Python x,y对的转置平坦矩阵
我得到了一个代表x和y值的十进制数的一维数组 我需要转换给定的一维数组,就像它是一个矩阵一样 我当前的代码可以做到这一点,但并不完全符合我的要求:Python x,y对的转置平坦矩阵,python,matrix,transpose,Python,Matrix,Transpose,我得到了一个代表x和y值的十进制数的一维数组 我需要转换给定的一维数组,就像它是一个矩阵一样 我当前的代码可以做到这一点,但并不完全符合我的要求: to_transpose = [0.914, 0.639, 0.058, 0.760, 0.926, 0.475, 0.255, 0.671, 0.195, 0.966, 0.336, 0.841, 0.279, 0.341, 0.591, 0.638, 0.520, 0.225] ma
to_transpose = [0.914, 0.639, 0.058, 0.760, 0.926, 0.475,
0.255, 0.671, 0.195, 0.966, 0.336, 0.841,
0.279, 0.341, 0.591, 0.638, 0.520, 0.225]
matrix_width = 6
matrix_height = 3
# INITIALIZE AN EMPTY LIST
transposed_list = [None] * matrix_width * matrix_height
for w in range(matrix_width):
for h in range(matrix_height):
transposed_list[w * matrix_height + h] = to_transpose[h * matrix_width + w]
这段代码正确地转换了矩阵,但不是我想要的格式
由于这是一个x,y值的数组,因此所需的输出如下所示:
correct_output=[0.914, 0.639, 0.255, 0.671, 0.279, 0.341,
0.058, 0.760, 0.195, 0.966, 0.591, 0.638,
0.926, 0.475, 0.336, 0.841, 0.520, 0.225]
在正确的输出中,转置中每2位小数被视为1
numpy
处理与矩阵操作相关的所有内容,例如重塑或转置,这两个方面您都需要:
>>> import numpy as np
>>> to_transpose = [0.914, 0.639, 0.058, 0.760, 0.926, 0.475,
0.255, 0.671, 0.195, 0.966, 0.336, 0.841,
0.279, 0.341, 0.591, 0.638, 0.520, 0.225]
>>> np.array(to_transpose).reshape((3,3,2)).transpose(1,0,2).ravel()
array([ 0.914, 0.639, 0.255, 0.671, 0.279, 0.341,
0.058, 0.76 , 0.195, 0.966, 0.591, 0.638,
0.926, 0.475, 0.336, 0.841, 0.52 , 0.225])
把它分解一下:
将列表转换为np.array
数组
- 然后将其
重塑为
,即元组的3x3矩阵3x3x2
- 然后对其进行转置,交换第一(0)和第二(1)个轴,并保持第三(2)个轴的位置
- 最后用
ravel
重新生成该元素,以_transpose
:
>>> list(np.array(list(range(w*h))).reshape((3,3,2)).transpose(1,0,2).ravel())
[0, 1, 6, 7, 12, 13, 2, 3, 8, 9, 14, 15, 4, 5, 10, 11, 16, 17]
>>> [i%2 + (i//2 * w % (w*h)) + 2 * (i//(h*2)) for i in range(w*h)]
[0, 1, 6, 7, 12, 13, 2, 3, 8, 9, 14, 15, 4, 5, 10, 11, 16, 17]
>>> [to_transpose[i%2 + (i//2 * w % (w*h)) + 2 * (i//(h*2))] for i in range(w*h)]
[0.914, 0.639, 0.255, 0.671, 0.279, 0.341,
0.058, 0.76, 0.195, 0.966, 0.591, 0.638,
0.926, 0.475, 0.336, 0.841, 0.52, 0.225]
当然,您也可以使用常规循环(而不是列表理解)和其他语言执行相同的操作。基本上,添加索引的三项中的每一项都对应于矩阵的一个维度,老实说,我更多的是通过猜测而不是真正理解发生了什么。不用说,
numpy
-解决方案更干净。如果您坚持不使用numpy
,我建议您先分离x、y值,或者将它们组合在一起:
to_transpose = [0.914, 0.639, 0.058, 0.760, 0.926, 0.475,
0.255, 0.671, 0.195, 0.966, 0.336, 0.841,
0.279, 0.341, 0.591, 0.638, 0.520, 0.225]
a = to_transpose
rows = 3
cols = 3
tot = rows*cols
# separated
x, y = a[::2], a[1::2]
xt = [x[i+j] for i in range(0, rows) for j in range(0, tot, cols)]
yt = [y[i+j] for i in range(0, rows) for j in range(0, tot, cols)]
transposed = [e for t in zip(xt, yt) for e in t]
# tupled
xy = [(i,j) for i,j in zip(a[::2], a[1::2])]
xyt = [xy[i+j] for i in range(0, rows) for j in range(0, tot, cols)]
