cols - b_cols
for idx_row in np.arange(row_diff):
for idx_col in np.arange(col_diff):
row_indexes = [idx + idx_row for idx in np.arange(b_rows)]
col_indexes = [idx + idx_col for idx in np.arange(b_cols)]
submatrix_indexes = np.ix_(row_indexes, col_indexes)
a_submatrix = a[submatrix_indexes]
are_equal = np.allclose(a_submatrix, b) # allclose is used for floating point numbers, if they
# are close while comparing, they are considered equal.
# Useful if your matrices come from operations that produce
# floating point numbers.
# You might want to fine tune the parameters to allclose()
if (are_equal):
return[idx_col, idx_row]
return None
Using the function above you can run the following example:
large_mtx = np.array([[1, 2, 3, 7, 4, 2, 6],
[4, 5, 6, 2, 1, 3, 11],
[10, 4, 2, 1, 3, 7, 6],
[4, 2, 1, 3, 7, 6, -3],
[5, 6, 2, 1, 3, 11, -1],
[0, 0, -1, 5, 4, -1, 2],
[10, 4, 2, 1, 3, 7, 6],
[10, 4, 2, 1, 3, 7, 6]
])
# Example 1: An intersection at column 2 and row 1 of large_mtx
small_mtx_1 = np.array([[4, 2], [2,1]])
intersect = find_submatrix(large_mtx, small_mtx_1)
print "Example 1, intersection (col,row): " + str(intersect)
# Example 2: No intersection
small_mtx_2 = np.array([[-14, 2], [2,1]])
intersect = find_submatrix(large_mtx, small_mtx_2)
print "Example 2, intersection (col,row): " + str(intersect)
Which would print:
Example 1, intersection: [1, 2]
Example 2, intersection: None

Here's a solution using the as_strided() function from stride_tricks module
import numpy as np
from numpy.lib.stride_tricks import as_strided
# field_array (I modified it to have two matching arrays)
A = np.array([[ 24, 25, 26, 27, 28, 29, 30, 31, 23],
[ 33, 0, 1, 2, 3, 38, 39, 40, 32],
[-39, 4, 5, 6, 7, -34, -33, -32, -40],
[-30, 8, 9, 10, 11, -25, -24, -23, -31],
[-21, -20, -19, -18, -17, -16, -15, -14, -22],
[-12, -11, -10, -9, -8, -7, -6, -5, -13],
[ -3, -2, -1, 0, 1, 2, 3, 4, -4],
[ 6, 7, 8, 4, 5, 6, 7, 13, 5],
[ 15, 16, 17, 8, 9, 10, 11, 22, 14]])
# match_array
B = np.arange(12).reshape(3,4)
# Window view of A
A_w = as_strided(A, shape=(A.shape[0] - B.shape[0] + 1,
A.shape[1] - B.shape[1] + 1,
B.shape[0], B.shape[1]),
strides=2*A.strides).reshape(-1, B.shape[0], B.shape[1])
match = (A_w == B).all(axis=(1,2))
We can also find the indices of the first element of each matching block in A
where = np.where(match)[0]
ind_flat = where + (B.shape[1] - 1)*(np.floor(where/(A.shape[1] - B.shape[1] + 1)).astype(int))
ind = [tuple(row) for row in np.array(np.unravel_index(ind_flat, A.shape)).T]
Result
print(match.any())
True
print(ind)
[(1, 1), (6, 3)]

numpy find two dimensional array within another [duplicate]

I have a large NumPy.array field_array and a smaller array match_array, both consisting of int values. Using the following example, how can I check if any match_array-shaped segment of field_array contains values that exactly correspond to the ones in match_array?
import numpy
raw_field = ( 24, 25, 26, 27, 28, 29, 30, 31, 23, \
33, 34, 35, 36, 37, 38, 39, 40, 32, \
-39, -38, -37, -36, -35, -34, -33, -32, -40, \
-30, -29, -28, -27, -26, -25, -24, -23, -31, \
-21, -20, -19, -18, -17, -16, -15, -14, -22, \
-12, -11, -10, -9, -8, -7, -6, -5, -13, \
-3, -2, -1, 0, 1, 2, 3, 4, -4, \
6, 7, 8, 4, 5, 6, 7, 13, 5, \
15, 16, 17, 8, 9, 10, 11, 22, 14)
field_array = numpy.array(raw_field, int).reshape(9,9)
match_array = numpy.arange(12).reshape(3,4)
These examples ought to return True since the pattern described by match_array aligns over [6:9,3:7].

