在Python中获取矩阵/列表中的所有对角线
我正在寻找一种Pythonic方法来获得（正方形）矩阵的所有对角线，表示为列表列表 假设我有以下矩阵：在Python中获取矩阵/列表中的所有对角线,python,matrix,diagonal,Python,Matrix,Diagonal,我正在寻找一种Pythonic方法来获得（正方形）矩阵的所有对角线，表示为列表列表 假设我有以下矩阵： matrix = [[-2, 5, 3, 2], [ 9, -6, 5, 1], [ 3, 2, 7, 3], [-1, 8, -4, 8]] 那么大对角线就很容易了： l = len(matrix[0]) print [matrix[i][i] for i in range(l)] #
matrix = [[-2, 5, 3, 2],
[ 9, -6, 5, 1],
[ 3, 2, 7, 3],
[-1, 8, -4, 8]]
l = len(matrix[0])
print [matrix[i][i] for i in range(l)] # [-2, -6, 7, 8]
print [matrix[l-1-i][i] for i in range(l-1,-1,-1)] # [ 2, 5, 2, -1]
[[-2], [9, 5], [3,-6, 3], [-1, 2, 5, 2], [8, 7, 1], [-4, 3], [8],
[2], [3,1], [5, 5, 3], [-2, -6, 7, 8], [9, 2, -4], [3, 8], [-1]]
从向上向右倾斜的对角线开始 如果（x，y）是矩阵内的直角坐标，则需要在坐标方案（p，q）之间进行变换，其中p是对角线的编号，q是沿对角线的索引。（因此p=0是[-2]对角线，p=1是[9,5]对角线，p=2是[3，-6,3]对角线，依此类推。） 要将a（p，q）转换为a（x，y），可以使用：
x = q
y = p - q
x = N - 1 - q
y = p - q
很抱歉，我没有用Python编写此代码。：-） 从向上向右倾斜的对角线开始 如果（x，y）是矩阵内的直角坐标，则需要在坐标方案（p，q）之间进行变换，其中p是对角线的编号，q是沿对角线的索引。（因此p=0是[-2]对角线，p=1是[9,5]对角线，p=2是[3，-6,3]对角线，依此类推。） 要将a（p，q）转换为a（x，y），可以使用：
x = q
y = p - q
x = N - 1 - q
y = p - q
很抱歉，我没有用Python编写此代码。：-） 也许有比下面更好的方法，但我还不太熟悉：
import numpy as np
matrix = np.array(
[[-2, 5, 3, 2],
[ 9, -6, 5, 1],
[ 3, 2, 7, 3],
[-1, 8, -4, 8]])
diags = [matrix[::-1,:].diagonal(i) for i in range(-3,4)]
diags.extend(matrix.diagonal(i) for i in range(3,-4,-1))
print [n.tolist() for n in diags]
编辑：更新以概括任何矩阵大小
import numpy as np
# Alter dimensions as needed
x,y = 3,4
# create a default array of specified dimensions
a = np.arange(x*y).reshape(x,y)
print a
print
# a.diagonal returns the top-left-to-lower-right diagonal "i"
# according to this diagram:
#
# 0 1 2 3 4 ...
# -1 0 1 2 3
# -2 -1 0 1 2
# -3 -2 -1 0 1
# :
#
# You wanted lower-left-to-upper-right and upper-left-to-lower-right diagonals.
#
# The syntax a[slice,slice] returns a new array with elements from the sliced ranges,
# where "slice" is Python's [start[:stop[:step]] format.
# "::-1" returns the rows in reverse. ":" returns the columns as is,
# effectively vertically mirroring the original array so the wanted diagonals are
# lower-right-to-uppper-left.
#
# Then a list comprehension is used to collect all the diagonals. The range
# is -x+1 to y (exclusive of y), so for a matrix like the example above
# (x,y) = (4,5) = -3 to 4.
diags = [a[::-1,:].diagonal(i) for i in range(-a.shape[0]+1,a.shape[1])]
# Now back to the original array to get the upper-left-to-lower-right diagonals,
# starting from the right, so the range needed for shape (x,y) was y-1 to -x+1 descending.
diags.extend(a.diagonal(i) for i in range(a.shape[1]-1,-a.shape[0],-1))
# Another list comp to convert back to Python lists from numpy arrays,
# so it prints what you requested.
print [n.tolist() for n in diags]
也许有比下面更好的方法，但我还不太熟悉：
import numpy as np
matrix = np.array(
[[-2, 5, 3, 2],
[ 9, -6, 5, 1],
[ 3, 2, 7, 3],
[-1, 8, -4, 8]])
diags = [matrix[::-1,:].diagonal(i) for i in range(-3,4)]
diags.extend(matrix.diagonal(i) for i in range(3,-4,-1))
print [n.tolist() for n in diags]
编辑：更新以概括任何矩阵大小
import numpy as np
# Alter dimensions as needed
x,y = 3,4
# create a default array of specified dimensions
a = np.arange(x*y).reshape(x,y)
