问题描述
我正在尝试根据彼此之间的距离删除数组中的某些项目.
I am trying to remove some items inside an array based on the distance from each other.
我在具有元组的数组中有 104 个项目:
I have 104 items in an array with tuples:
points = [(910, 28), (914, 29), (919, 30), (915, 32), (766, 73), (777, 75), (768, 80), (1227, 117), (1224, 119), (1228, 120), (1224, 121), (1228, 122), (1221, 124), (1218, 126), (486, 147), (482, 150), (474, 153), (905, 182), (908, 184), (904, 186), (901, 187), (913, 188), (909, 190), (903, 193), (187, 213), (186, 214), (189, 215), (611, 262), (617, 264), (619, 265), (610, 268), (1231, 272), (1228, 274), (1228, 276), (1232, 278), (1223, 282), (486, 306), (477, 309), (463, 312), (470, 313), (486, 315), (473, 319), (764, 376), (773, 379), (770, 383), (795, 386), (778, 390), (631, 412), (626, 414), (624, 416), (626, 418), (1218, 434), (1217, 435), (1218, 436), (1219, 437), (1220, 438), (1222, 439), (1225, 440), (1226, 442), (480, 460), (478, 463), (1071, 466), (1062, 468), (1067, 469), (1072, 470), (339, 485), (343, 487), (345, 489), (346, 490), (350, 492), (343, 495), (352, 497), (930, 505), (929, 508), (929, 513), (199, 535), (197, 537), (203, 539), (201, 542), (771, 547), (774, 547), (773, 548), (772, 549), (776, 549), (776, 550), (629, 576), (628, 579), (631, 580), (625, 583), (1237, 586), (1218, 590), (1226, 593), (1223, 595), (1227, 599), (639, 732), (643, 733), (643, 734), (204, 875), (209, 877), (210, 879), (210, 880), (210, 882), (210, 884), (204, 887)]
我的代码如下:
print len(points)
w = 152
h = 157
for pt in points:
for fpt in points:
if pt == fpt:
continue
else:
distX = abs(pt[0] - fpt[0])
distY = abs(pt[1] - fpt[1])
dist = distX + distY
if distX < w / 2 and distY < h / 2:
points.remove(fpt)
print len(points)
当我运行它时，它会删除很多项目，只剩下 33 ，但这是不正确的.如果我再次运行它，则会留下正确的数字 22 .
When I run it, it removes lots of items and it leaves me with only 33, but it is not correct. If I run it again it leaves me the correct number which is 22.
我确定自己做错了什么，有人可以帮助我找到并纠正我的错误吗?
I am sure that I did something wrong, can anyone help me locate and correct my mistake?
推荐答案
在迭代列表的同时删除列表项不是一个好习惯.例如，
Removing list items in place while iterating the list is not good practice.For instance,
if points has only 3 elements [a, b, c],
a is adjacent to b,
b is adjacent to c,
but a is not adjacent to c,
Correct result is [a] after the algorithm runs. But with your algorithm,
after the first iteration, b is removed; points = [a, c]
c can never be removed, since a and c are not adjacent.
您要寻找的是联合查找集，这是一个实现: http://code.activestate.com/recipes/215912-union -find-data-structure/
What you are looking for is a Union-find-set, here is a implementation:http://code.activestate.com/recipes/215912-union-find-data-structure/
这是Union-Find的一个更幼稚的实现，供您参考:集合并集查找算法
And here is a more naive implementation of Union-Find, for your reference:A set union find algorithm
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