按一定间隔对列表进行切片，然后合并

Question

如何在每个x索引处"列化"列表？

我想通过获取起始列表，然后创建两个新列表，合并它们，然后添加不可被x整除的余数来实现这一点。

例如，间隔为2:

start = [
    [1, 'one'],
    [2, 'two'],
    [3, 'three'],
    [4, 'four'],
    [5, 'five'],
    [6, 'six'],
    [7, 'seven'],
    [8, 'eight'],
    [9, 'nine'],
]
expected = [
    [1, 'one', 3, 'three'],
    [2, 'two', 4, 'four'],
    # page break
    [5, 'five', 7, 'seven'],
    [6, 'six', 8, 'eight'],
    # page break
    [9, 'nine'],
]

只是想知道是否有一种快速的方法？

Answer 1

我同意关于这是一种奇怪的"列化"方式的评论。然而，这里有一个函数可以实现您所描述的功能：

def columnize(A, interval=2):
    ans = []
    for i in range(0,len(A), interval*2):
        for j in range(min(interval, len(A)-i)):
            ans.append(A[i+j] + (A[i+j+interval] if i+j+interval < len(A) else []))
    return ans

Answer 2

你在找这样的东西吗？列平方矩阵？

start = [
    [1, 'one'],
    [2, 'two'],
    [3, 'three'],
    [4, 'four'],
    [5, 'five'],
    [6, 'six'],
    [7, 'seven'],
    [8, 'eight'],
    [9, 'nine'],
]
expected = [
    [1, 'one', 3, 'three'],
    [2, 'two', 4, 'four'],
    # page break
    [5, 'five', 7, 'seven'],
    [6, 'six', 8, 'eight'],
    # page break
    [9, 'nine'],
]
a = 2
r = a*a
ans = []
for i in range(0, len(start), r):
    l_tmp = start[i:i+r]
    if l_tmp[::a]:
        ans.append([item for sublist in l_tmp[::a] for item in sublist])
    if l_tmp[1::a]:
        ans.append([item for sublist in l_tmp[1::a] for item in sublist])
    # You can easily add page break here
print(ans)