问题描述
我有一个列表列表，其中包含角度集的下限和上限
I have a list of lists which contains the lower and upper limit of sets of angles
类似
[[1,22],[2,24]...[359,15],[360,21]]
[[1,22],[2,24]...[359,15],[360,21]]
总共 360 个元素
现在我想检查从 1 到 360 的每个角度列表中包含该角度的元素
Now I want to check for each angle from 1 to 360 the elements in the list that contains that angle
我正在考虑使用下限和上限来创建列表的所有元素 range 或 np.arange(lower,upper) 并检查是否角度包含在内，但是np.arange在lower高于upper时生成空列表
I was thinking about using the lower and upper limits to create all the elements of the list with range or np.arange(lower,upper) and check if the angle is contained, but np.arange generates empty list when lower is higher than upper
for i in range(1,361):
sel=[]
for coe in coef:
if i in range(coe[0],coe[1]):
sel.append(coe)
我也尝试了 itertools 和 cycle 但我不知道在这种情况下如何应用它
I tried also with itertools and cycle but i do not see how it can be applied in this case
推荐答案
执行此操作的标准方法是使用模数.这个答案 by sschuberth 展示了如何在 C/C++ 中做到这一点.
The standard way to do this is to use modulus. This answer by sschuberth shows how to do this in C/C++.
这是一个 Python 实现和测试.Python 中的代码更简单，因为在 Python 中 a % b 总是与 b 具有相同的符号.
Here's a Python implementation and test. The code is simpler in Python, because in Python a % b always has the same sign as b.
测试代码以 60 的步长循环从 0 到 360 度的所有角度对，对于 a 和 b，测试所有角度 x 从 0 到 360 度，步长为 30.扇区从 a 开始并扫过 b.因此扇区 (60, 120) 包含 60°，但 (120, 60) 包含 300°.
The test code loops through all pairs of angles from 0 to 360 degrees in steps of 60, for a and b, testing all angles x from 0 to 360 degrees in steps of 30. The sector starts at a and sweeps through to b. Thus the sector (60, 120) contains 60°, but (120, 60) contains 300°.
如果 x 在扇区内(包括端点)，它将被添加到 result 列表中.
If x is within the sector (including the endpoints) it gets added to the result list.
def in_angle_interval(x, a, b):
return (x - a) % 360 <= (b - a) % 360
# test
for a in range(0, 420, 60):
for b in range(0, 420, 60):
result = [x for x in range(0, 390, 30) if in_angle_interval(x, a, b)]
print('{:3}-{:3} {}'.format(a, b, result))
输出
0- 0 [0, 360]
0- 60 [0, 30, 60, 360]
0-120 [0, 30, 60, 90, 120, 360]
0-180 [0, 30, 60, 90, 120, 150, 180, 360]
0-240 [0, 30, 60, 90, 120, 150, 180, 210, 240, 360]
0-300 [0, 30, 60, 90, 120, 150, 180, 210, 240, 270, 300, 360]
0-360 [0, 360]
60- 0 [0, 60, 90, 120, 150, 180, 210, 240, 270, 300, 330, 360]
60- 60 [60]
60-120 [60, 90, 120]
60-180 [60, 90, 120, 150, 180]
60-240 [60, 90, 120, 150, 180, 210, 240]
60-300 [60, 90, 120, 150, 180, 210, 240, 270, 300]
60-360 [0, 60, 90, 120, 150, 180, 210, 240, 270, 300, 330, 360]
120- 0 [0, 120, 150, 180, 210, 240, 270, 300, 330, 360]
120- 60 [0, 30, 60, 120, 150, 180, 210, 240, 270, 300, 330, 360]
