我有2个带有网格像素坐标的数组。
[402, 401, 356, 355, 356, 355, 356, 386, 354, 355, 356, 386, 354, 355, 386, 354, 287, 288, 289, 290, 291, 292, 287, 288, 290, 291, 292, 293, 292, 293, 294, 295, 293, 294, 295, 296, 294, 295, 296, 146, 145, 146, 167, 168, 162, 163, 164, 165, 166, 167, 168, 169, 170, 160, 161, 162, 165, 166, 167, 168, 169, 170, 162, 163, 166, 167, 168, 169, 170, 163, 167, 168, 169, 170, 168, 169, 169, 170, 170, 310, 310, 311, 312, 313, 314, 315, 316, 317, 311, 312, 313, 314, 315, 316, 317, 311, 312, 313, 314, 315, 316, 311, 312, 313, 314, 315, 316, 311, 312, 313, 314, 274, 275, 275, 274, 275, 274, 275, 273, 274, 275, 274, 275, 274, 272, 273, 272, 273, 272, 273, 272, 273, 272, 273, 271, 272, 271]
[242, 243, 257, 258, 258, 259, 259, 259, 260, 260, 260, 260, 261, 261, 261, 262, 284, 284, 284, 284, 284, 284, 285, 285, 285, 285, 285, 285, 286, 286, 286, 286, 287, 287, 287, 287, 288, 288, 288, 326, 327, 327, 337, 337, 338, 338, 338, 338, 338, 338, 338, 338, 338, 339, 339, 339, 339, 339, 339, 339, 339, 339, 340, 340, 340, 340, 340, 340, 340, 341, 341, 341, 341, 341, 342, 342, 343, 343, 344, 382, 383, 383, 383, 383, 383, 383, 383, 383, 384, 384, 384, 384, 384, 384, 384, 385, 385, 385, 385, 385, 385, 386, 386, 386, 386, 386, 386, 387, 387, 387, 387, 416, 416, 417, 418, 418, 419, 419, 420, 420, 420, 421, 423, 427, 428, 428, 429, 429, 430, 430, 431, 431, 432, 432, 433, 433, 434]
数据中大约有8个点，其中数字相互连接。主要问题是，如何找到这些景点。我们想为每个包含数字的地点创建一个数组。我认为这是一个简单的问题，但仍然没有运气找出如何编程。
我尝试过while和for循环，但结果却什么都没有。最好的方法是什么？
答案 0 :(得分：1)
这是一种天真的做法：
from itertools import product
def neighbours(xs,ys):
xys = list(zip(xs,ys))
for x,y in product(range(max(xs)),range(max(ys))):
if (x,y) in xys:
if (x+1,y) in xys:
yield [(x,y),(x+1,y)]
if (x,y+1) in xys:
yield [(x,y),(x,y+1)]
对于您的输入，这会生成[[(287, 287), (288, 287)], [(287, 287), (287, 288)], [(288, 288), (289, 288)], [(386, 386), (386, 387)], ...]形式的内容。
要获取8个元组的列表作为输出，请将yield [...]语句更改为每个列表项的单独yield语句。
答案 1 :(得分：1)
嗯，我不确定为什么它会提出额外的位置，但它们看起来在你的列表中匹配。我得到175个点！ （当然2-4,4-4,3-5,4-6,5-6看起来与你的名单相符）。也许我正在使用不同的标准。这是您可以随意破解的代码：
import numpy as np
a = np.array([[402, 401, 356, 355, 356, 355, 356, 386, 354, 355, 356, 386, 354, 355, 386, 354, 287, 288, 289, 290, 291, 292, 287, 288, 290, 291, 292, 293, 292, 293, 294, 295, 293, 294, 295, 296, 294, 295, 296, 146, 145, 146, 167, 168, 162, 163, 164, 165, 166, 167, 168, 169, 170, 160, 161, 162, 165, 166, 167, 168, 169, 170, 162, 163, 166, 167, 168, 169, 170, 163, 167, 168, 169, 170, 168, 169, 169, 170, 170, 310, 310, 311, 312, 313, 314, 315, 316, 317, 311, 312, 313, 314, 315, 316, 317, 311, 312, 313, 314, 315, 316, 311, 312, 313, 314, 315, 316, 311, 312, 313, 314, 274, 275, 275, 274, 275, 274, 275, 273, 274, 275, 274, 275, 274, 272, 273, 272, 273, 272, 273, 272, 273, 272, 273, 271, 272, 271],