print a
print
# a.diagonal returns the top-left-to-lower-right diagonal "i"
# according to this diagram:
#
# 0 1 2 3 4 ...
# -1 0 1 2 3
# -2 -1 0 1 2
# -3 -2 -1 0 1
# :
#
# You wanted lower-left-to-upper-right and upper-left-to-lower-right diagonals.
#
# The syntax a[slice,slice] returns a new array with elements from the sliced ranges,
# where "slice" is Python's [start[:stop[:step]] format.
# "::-1" returns the rows in reverse. ":" returns the columns as is,
# effectively vertically mirroring the original array so the wanted diagonals are
# lower-right-to-uppper-left.
#
# Then a list comprehension is used to collect all the diagonals. The range
# is -x+1 to y (exclusive of y), so for a matrix like the example above
# (x,y) = (4,5) = -3 to 4.
diags = [a[::-1,:].diagonal(i) for i in range(-a.shape[0]+1,a.shape[1])]
# Now back to the original array to get the upper-left-to-lower-right diagonals,
# starting from the right, so the range needed for shape (x,y) was y-1 to -x+1 descending.
diags.extend(a.diagonal(i) for i in range(a.shape[1]-1,-a.shape[0],-1))
# Another list comp to convert back to Python lists from numpy arrays,
# so it prints what you requested.
print [n.tolist() for n in diags]
def get_rows(grid):
return [[c for c in r] for r in grid]
def get_cols(grid):
return zip(*grid)
def get_rows(grid):
return [[c for c in r] for r in grid]
def get_cols(grid):
return zip(*grid)
这仅适用于宽度和高度相等的矩阵。 但它也不依赖任何第三方
matrix = [[11, 2, 4],[4, 5, 6],[10, 8, -12]]
# only works for diagnoals of equal width and height
def forward_diagonal(matrix):
if not isinstance(matrix, list):
raise TypeError("Must be of type list")
results = []
x = 0
for k, row in enumerate(matrix):
# next diag is (x + 1, y + 1)
for i, elm in enumerate(row):
if i == 0 and k == 0:
results.append(elm)
break
if (x + 1 == i):
results.append(elm)
x = i
break
return results
print 'forward diagnoals', forward_diagonal(matrix)
这仅适用于宽度和高度相等的矩阵。 但它也不依赖任何第三方
matrix = [[11, 2, 4],[4, 5, 6],[10, 8, -12]]
# only works for diagnoals of equal width and height
def forward_diagonal(matrix):
if not isinstance(matrix, list):
raise TypeError("Must be of type list")
results = []
x = 0
for k, row in enumerate(matrix):
# next diag is (x + 1, y + 1)
for i, elm in enumerate(row):
if i == 0 and k == 0:
results.append(elm)
break
if (x + 1 == i):
results.append(elm)
x = i
break
return results
print 'forward diagnoals', forward_diagonal(matrix)
对于这个问题，我遇到了另一个有趣的解决方案。 通过查看x和y的组合，可以立即发现行、列、前向和后向对角线
Column = x Row = y F-Diag = x+y B-Diag = x-y B-Diag` = x-y-MIN
| 0 1 2 | 0 1 2 | 0 1 2 | 0 1 2 | 0 1 2
--|--------- --|--------- --|--------- --|--------- --|---------
0 | 0 1 2 0 | 0 0 0 0 | 0 1 2 0 | 0 1 2 0 | 2 3 4
1 | 0 1 2 1 | 1 1 1 1 | 1 2 3 1 |-1 0 1 1 | 1 2 3
2 | 0 1 2 2 | 2 2 2 2 | 2 3 4 2 |-2 -1 0 2 | 0 1 2
test = [[1,2,3],[4,5,6],[7,8,9],[10,11,12]]
max_col = len(test[0])
max_row = len(test)
cols = [[] for _ in range(max_col)]
rows = [[] for _ in range(max_row)]
fdiag = [[] for _ in range(max_row + max_col - 1)]
bdiag = [[] for _ in range(len(fdiag))]
min_bdiag = -max_row + 1
for x in range(max_col):
for y in range(max_row):
cols[x].append(test[y][x])
rows[y].append(test[y][x])
fdiag[x+y].append(test[y][x])
bdiag[x-y-min_bdiag].append(test[y][x])
print(cols)
print(rows)
print(fdiag)
print(bdiag)
