python - 如何打印特定的 numpy 数组对

Question

我想在一个 numpy 数组中制作特定的对，并用一个简单的打印函数显示我想要的内容。我有两个数组:

points=np.arange(1,15)

然后我有另一个数组:

repetition= np.array([4, 4, 2, 2, 1, 1])

现在，我想打印以下对(我只是写了注释来显示我想要的):

1 5 # first value of points and (1+4)th value of point
2 6 # second value of points and (2+4)th value of point
3 7 # third value of points and (3+4)th value of point
4 8 # fourth value of points and (3+4)th value of point
7 9 # seventh value of points and (6+2)th value of point
8 10 # eighth value of points and (8+2)th value of point
9 11 # ninth value of points and (9+2)th value of point
10 12 # tenth value of points and (10+2)th value of point
12 13 # twelfth value of points and (11+2)th value of point
13 14 # thirteenth value of points and (13+1)th value of point

我尝试了下面的代码，但它没有给我预期的结果:

for m, n in zip (points, repetition):
    print (m, m+n)

在图中，我想象了我的问题，其中红线显示了我的对。我非常感谢提前提供的任何帮助。

Answer 1

您实际上可以在 Python 中完成这一切。由于您的图片似乎表明数组参差不齐，因此在 numpy 中执行此操作可能有点棘手。

from itertools import islice, zip_longest, count, tee

def pairwise(iterable):  # as per itertools recipes
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return zip(a, b)

def idx_pairs(repetitions):
    it = count()
    z = [list(islice(it, n))[::-1] for n in repetitions] 
    idx = sorted([
        (i, j) for seq in zip_longest(*z)
        for i, j in pairwise([k for k in seq if k is not None])])
    return idx

[(points[i], points[j]) for i, j in idx_pairs(repetition)]

输出:

[(1, 5),
 (2, 6),
 (3, 7),
 (4, 8),
 (7, 9),
 (8, 10),
 (9, 11),
 (10, 12),
 (12, 13),
 (13, 14)]

为了更好地理解这些步骤，我建议检查:

• z
• idx
• 列表(zip_longest(*z))

特别是后者(直接在点上而不是点索引上完成)显示了与 OP 的绘图非常相似的内容:

it = iter(points)
z = [list(islice(it, n))[::-1] for n in repetition]
list(zip_longest(*z))
# out:
[(4, 8, 10, 12, 13, 14),
 (3, 7, 9, 11, None, None),
 (2, 6, None, None, None, None),
 (1, 5, None, None, None, None)]

顺便说一句，看看当 repetition 列表不是单调递减时会发生什么很有趣:

repetition = np.array([4, 2, 4, 2, 1, 1])

it = iter(points)
z = [list(islice(it, n))[::-1] for n in repetition]
list(zip_longest(*z))

# out:
[(4, 6, 10, 12, 13, 14),
 (3, 5, 9, 11, None, None),
 (2, None, 8, None, None, None),
 (1, None, 7, None, None, None)]

我相信这样一个repetition的正确输出应该是:

[(1, 7),
 (2, 8),
 (3, 5),
 (4, 6),
 (5, 9),
 (6, 10),
 (9, 11),
 (10, 12),
 (12, 13),
 (13, 14)]

为了好玩，点可以包含任何东西；这是一个字符串示例:

points = [
    'foo', 'bar', 'hi', 'hello', 'world',
    'fuzz', 'ball', 'apple', 'tom', 'nancy',
    'fred', 'james', 'mary', 'bob', 'lisa', 
]
repetition = np.array([4, 2, 4, 2, 1, 1])
[(points[i], points[j]) for i, j in idx_pairs(repetition)]

# out:
[('foo', 'ball'),
 ('bar', 'apple'),
 ('hi', 'world'),
 ('hello', 'fuzz'),
 ('world', 'tom'),
 ('fuzz', 'nancy'),
 ('tom', 'fred'),
 ('nancy', 'james'),
 ('james', 'mary'),
 ('mary', 'bob')]