120-120 [120]
120-180 [120, 150, 180]
120-240 [120, 150, 180, 210, 240]
120-300 [120, 150, 180, 210, 240, 270, 300]
120-360 [0, 120, 150, 180, 210, 240, 270, 300, 330, 360]
180- 0 [0, 180, 210, 240, 270, 300, 330, 360]
180- 60 [0, 30, 60, 180, 210, 240, 270, 300, 330, 360]
180-120 [0, 30, 60, 90, 120, 180, 210, 240, 270, 300, 330, 360]
180-180 [180]
180-240 [180, 210, 240]
180-300 [180, 210, 240, 270, 300]
180-360 [0, 180, 210, 240, 270, 300, 330, 360]
240- 0 [0, 240, 270, 300, 330, 360]
240- 60 [0, 30, 60, 240, 270, 300, 330, 360]
240-120 [0, 30, 60, 90, 120, 240, 270, 300, 330, 360]
240-180 [0, 30, 60, 90, 120, 150, 180, 240, 270, 300, 330, 360]
240-240 [240]
240-300 [240, 270, 300]
240-360 [0, 240, 270, 300, 330, 360]
300- 0 [0, 300, 330, 360]
300- 60 [0, 30, 60, 300, 330, 360]
300-120 [0, 30, 60, 90, 120, 300, 330, 360]
300-180 [0, 30, 60, 90, 120, 150, 180, 300, 330, 360]
300-240 [0, 30, 60, 90, 120, 150, 180, 210, 240, 300, 330, 360]
300-300 [300]
300-360 [0, 300, 330, 360]
360- 0 [0, 360]
360- 60 [0, 30, 60, 360]
360-120 [0, 30, 60, 90, 120, 360]
360-180 [0, 30, 60, 90, 120, 150, 180, 360]
360-240 [0, 30, 60, 90, 120, 150, 180, 210, 240, 360]
360-300 [0, 30, 60, 90, 120, 150, 180, 210, 240, 270, 300, 360]
360-360 [0, 360]
<小时>
这是使用问题中的数据进行的测试.
And here's a test using the data in the question.
coef = [[1,22], [2,24], [359,15], [360,21]]
print(coef)
for x in range(0, 361):
sel = [coe for coe in coef if in_angle_interval(x, coe[0], coe[1])]
if sel:
print('{:3} {}'.format(x, sel))
输出
[[1, 22], [2, 24], [359, 15], [360, 21]]
0 [[359, 15], [360, 21]]
1 [[1, 22], [359, 15], [360, 21]]
2 [[1, 22], [2, 24], [359, 15], [360, 21]]
3 [[1, 22], [2, 24], [359, 15], [360, 21]]
4 [[1, 22], [2, 24], [359, 15], [360, 21]]
5 [[1, 22], [2, 24], [359, 15], [360, 21]]
6 [[1, 22], [2, 24], [359, 15], [360, 21]]
7 [[1, 22], [2, 24], [359, 15], [360, 21]]
8 [[1, 22], [2, 24], [359, 15], [360, 21]]
9 [[1, 22], [2, 24], [359, 15], [360, 21]]
10 [[1, 22], [2, 24], [359, 15], [360, 21]]
11 [[1, 22], [2, 24], [359, 15], [360, 21]]
12 [[1, 22], [2, 24], [359, 15], [360, 21]]
13 [[1, 22], [2, 24], [359, 15], [360, 21]]
14 [[1, 22], [2, 24], [359, 15], [360, 21]]
15 [[1, 22], [2, 24], [359, 15], [360, 21]]
16 [[1, 22], [2, 24], [360, 21]]
17 [[1, 22], [2, 24], [360, 21]]
18 [[1, 22], [2, 24], [360, 21]]
19 [[1, 22], [2, 24], [360, 21]]
20 [[1, 22], [2, 24], [360, 21]]
21 [[1, 22], [2, 24], [360, 21]]
22 [[1, 22], [2, 24]]
23 [[2, 24]]
24 [[2, 24]]
359 [[359, 15]]
360 [[359, 15], [360, 21]]
这篇关于通过使用一系列角度python创建一个圆形列表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，WP2