[242, 243, 257, 258, 258, 259, 259, 259, 260, 260, 260, 260, 261, 261, 261, 262, 284, 284, 284, 284, 284, 284, 285, 285, 285, 285, 285, 285, 286, 286, 286, 286, 287, 287, 287, 287, 288, 288, 288, 326, 327, 327, 337, 337, 338, 338, 338, 338, 338, 338, 338, 338, 338, 339, 339, 339, 339, 339, 339, 339, 339, 339, 340, 340, 340, 340, 340, 340, 340, 341, 341, 341, 341, 341, 342, 342, 343, 343, 344, 382, 383, 383, 383, 383, 383, 383, 383, 383, 384, 384, 384, 384, 384, 384, 384, 385, 385, 385, 385, 385, 385, 386, 386, 386, 386, 386, 386, 387, 387, 387, 387, 416, 416, 417, 418, 418, 419, 419, 420, 420, 420, 421, 423, 427, 428, 428, 429, 429, 430, 430, 431, 431, 432, 432, 433, 433, 434]])
a = a.T
b = np.where(
(abs(np.subtract.outer(a[:,0], a[:,0])) <= 1.0) &
(abs(np.subtract.outer(a[:,1], a[:,1])) <= 1.0))
non_dup = np.where(b[0] < b[1]) # remove double count and 'self' overlaps
b = (b[0][non_dup], b[1][non_dup]) # remake slimmed down index
def find_group(group, start):
st_ix = np.where(b[0] == start)
for i in st_ix[0]:
dest = b[1][i]
group.add(start)
group.add(dest) # is this needed?
find_group(group, dest)
group_list = []
for i in b[0]: # NB this part is slow - it would be better to eliminate entries already dealt with - you can figure that out!
gp = set()
find_group(gp, i)
add_group = True
for old_gp in group_list:
if len(gp.intersection(old_gp)) > 0: # hopefully this has fixed it
old_gp = old_gp.union(gp)
add_group = False
break
if add_group:
group_list.append(gp)
xy_list = []
for gp in group_list:
xy_list.append(np.array([a[i] for i in gp]))
print(xy_list)
如上所述修改EDIT以创建注释中讨论的连续像素的numpy数组列表。结果在此
[array([[402, 242],
[401, 243]]),
array([[356, 257],
[355, 258],
[356, 258],
[355, 259],
[356, 259],
[354, 260],
[355, 260],
[356, 260],
[354, 261],
[355, 261],
[354, 262]]),
array([[386, 260],
[386, 261],
[386, 259]]),
array([[287, 284],
[288, 284],
[289, 284],
[290, 284],
[291, 284],
[292, 284],
[287, 285],
[288, 285],
[290, 285],
[291, 285],
[292, 285],
[293, 285],
[292, 286],
[293, 286],
[294, 286],
[295, 286],
[293, 287],
[294, 287],
[295, 287],
[296, 287],
[294, 288],
[295, 288],
[296, 288]]),
array([[145, 327],
[146, 327],
[146, 326]]),
array([[167, 337],
[168, 337],
[166, 338],
[167, 338],
[168, 338],
[169, 338],
[170, 338],
[165, 339],
[166, 339],
[167, 339],
[168, 339],
[169, 339],
[170, 339],
[166, 340],
[167, 340],
[168, 340],
[169, 340],
[170, 340],
[167, 341],
[168, 341],
[169, 341],
[170, 341],
[168, 342],
[169, 342],
[169, 343],
[170, 343],
[170, 344]]),
array([[163, 341],
[160, 339],
[161, 339],
[162, 339],
[162, 340],
[163, 340]]),
array([[310, 382],
[310, 383],
[311, 383],
[312, 383],
[313, 383],
[314, 383],
[315, 383],
[316, 383],
[317, 383],
[311, 384],
[312, 384],
[313, 384],
[314, 384],
[315, 384],
[316, 384],
[317, 384],
[311, 385],
[312, 385],
[313, 385],
[314, 385],
[315, 385],
[316, 385],
[311, 386],
[312, 386],
[313, 386],
[314, 386],
[315, 386],
[316, 386],
[311, 387],
[312, 387],
[313, 387],
[314, 387]]),
array([[274, 416],
[275, 416],
[275, 417],
[274, 418],
[275, 418],
[274, 419],
[275, 419],
[273, 420],
[274, 420],
[275, 420],
[274, 421]]),
array([[272, 430],
[273, 430],
[272, 431],
[273, 431],
[272, 432],
[273, 432],
[271, 433],
[272, 433],
[271, 434],
[274, 427],
[273, 428],
[272, 429],
[273, 429]])]