Approach #1
This approach derives from a solution to Implement Matlab's im2col 'sliding' in python that was designed to rearrange sliding blocks from a 2D array into columns. Thus, to solve our case here, those sliding blocks from field_array could be stacked as columns and compared against column vector version of match_array.
Here's a formal definition of the function for the rearrangement/stacking -
def im2col(A,BLKSZ):
# Parameters
M,N = A.shape
col_extent = N - BLKSZ[1] + 1
row_extent = M - BLKSZ[0] + 1
# Get Starting block indices
start_idx = np.arange(BLKSZ[0])[:,None]*N + np.arange(BLKSZ[1])
# Get offsetted indices across the height and width of input array
offset_idx = np.arange(row_extent)[:,None]*N + np.arange(col_extent)
# Get all actual indices & index into input array for final output
return np.take (A,start_idx.ravel()[:,None] + offset_idx.ravel())
To solve our case, here's the implementation based on im2col -
# Get sliding blocks of shape same as match_array from field_array into columns
# Then, compare them with a column vector version of match array.
col_match = im2col(field_array,match_array.shape) == match_array.ravel()[:,None]
# Shape of output array that has field_array compared against a sliding match_array
out_shape = np.asarray(field_array.shape) - np.asarray(match_array.shape) + 1
# Now, see if all elements in a column are ONES and reshape to out_shape.
# Finally, find the position of TRUE indices
R,C = np.where(col_match.all(0).reshape(out_shape))
The output for the given sample in the question would be -
In [151]: R,C
Out[151]: (array([6]), array([3]))
Approach #2
Given that opencv already has template matching function that does square of differences, you can employ that and look for zero differences, which would be your matching positions. So, if you have access to cv2 (opencv module), the implementation would look something like this -
import cv2
from cv2 import matchTemplate as cv2m
M = cv2m(field_array.astype('uint8'),match_array.astype('uint8'),cv2.TM_SQDIFF)
R,C = np.where(M==0)
giving us -
In [204]: R,C
Out[204]: (array([6]), array([3]))
Benchmarking
This section compares runtimes for all the approaches suggested to solve the question. The credit for the various methods listed in this section goes to their contributors.
Method definitions -
def seek_array(search_in, search_for, return_coords = False):
si_x, si_y = search_in.shape
sf_x, sf_y = search_for.shape
for y in xrange(si_y-sf_y+1):
for x in xrange(si_x-sf_x+1):
if numpy.array_equal(search_for, search_in[x:x+sf_x, y:y+sf_y]):
return (x,y) if return_coords else True
return None if return_coords else False
def skimage_based(field_array,match_array):
windows = view_as_windows(field_array, match_array.shape)
return (windows == match_array).all(axis=(2,3)).nonzero()
def im2col_based(field_array,match_array):
col_match = im2col(field_array,match_array.shape)==match_array.ravel()[:,None]
out_shape = np.asarray(field_array.shape) - np.asarray(match_array.shape) + 1
return np.where(col_match.all(0).reshape(out_shape))
def cv2_based(field_array,match_array):
M = cv2m(field_array.astype('uint8'),match_array.astype('uint8'),cv2.TM_SQDIFF)
return np.where(M==0)
Runtime tests -
Case # 1 (Sample data from question):
In [11]: field_array
Out[11]:
array([[ 24, 25, 26, 27, 28, 29, 30, 31, 23],
[ 33, 34, 35, 36, 37, 38, 39, 40, 32],
[-39, -38, -37, -36, -35, -34, -33, -32, -40],
[-30, -29, -28, -27, -26, -25, -24, -23, -31],
[-21, -20, -19, -18, -17, -16, -15, -14, -22],
[-12, -11, -10, -9, -8, -7, -6, -5, -13],
[ -3, -2, -1, 0, 1, 2, 3, 4, -4],
[ 6, 7, 8, 4, 5, 6, 7, 13, 5],
[ 15, 16, 17, 8, 9, 10, 11, 22, 14]])
In [12]: match_array
Out[12]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
In [13]: %timeit seek_array(field_array, match_array, return_coords = False)
1000 loops, best of 3: 465 µs per loop
In [14]: %timeit skimage_based(field_array,match_array)
10000 loops, best of 3: 97.9 µs per loop
In [15]: %timeit im2col_based(field_array,match_array)
10000 loops, best of 3: 74.3 µs per loop
In [16]: %timeit cv2_based(field_array,match_array)
10000 loops, best of 3: 30 µs per loop
Case #2 (Bigger random data):
In [17]: field_array = np.random.randint(0,4,(256,256))
In [18]: match_array = field_array[100:116,100:116].copy()
In [19]: %timeit seek_array(field_array, match_array, return_coords = False)
1 loops, best of 3: 400 ms per loop
In [20]: %timeit skimage_based(field_array,match_array)
10 loops, best of 3: 54.3 ms per loop
In [21]: %timeit im2col_based(field_array,match_array)
10 loops, best of 3: 125 ms per loop
In [22]: %timeit cv2_based(field_array,match_array)
100 loops, best of 3: 4.08 ms per loop

