Question

如果我有一个说'n'元素的列表（每个元素是一个字节），它代表一个矩形的2d矩阵，我怎么能把它分成说w * h的矩形，从列表的第一个元素开始，只使用python标准函数

例如

l =  
 [ 1,2,3,4,5,6,7,8,9,10,  
   11,12,13,14,15....20.  
   21,22,23,24,25....30  
   .....    
   .................200]

这些是在1d列表中

如果我们选择2 * 3（w * h）的矩形第一个将包含1,2,11,12,21,22 第二个将包含3,4,13,14,23,24等等，直到结束

由于

Answer 1

width = 6
height = 4
xs = range(1,25)
w = 3
h = 2

def subrect(x,y):
    pos = y*h*width+x*w
    return [xs[(pos+row*width):(pos+row*width+w)] for row in range(h)]

print [subrect(x,y) for y in range(height / h) for x in range(width / w)]

按如下方式拆分矩阵：

 1  2  3     4  5  6
 7  8  9    10 11 12

13 14 15    16 17 18
19 20 21    22 23 24

编辑：或者您提供的示例......

width = 10
height = 20
xs = range(1,201)
w = 2
h = 3

Answer 2

请注意，您的问题指定输入列表是1D，但不指示每个逻辑行的项目数;你似乎神奇地暗示它应该是每行10件。

因此，给定1D列表，每行逻辑项的计数，请求的tile的宽度和高度，您可以这样做：

def gettiles(list1d, row_items, width, height):
    o_row= 0
    row_count, remainder= divmod(len(list1d), row_items)
    if remainder != 0:
        raise RuntimeError("item count not divisible by %d" % row_items)
    if row_count % height != 0:
        raise RuntimeError("row count not divisible by height %d" % height)
    if row_items % width != 0:
        raise RuntimeError("row width not divisible by %d" % width)
    for o_row in xrange(0, row_count, height):
        for o_col in xrange(0, row_items, width):
            result= []
            top_left_index= o_row*row_items + o_col
            for off_row in xrange(height):
                for off_col in xrange(width):
                    result.append(list1d[top_left_index + off_row*row_items + off_col])
            yield result

>>> import pprint
>>> pprint.pprint(list(gettiles(range(100), 10, 2, 5)))
[[0, 1, 10, 11, 20, 21, 30, 31, 40, 41],
 [2, 3, 12, 13, 22, 23, 32, 33, 42, 43],
 [4, 5, 14, 15, 24, 25, 34, 35, 44, 45],
 [6, 7, 16, 17, 26, 27, 36, 37, 46, 47],
 [8, 9, 18, 19, 28, 29, 38, 39, 48, 49],
 [50, 51, 60, 61, 70, 71, 80, 81, 90, 91],
 [52, 53, 62, 63, 72, 73, 82, 83, 92, 93],
 [54, 55, 64, 65, 74, 75, 84, 85, 94, 95],
 [56, 57, 66, 67, 76, 77, 86, 87, 96, 97],
 [58, 59, 68, 69, 78, 79, 88, 89, 98, 99]]

Answer 3

或者这很简单。

def genMatrix(rows, cols, mylist):
   for x in xrange(rows):
      yield mylist[x*cols:x*cols+cols]

结果

>>> L = [1,1,1,1,2,2,2,2]
>>> list(genMatrix(2, 4, L))
[[1, 1, 1, 1], [2, 2, 2, 2]]
>>> L = [1,1,1,1,2,2,2,2,3,3,3,3]
>>> list(genMatrix(3, 4, L))
[[1, 1, 1, 1], [2, 2, 2, 2], [3, 3, 3, 3]]

Answer 4

这是一个建议（可能非常低效，但似乎有效）：

def rect_slice(seq, cols, width, height):
    rows = len(seq) // cols
    for i in xrange(0, rows - rows % height, height):
        for j in xrange(0, cols - cols % width, width):
            yield [seq[k * cols + l] for k in xrange(i, i + height) for l in xrange(j, j + width)]

print list(rect_slice(range(1, 201), 10, 2, 3))

Answer 5

如果我理解正确，您[1,1,1,1,2,2,2,2]并希望[[1,1,1,1], [2,2,2,2]]？如果是这样，那就是：

L = [1,1,1,1,2,2,2,2]
w = 4
matrix = [L[:w], L[w:]]

至少是2d。

或者你可以写一个更通用的解决方案：

def genMatrix(rows, cols, mylist):
   matrix = []
   for x in xrange(rows):
      row = []
      for y in xrange(cols):
         row.append(mylist[x*cols])
      matrix.append(row)
   return matrix

print genMatrix(2, 4, L) # => [[1,1,1,1], [2,2,2,2]]
L = [1,1,1,1,2,2,2,2,3,3,3,3]
print getnMatrix(3, 4, L) # => [[1,1,1,1], [2,2,2,2], [3,3,3,3]]

切片python列表

5 个答案: