如何打印特定的numpy数组对

Question

我想在一个 numpy 数组中创建特定的对，并用一个简单的打印函数展示我想要的东西。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（我刚刚写了评论以显示我想要的）：

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

我尝试了以下代码，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，我可视化了我的问题，其中红线显示了我的配对。我非常感谢您提前提供的任何帮助。

Answer 1

您实际上可以在 Python 中完成所有操作。由于您的图片似乎显示参差不齐的数组，因此在 numpy 中执行此操作可能会有点棘手。

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
• list(zip_longest(*z))

尤其是后者（直接在点上完成，而不是在点索引上），显示的内容与 OP 的绘图非常相似：

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)]

我相信这样的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，积分可以包含任何东西；下面是一个字符串示例：

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')]

Answer 2

只有在@Pierre D. 的解释之后，我才真正理解了这个问题。谢谢。我更正了我之前的代码，并在里面放了一些打印功能以简化解释。

points=np.arange(1,15)
repetition= np.array([4, 4, 2, 2, 1, 1])

L=[]
k=0
for r in repetition:
    last= k+r
    L.append(points[k:last])
    k= last

print(L,"\n")

lmax= len(max(L,key=len))
L=[ np.concatenate((np.full(lmax-len(l),None),l))  for l in L ]
print(*L,"\n",sep="\n")

print(*zip(*L),"\n",sep="\n")

L= [ [ e for e in l if e] for l in zip(*L) ]
print(*L,"\n",sep="\n")

P= sorted([ p  for l in L for p in zip(l,l[1:]) ])
print(*P,"\n",sep="\n")