There's no such search function built in to NumPy, but it is certainly possible to do in NumPy
As long as your arrays are not too massive*, you could use a rolling window approach:
from skimage.util import view_as_windows
windows = view_as_windows(field_array, match_array.shape)
The function view_as_windows is written purely in NumPy so if you don't have skimage you can always copy the code from here.
Then to see if the sub-array appears in the larger array, you can write:
>>> (windows == match_array).all(axis=(2,3)).any()
True
To find the indices of where the top-left corner of the sub-array matches, you can write:
>>> (windows == match_array).all(axis=(2,3)).nonzero()
(array([6]), array([3]))
This approach should also work for arrays of higher dimensions.
*although the array windows takes up no additional memory (only the strides and shape are changed to create a new view of the data), writing windows == match_array creates a boolean array of size (7, 6, 3, 4) which is 504 bytes of memory. If you're working with very large arrays, this approach might not be feasible.

One solution is to search the entire search_in array block-at-a-time (a 'block' being a search_for-shaped slice) until either a matching segment is found or the search_for array is exhausted. I can use it to get coordinates for the matching block, or just a bool result by sending True or False for the return_coords optional argument...
def seek_array(search_in, search_for, return_coords = False):
"""Searches for a contiguous instance of a 2d array `search_for` within a larger `search_in` 2d array.
If the optional argument return_coords is True, the xy coordinates of the zeroeth value of the first matching segment of search_in will be returned, or None if there is no matching segment.
If return_coords is False, a boolean will be returned.
* Both arrays must be sent as two-dimensional!"""
si_x, si_y = search_in.shape
sf_x, sf_y = search_for.shape
for y in xrange(si_y-sf_y+1):
for x in xrange(si_x-sf_x+1):
if numpy.array_equal(search_for, search_in[x:x+sf_x, y:y+sf_y]):
return (x,y) if return_coords else True # don't forget that coordinates are transposed when viewing NumPy arrays!
return None if return_coords else False
I wonder if NumPy doesn't already have a function that can do the same thing, though...

