将列表移至其他列表一半大小以获取公共元素

Question

I have two lists s1 and s2 . I need to slide the second list half its size on left and return the matching elements in first list. Then I need to move the second list one position and get the common elements as shown in the Answer below. I am plaaning to perforn some operation on each pair of sublists returned. What is the good way to accomplish this.

s1 = [11, 12, 13, 14, 15, 16, 17, 18]
s2 = [21, 22, 23, 24, 25, 25] 


11 12 13 14 15 16 17 18 
21 22 23 24 25 25 

Answer:

11 12 13 
24 25 25 

11 12 13 14 
23 24 25 25 

11 12 13 14 15 
22 23 24 25 25 

11 12 13 14 15 16 
21 22 23 24 25 25 

12 13 14 15 16 17 
21 22 23 24 25 25 

13 14 15 16 17 18 
21 22 23 24 25 25 


14 15 16 17 18 
21 22 23 24 25 


15 16 17 18 
21 22 23 24 


16 17 18 
21 22 23 

Answer 1

Hope you get the idea

s1 = [11, 12, 13, 14, 15, 16, 17, 18]
s2 = [21, 22, 23, 24, 25, 25] 

l1 = len(s1)
l2 = len(s2)
l = min(l1,l2)+1

k = l2/2;

for i in range(k,l2+1):
    print s1[:i]
    print s2[-i:]
    print

for i in range(1,l1-l2):
    print s1[i:i+l2]
    print s2[:]
    print


for i in reversed(range(k,l)):
    print s1[-i:]
    print s2[:i]
    print

Answer 2

It doesn't do any bounds-checking, but this should do the trick:

def slide(s1, s2, size):
    start = size - len(s1)
    finish = len(s2) - size + 1
    for offset in xrange(start, finish):
        m = max(-offset, 0)
        n = max( offset, 0)
        l = min(len(s1) - m, len(s2) - n)

        yield s1[m:m+l], s2[n:n+l]

Or, if you'd prefer some zip trickery:

def slide(s1, s2, size):
    start = size - len(s1)
    finish = len(s2) - size + 1
    for offset in xrange(start, finish):
        m = max(-offset, 0)
        n = max( offset, 0)
        yield zip(*zip(s1[m:], s2[n:]))

Here's a test:

>>> s1 = [11, 12, 13, 14, 15, 16, 17, 18]
>>> s2 = [21, 22, 23, 24, 25, 25]
>>> list(slide(s1, s2, 3))
[([16, 17, 18], [21, 22, 23]),
 ([15, 16, 17, 18], [21, 22, 23, 24]),
 ([14, 15, 16, 17, 18], [21, 22, 23, 24, 25]),
 ([13, 14, 15, 16, 17, 18], [21, 22, 23, 24, 25, 25]),
 ([12, 13, 14, 15, 16, 17], [21, 22, 23, 24, 25, 25]),
 ([11, 12, 13, 14, 15, 16], [21, 22, 23, 24, 25, 25]),
 ([11, 12, 13, 14, 15], [22, 23, 24, 25, 25]),
 ([11, 12, 13, 14], [23, 24, 25, 25]),
 ([11, 12, 13], [24, 25, 25])]

Note that this produces the pairs in reverse order.