transposed = [e for t in xyt for e in t]
很明显,即使这样做有效,
numpy
对于此类操作来说将是一个更好的工具。我发现它可以按照您希望的方式工作,只需通过矩阵一次,并且不使用任何库
#!/bin/python3
to_transpose = [
0.914, 0.639, 0.058, 0.760, 0.926, 0.475,
0.255, 0.671, 0.195, 0.966, 0.336, 0.841,
0.279, 0.341, 0.591, 0.638, 0.520, 0.225
]
matrix_width = 6
matrix_height = 3
to_transpose_with_pairs = [(to_transpose[2 * i], to_transpose[2 * i + 1]) for i in range(len(to_transpose) // 2)]
# [
# (0.914, 0.639), (0.058, 0.76), (0.926, 0.475),
# (0.255, 0.671), (0.195, 0.966), (0.336, 0.841),
# (0.279, 0.341), (0.591, 0.638), (0.52, 0.225)
# ]
to_transpose_as_matrix = [None for _ in range(matrix_height)]
for row in range(matrix_height):
start = row * matrix_width // 2
end = start + matrix_width // 2
to_transpose_as_matrix[row] = to_transpose_with_pairs[start:end]
# [
# [(0.914, 0.639), (0.058, 0.76), (0.926, 0.475)],
# [(0.255, 0.671), (0.195, 0.966), (0.336, 0.841)],
# [(0.279, 0.341), (0.591, 0.638), (0.52, 0.225)]
# ]
transposed_as_matrix = list(map(list, zip(*to_transpose_as_matrix)))
# [
# [(0.914, 0.639), (0.255, 0.671), (0.279, 0.341)],
# [(0.058, 0.76), (0.195, 0.966), (0.591, 0.638)],
# [(0.926, 0.475), (0.336, 0.841), (0.52, 0.225)]
# ]
transposed_with_pairs = [pair for row in transposed_as_matrix for pair in row]
# [
# (0.914, 0.639), (0.255, 0.671), (0.279, 0.341),
# (0.058, 0.76), (0.195, 0.966), (0.591, 0.638),
# (0.926, 0.475), (0.336, 0.841), (0.52, 0.225)
# ]
transposed = [val for pair in transposed_with_pairs for val in pair]
# [
# 0.914, 0.639, 0.255, 0.671, 0.279, 0.341,
# 0.058, 0.76, 0.195, 0.966, 0.591, 0.638,
# 0.926, 0.475, 0.336, 0.841, 0.52, 0.225
# ]
to_transpose = [0.914, 0.639, 0.058, 0.760, 0.926, 0.475,
0.255, 0.671, 0.195, 0.966, 0.336, 0.841,
0.279, 0.341, 0.591, 0.638, 0.520, 0.225]
matrix_width = 6
matrix_height = 3
# INITIALIZE AN EMPTY LIST
transposed_list = [None] * matrix_width * matrix_height
for w in range(0, matrix_width, 2):
for h in range(matrix_height):
transposed_list[w * matrix_height + (2 * h)] = to_transpose[h * matrix_width + w]
transposed_list[(w * matrix_height + (2 * h)) + 1] = to_transpose[(h * matrix_width + w) + 1]
print(transposed_list)
我做了三件事:
- 使用步长值作为2,使w跳过奇数值
- 添加了一个额外的行,其中奇数索引元素获得y的值
- 最后,将h的值加倍,以便每个替换元素都得到一个 新价值
m=to_transpose # for easy writing only
[ n for l in zip( *[[(m[i],m[i+1]) for i in range(k,k+6,2)] for k in range(0,len(m),6)] ) for t in l for n in t ]
改用numpy。您要求“无外部库”,但
numpy
将非常适合:np.array(to_transpose).restrape((6,3)).T.ravel()
鉴于每两个值都属于同一个值,它可能更像是restrape(3,3)
或(2,3,3)
@mkrieger1同意,我一开始并没有看到两个值是成对的。@Dominicksarle如果您将列表预处理为[(0.914,0.639),(0.058,0.760),…]
,那么您的方法会起作用。尝试列出列表我会接受这个答案,因为它是正确和快速的。非常感谢。但是,您是否知道一种不使用numpy的方法(在for循环中)?