Question

我有一个带有矢量点的列表，我希望根据嵌套列表的第二个值y位置进行拆分。所以，points[0][1]。示例列表包含两个唯一的y高度：920和940。但是列表可以轻松达到10个独特的高度。

points = [(418, 920), (558, 920), (726, 920), (858, 920), (906, 920), (1042, 920), (418, 940), (558, 940), (734, 940), (865, 940), (899, 940), (1034, 940)]

# Desired result:
newPoints = [ [ [x,920], [x,920]  ], [ [x,940], [x,940] ] ]

Answer 1

这是collections .defaultdict的来源：

from collections import defaultdict
new_points = defaultdict(list)
for point in points:
    new_points[point[1]].append(point)

defaultdict(<class 'list'>, {
920: [(418, 920), (558, 920), (726, 920), (858, 920), (906, 920), (1042, 920)], 
940: [(418, 940), (558, 940), (734, 940), (865, 940), (899, 940), (1034, 940)]
})

如果您需要这些列表，可以执行new_points.values()或list(new_points.values())

[[(418, 920), (558, 920), (726, 920), (858, 920), (906, 920), (1042, 920)],
 [(418, 940), (558, 940), (734, 940), (865, 940), (899, 940), (1034, 940)]]

Answer 2

我在其x值列表中创建了y值的字典，然后迭代y值，为每个值创建子列表，并迭代x值，为每个值创建一个子列表。 / p>

points = [(418, 920), (558, 920), (726, 920), (858, 920), (906, 920), (1042, 920), (418, 940), (558, 940), (734, 940), (865, 940), (899, 940), (1034, 940)]

y_dict = dict()

for point in points:
    x = point[0]
    y = point[1]
    if y in y_dict.keys():
        #already found this y
        y_dict[y].append(x)
    else:
        #haven't found this y
        y_dict[y] = [x]

newPoints = []

for y in y_dict.keys():
    sublist = []
    for x in y_dict[y]:
        sublist.append([x, y])
    newPoints.append(sublist)

print(newPoints)

然后结果是：

[[[418, 920], [558, 920], [726, 920], [858, 920], [906, 920], [1042, 920]], [[418, 940], [558, 940], [734, 940], [865, 940], [899, 940], [1034, 940]]]

基于唯一值拆分嵌套列表

2 个答案: