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我认为这些要求可以在numpy中没有显式循环的情况下得到满足。
import numpy as np
np.random.seed(1234) # Make random array reproduceable
arr = np.random.randint( 0, 2, size = (10,10))
leftshifted = arr[ :, 1:] # Shift arr 1 col left, shape = (10, 9)
downshifted = arr[ 1: ] # Shift arr 1 row down, shape = ( 9, 10)
hrows, hcols = np.where( arr[ :, :-1 ] & leftshifted )
# arr[ :,:-1 ] => ignore last column for the comparison
# returns rows and columns where arr and leftshifted = 1
# i.e. where two adjacent columns in a row are 1
vrows, vcols = np.where( arr[ :-1 ] & downshifted )
# arr[ :-1 ] => ignore last row for the comparison
# returns rows and columns where arr and downshifted = 1
# i.e. where two adjacent rows in a column are 1
print(arr, '\n')
# [[1 1 0 1 0 0 0 1 1 1]
# [1 1 0 0 1 0 0 0 0 0]
# [0 0 0 0 1 0 1 1 0 0]
# [1 0 0 1 0 1 0 0 0 1]
# [1 1 0 1 1 0 1 0 1 0]
# [1 1 1 1 0 1 0 1 1 0]
# [0 1 0 0 1 1 1 0 0 0]
# [1 1 1 1 1 1 1 0 1 0]
# [1 0 1 0 0 0 0 0 0 0]
# [0 1 1 1 0 1 0 0 1 1]]
print('Row indices :', hrows)
print('Col start ix :', hcols)
print('Col end ix :', hcols+1)
# Row indices : [0 0 0 1 2 4 4 5 5 5 5 6 6 7 7 7 7 7 7 9 9 9]
# Col start ix : [0 7 8 0 6 0 3 0 1 2 7 4 5 0 1 2 3 4 5 1 2 8]
# Col end ix : [1 8 9 1 7 1 4 1 2 3 8 5 6 1 2 3 4 5 6 2 3 9]
print('\nStart Row:', vrows, '\nEnd Row :',vrows+1, '\nColumn :', vcols)
# Start Row: [0 0 1 3 3 4 4 4 4 5 5 6 6 6 6 7 7 8]
# End Row : [1 1 2 4 4 5 5 5 5 6 6 7 7 7 7 8 8 9]
# Column : [0 1 4 0 3 0 1 3 8 1 5 1 4 5 6 0 2 2]
一个元素可以有多个对吗?在上面,它可以是。两个对角线接触的元素算作一对吗?如果是这样,还需要一个shifted_left_and_down数组。
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