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]]
[[1, 4, 7, 10], [2, 5, 8, 11], [3, 6, 9, 12]]
[[1], [2, 4], [3, 5, 7], [6, 8, 10], [9, 11], [12]]
[[10], [7, 11], [4, 8, 12], [1, 5, 9], [2, 6], [3]]
对于这个问题，我遇到了另一个有趣的解决方案。 通过查看x和y的组合，可以立即发现行、列、前向和后向对角线
Column = x Row = y F-Diag = x+y B-Diag = x-y B-Diag` = x-y-MIN
| 0 1 2 | 0 1 2 | 0 1 2 | 0 1 2 | 0 1 2
--|--------- --|--------- --|--------- --|--------- --|---------
0 | 0 1 2 0 | 0 0 0 0 | 0 1 2 0 | 0 1 2 0 | 2 3 4
1 | 0 1 2 1 | 1 1 1 1 | 1 2 3 1 |-1 0 1 1 | 1 2 3
2 | 0 1 2 2 | 2 2 2 2 | 2 3 4 2 |-2 -1 0 2 | 0 1 2
test = [[1,2,3],[4,5,6],[7,8,9],[10,11,12]]
max_col = len(test[0])
max_row = len(test)
cols = [[] for _ in range(max_col)]
rows = [[] for _ in range(max_row)]
fdiag = [[] for _ in range(max_row + max_col - 1)]
bdiag = [[] for _ in range(len(fdiag))]
min_bdiag = -max_row + 1
for x in range(max_col):
for y in range(max_row):
cols[x].append(test[y][x])
rows[y].append(test[y][x])
fdiag[x+y].append(test[y][x])
bdiag[x-y-min_bdiag].append(test[y][x])
print(cols)
print(rows)
print(fdiag)
print(bdiag)
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]]
[[1, 4, 7, 10], [2, 5, 8, 11], [3, 6, 9, 12]]
[[1], [2, 4], [3, 5, 7], [6, 8, 10], [9, 11], [12]]
[[10], [7, 11], [4, 8, 12], [1, 5, 9], [2, 6], [3]]
我最近重新发明了这个轮子。下面是一种易于重用/扩展的方法，用于在正方形列表中查找对角线：
def get_diagonals(grid, bltr = True):
dim = len(grid)
assert dim == len(grid[0])
return_grid = [[] for total in xrange(2 * len(grid) - 1)]
for row in xrange(len(grid)):
for col in xrange(len(grid[row])):
if bltr: return_grid[row + col].append(grid[col][row])
else: return_grid[col - row + (dim - 1)].append(grid[row][col])
return return_grid
bltr=True00 # row + col == 0
10 01 # row + col == 1
20 11 02 # row + col == 2
30 21 12 03 # row + col == 3
31 22 13 # row + col == 4
32 23 # row + col == 5
33 # row + col == 6
30 # (col - row) == -3
20 31 # (col - row) == -2
10 21 32 # (col - row) == -1
00 11 22 33 # (col - row) == 0
01 12 23 # (col - row) == +1
02 13 # (col - row) == +2
03 # (col - row) == +3
00 # row + col == 0
10 01 # row + col == 1
20 11 02 # row + col == 2
30 21 12 03 # row + col == 3
31 22 13 # row + col == 4
32 23 # row + col == 5
33 # row + col == 6
30 # (col - row) == -3
20 31 # (col - row) == -2
10 21 32 # (col - row) == -1
00 11 22 33 # (col - row) == 0
01 12 23 # (col - row) == +1
02 13 # (col - row) == +2
03 # (col - row) == +3
使用OP的输入矩阵。我最近重新发明了这个轮子。下面是一种易于重用/扩展的方法，用于在正方形列表中查找对角线：
def get_diagonals(grid, bltr = True):
dim = len(grid)
assert dim == len(grid[0])
return_grid = [[] for total in xrange(2 * len(grid) - 1)]
for row in xrange(len(grid)):
for col in xrange(len(grid[row])):
if bltr: return_grid[row + col].append(grid[col][row])
else: return_grid[col - row + (dim - 1)].append(grid[row][col])
return return_grid
bltr=True00 # row + col == 0
10 01 # row + col == 1
20 11 02 # row + col == 2
30 21 12 03 # row + col == 3
31 22 13 # row + col == 4
32 23 # row + col == 5
33 # row + col == 6
30 # (col - row) == -3
20 31 # (col - row) == -2
10 21 32 # (col - row) == -1
00 11 22 33 # (col - row) == 0
01 12 23 # (col - row) == +1
02 13 # (col - row) == +2
03 # (col - row) == +3
00 # row + col == 0
10 01 # row + col == 1
20 11 02 # row + col == 2
30 21 12 03 # row + col == 3
31 22 13 # row + col == 4
32 23 # row + col == 5