To add to the answers already posted, I'd like to add one that takes into account errors due to floating point precision in case that matrices come from, let's say, image processing for instance, where numbers are subject to floating point operations.
You can recurse the indexes of the larger matrix, searching for the smaller matrix. Then you can extract a submatrix of the larger matrix matching the size of the smaller matrix.
You have a match if the contents of both, the submatrix of 'large' and the 'small' matrix match.
The following example shows how to return the first indexes of the location in the large matrix found to match. It would be trivial to extend this function to return an array of locations found to match if that's the intent.
import numpy as np
def find_submatrix(a, b):
""" Searches the first instance at which 'b' is a submatrix of 'a', iterates
rows first. Returns the indexes of a at which 'b' was found, or None if
'b' is not contained within 'a'"""
a_rows=a.shape[0]
a_cols=a.shape[1]
b_rows=b.shape[0]
b_cols=b.shape[1]
row_diff = a_rows - b_rows
col_diff = a_cols - b_cols
for idx_row in np.arange(row_diff):
for idx_col in np.arange(col_diff):
row_indexes = [idx + idx_row for idx in np.arange(b_rows)]
col_indexes = [idx + idx_col for idx in np.arange(b_cols)]
submatrix_indexes = np.ix_(row_indexes, col_indexes)
a_submatrix = a[submatrix_indexes]
are_equal = np.allclose(a_submatrix, b) # allclose is used for floating point numbers, if they
# are close while comparing, they are considered equal.
# Useful if your matrices come from operations that produce
# floating point numbers.
# You might want to fine tune the parameters to allclose()
if (are_equal):
return[idx_col, idx_row]
return None
Using the function above you can run the following example:
large_mtx = np.array([[1, 2, 3, 7, 4, 2, 6],
[4, 5, 6, 2, 1, 3, 11],
[10, 4, 2, 1, 3, 7, 6],
[4, 2, 1, 3, 7, 6, -3],
[5, 6, 2, 1, 3, 11, -1],
[0, 0, -1, 5, 4, -1, 2],
[10, 4, 2, 1, 3, 7, 6],
[10, 4, 2, 1, 3, 7, 6]
])
# Example 1: An intersection at column 2 and row 1 of large_mtx
small_mtx_1 = np.array([[4, 2], [2,1]])
intersect = find_submatrix(large_mtx, small_mtx_1)
print "Example 1, intersection (col,row): " + str(intersect)
# Example 2: No intersection
small_mtx_2 = np.array([[-14, 2], [2,1]])
intersect = find_submatrix(large_mtx, small_mtx_2)
print "Example 2, intersection (col,row): " + str(intersect)
Which would print:
Example 1, intersection: [1, 2]
Example 2, intersection: None

Here's a solution using the as_strided() function from stride_tricks module
import numpy as np
from numpy.lib.stride_tricks import as_strided
# field_array (I modified it to have two matching arrays)
A = np.array([[ 24, 25, 26, 27, 28, 29, 30, 31, 23],
[ 33, 0, 1, 2, 3, 38, 39, 40, 32],
[-39, 4, 5, 6, 7, -34, -33, -32, -40],
[-30, 8, 9, 10, 11, -25, -24, -23, -31],
[-21, -20, -19, -18, -17, -16, -15, -14, -22],
[-12, -11, -10, -9, -8, -7, -6, -5, -13],
[ -3, -2, -1, 0, 1, 2, 3, 4, -4],
[ 6, 7, 8, 4, 5, 6, 7, 13, 5],
[ 15, 16, 17, 8, 9, 10, 11, 22, 14]])
# match_array
B = np.arange(12).reshape(3,4)
# Window view of A
A_w = as_strided(A, shape=(A.shape[0] - B.shape[0] + 1,
A.shape[1] - B.shape[1] + 1,
B.shape[0], B.shape[1]),
strides=2*A.strides).reshape(-1, B.shape[0], B.shape[1])
match = (A_w == B).all(axis=(1,2))
We can also find the indices of the first element of each matching block in A
where = np.where(match)[0]
ind_flat = where + (B.shape[1] - 1)*(np.floor(where/(A.shape[1] - B.shape[1] + 1)).astype(int))
ind = [tuple(row) for row in np.array(np.unravel_index(ind_flat, A.shape)).T]
Result
print(match.any())
True
print(ind)
[(1, 1), (6, 3)]

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