在Python中迭代N维

时间:2010-09-11 15:04:20

标签: python recursion iteration

我有一张地图,我们称之为M,其中包含通过N维映射的数据。

# If it was a 2d map, I could iterate it thusly:
start, size = (10, 10), (3, 3)
for x in range(start[0], start[0]+size[0]):
    for y in range(start[1], start[1]+size[1]):
        M.get((x, y))
# A 3d map would add a for z in ... and access it thusly
M.get((x, y, z))
# And so on.

我的问题是:如何创建一个可以产生正确迭代序列的迭代器?也就是说,给定start, size = (10, 10), (3, 3)它将产生2元组序列(10, 10), (10, 11), (10, 12), (11, 10), (11, 11)等。并且给定start, size = (10, 10, 10), (3, 3, 3)它将产生正确的3元组序列。

是的,我试过自己,但我的脑袋爆炸了。或者我无法证明花时间搞清楚,即使它很有趣。请选择:)

2 个答案:

答案 0 :(得分:8)

在Python 2.6 +中:

itertools.product(*[xrange(i, i+j) for i,j in zip(start, size)])

答案 1 :(得分:0)

自己做生成器表达式:

start, size = (10, 10), (3, 3)
values2=((x+xd,y+yd)
        for x,y in (start,)
        for xr,yr in (size,)
        for xd in range(xr)
        for yd in range(yr))

for x,y in values2:
    print x,y

start, size = (10, 10, 10), (3, 3, 3)
values3=((x+xd,y+yd, z+zd)
        for x,y,z in (start,)
        for xr,yr,zr in (size,)
        for xd in range(xr)
        for yd in range(yr)
        for zd in range(zr))

for x,y,z in values3:
    print x,y,z