33 # row + col == 6
30 # (col - row) == -3
20 31 # (col - row) == -2
10 21 32 # (col - row) == -1
00 11 22 33 # (col - row) == 0
01 12 23 # (col - row) == +1
02 13 # (col - row) == +2
03 # (col - row) == +3
使用OP的输入矩阵。基于Nemo上述答案的代码：
def print_diagonals(matrix):
n = len(matrix)
diagonals_1 = [] # lower-left-to-upper-right diagonals
diagonals_2 = [] # upper-left-to-lower-right diagonals
for p in range(2*n-1):
diagonals_1.append([matrix[p-q][q] for q in range(max(0, p - n + 1), min(p, n - 1) + 1)])
diagonals_2.append([matrix[n-p+q-1][q] for q in range(max(0, p - n + 1), min(p, n - 1) + 1)])
print("lower-left-to-upper-right diagonals: ", diagonals_1)
print("upper-left-to-lower-right diagonals: ", diagonals_2)
print_diagonals([
[1, 2, 1, 1],
[1, 1, 4, 1],
[1, 3, 1, 6],
[1, 7, 2, 5],
])
lower-left-to-upper-right diagonals: [[1], [1, 2], [1, 1, 1], [1, 3, 4, 1], [7, 1, 1], [2, 6], [5]]
upper-left-to-lower-right diagonals: [[1], [1, 7], [1, 3, 2], [1, 1, 1, 5], [2, 4, 6], [1, 1], [1]]
代码基于Nemo的上述回答：
def print_diagonals(matrix):
n = len(matrix)
diagonals_1 = [] # lower-left-to-upper-right diagonals
diagonals_2 = [] # upper-left-to-lower-right diagonals
for p in range(2*n-1):
diagonals_1.append([matrix[p-q][q] for q in range(max(0, p - n + 1), min(p, n - 1) + 1)])
diagonals_2.append([matrix[n-p+q-1][q] for q in range(max(0, p - n + 1), min(p, n - 1) + 1)])
print("lower-left-to-upper-right diagonals: ", diagonals_1)
print("upper-left-to-lower-right diagonals: ", diagonals_2)
print_diagonals([
[1, 2, 1, 1],
[1, 1, 4, 1],
[1, 3, 1, 6],
[1, 7, 2, 5],
])
lower-left-to-upper-right diagonals: [[1], [1, 2], [1, 1, 1], [1, 3, 4, 1], [7, 1, 1], [2, 6], [5]]
upper-left-to-lower-right diagonals: [[1], [1, 7], [1, 3, 2], [1, 1, 1, 5], [2, 4, 6], [1, 1], [1]]
肾盂法
tril = [np.flip(np.fliplr(mat).diagonal(n)) for n in xrange(mat.shape[0])][::-1]
trir = [np.flipud(mat).diagonal(n) for n in xrange(1, mat.shape[0])]
dia = tril + trir
[array([-2]),
array([9, 5]),
array([ 3, -6, 3]),
array([-1, 2, 5, 2]),
array([8, 7, 1]),
array([-4, 3]),
array([8])]
W, H = len(mat[0]), len(mat)
idx = range(W-1) + range(W-1, W*H, W)
rng = range(1, W) + range(H, 0, -1)
rng = map(lambda x: x if (x < min(W, H)) else min(W, H), rng)
dia = [[i + (W-1) * m for m in xrange(r)] for i, r in zip(idx, rng)]
arr2 = [e for row in zip(*mat[::-1]) for e in row] #Flatten and rotate the matrix by 90°
for d in dia[::-1]:
print [arr2[e] for e in d]
[2]
[3, 1]
[5, 5, 3]
[-2, -6, 7, 8]
[9, 2, -4]
[3, 8]
[-1]
tril = [np.flip(np.fliplr(mat).diagonal(n)) for n in xrange(mat.shape[0])][::-1]
trir = [np.flipud(mat).diagonal(n) for n in xrange(1, mat.shape[0])]
dia = tril + trir
[array([-2]),
array([9, 5]),
array([ 3, -6, 3]),
array([-1, 2, 5, 2]),
array([8, 7, 1]),
array([-4, 3]),
array([8])]
肾盂法
tril = [np.flip(np.fliplr(mat).diagonal(n)) for n in xrange(mat.shape[0])][::-1]
trir = [np.flipud(mat).diagonal(n) for n in xrange(1, mat.shape[0])]
dia = tril + trir
[array([-2]),
array([9, 5]),
array([ 3, -6, 3]),
array([-1, 2, 5, 2]),
array([8, 7, 1]),
array([-4, 3]),
array([8])]
W, H = len(mat[0]), len(mat)
idx = range(W-1) + range(W-1, W*H, W)
rng = range(1, W) + range(H, 0, -1)
rng = map(lambda x: x if (x < min(W, H)) else min(W, H), rng)
dia = [[i + (W-1) * m for m in xrange(r)] for i, r in zip(idx, rng)]