Get the position of elements by setting some criteria in numpy

Question

For a given 2d data array, how to get the position (index) of 7 and 11 in bold. Because only they are elements surrounded by the same value in neighboring

  import numpy as np
    data  = np.array([
        [0,1,2,3,4,7,6,7,8,9,10], 
        [3,3,3,4,7,7,7,8,11,12,11],  
        [3,3,3,5,7,**7**,7,9,11,11,11],
        [3,4,3,6,7,7,7,10,11,**11**,11],
        [4,5,6,7,7,9,10,11,11,11,11]
        ])

    print data

Answer 1

Using scipy, you can characterize points such as those that are both the maximum and minimum of its surroundings:

import numpy as np
import scipy.ndimage.filters as filters

def using_filters(data):
    return np.where(np.logical_and.reduce(
        [data == f(data, footprint=np.ones((3,3)), mode='constant', cval=np.inf)
         for f in (filters.maximum_filter, filters.minimum_filter)]))  

using_filters(data)
# (array([2, 3]), array([5, 9]))

Using only numpy, you can compare datawith 8 shifted fragments of yourself to find points that are equal:

def using_eight_shifts(data):
    h, w = data.shape
    data2 = np.empty((h+2, w+2))
    data2[(0,-1),:] = np.nan
    data2[:,(0,-1)] = np.nan
    data2[1:1+h,1:1+w] = data

    result = np.where(np.logical_and.reduce([
        (data2[i:i+h,j:j+w] == data)
        for i in range(3)
        for j in range(3)
        if not (i==1 and j==1)]))
    return result

, , NaN . data2[i:i+h,j:j+w].

, , data NaN , , .

( ) , data :

def using_quadratic_loop(data):
    return np.array([[i,j]
            for i in range(1,np.shape(data)[0]-1)
            for j in range(1,np.shape(data)[1]-1)
            if np.all(data[i-1:i+2,j-1:j+2]==data[i,j])]).T

:

using_filters            : 0.130
using_eight_shifts       : 0.340
using_quadratic_loop     : 18.794

---

, :

import timeit
import operator
import numpy as np
import scipy.ndimage.filters as filters
import matplotlib.pyplot as plt

data  = np.array([
    [0,1,2,3,4,7,6,7,8,9,10], 
    [3,3,3,4,7,7,7,8,11,12,11],  
    [3,3,3,5,7,7,7,9,11,11,11],
    [3,4,3,6,7,7,7,10,11,11,11],
    [4,5,6,7,7,9,10,11,11,11,11]
    ])

data = np.tile(data, (50,50))

def using_filters(data):
    return np.where(np.logical_and.reduce(
        [data == f(data, footprint=np.ones((3,3)), mode='constant', cval=np.inf)
         for f in (filters.maximum_filter, filters.minimum_filter)]))    


def using_eight_shifts(data):
    h, w = data.shape
    data2 = np.empty((h+2, w+2))
    data2[(0,-1),:] = np.nan
    data2[:,(0,-1)] = np.nan
    data2[1:1+h,1:1+w] = data

    result = np.where(np.logical_and.reduce([
        (data2[i:i+h,j:j+w] == data)
        for i in range(3)
        for j in range(3)
        if not (i==1 and j==1)]))
    return result


def using_quadratic_loop(data):
    return np.array([[i,j]
            for i in range(1,np.shape(data)[0]-1)
            for j in range(1,np.shape(data)[1]-1)
            if np.all(data[i-1:i+2,j-1:j+2]==data[i,j])]).T

np.testing.assert_equal(using_quadratic_loop(data), using_filters(data))
np.testing.assert_equal(using_eight_shifts(data), using_filters(data))

timing = dict()
for f in ('using_filters', 'using_eight_shifts', 'using_quadratic_loop'):
    timing[f] = timeit.timeit('{f}(data)'.format(f=f),
                              'from __main__ import data, {f}'.format(f=f),
                              number=10) 

for f, t in sorted(timing.items(), key=operator.itemgetter(1)):
    print('{f:25}: {t:.3f}'.format(f=f, t=t))

Answer 2

,

A = [(i,j) for i in range(1,data.shape[0]-1) for j in range(1,data.shape[1]-1) if all((data[i-1:i+2,j-1:j+2]==data[i,j]).flatten())]

EDIT:

([i, j], dtype = int64), :

A= [np.array([i,j], dtype=np.int64) for i in range(1,data.shape[0]-1) for j in range(1,data.shape[1]-1) if all((data[i-1:i+2,j-1:j+2]==data[i,j]).flatten())]

Answer 3

, :

for i in range(1,np.shape(data)[0]-1):
    for j in range(1,np.shape(data)[1]-1):
        if np.all(data[i-1:i+2,j-1:j+2]==data[i,j]):
            print np.array([i,j], dtype=np.int64)

Answer 4

displacements = [[-1, -1], [-1, 0], [-1, 1], [0, -1], [0, 1], [1, -1], [1, 0], [1, 1]]

for x in range(1, data.shape[0] - 1):
    for y in range(1, data.shape[1] - 1):
        if all((data[x][y] == data[x + a][y + b]) for a, b in displacements):
            print np.array([x, y], dtype=np.int64)

Not as succint as the other answers, but it clears and prints the correct output. I think it is a little easier to change / add offset values.

Oops, I didn’t understand that you wanted all 8 neighbors. Easy to fix though. :)