如何获得列表中两个相同元素之间的连续元素？

Question

For example I have a list l = ['.','b','w','w','w','b','.','.'] , how would I traverse this list and return the indices of the three w's that are between the b's.

I need to be able to handle cases such as l = ['.','b','w','w','w','w','.','.'] , where there is no second 'b', in this case nothing would be returned. Other cases that could be encountered are l = ['.','b','w','w','.','.','b','.'] , in this case nothing would be returned either. I need to be able to get the consecutive w's between the b's. What I have tried so far is this:

l = ['.','b','w','w','.','.','.','.']
q = []
loop = False
for i in range(len(l)):
    if l[i] == 'b' and l[i+1] == 'w':
        loop = True
    elif l[i] == 'w' and l[i+1] == 'w' and l[i+2] == '.':
        pass
    elif l[i] == 'w' and l[i-1] == 'b' and l[i+1] == 'w' and loop == True:
        q.append(i)
    elif l[i-1] == 'w' and l[i] == 'w' and l[i+1] == 'w' and loop == True:               q.append(i)
    elif l[i] == 'w' and l[i+1] == 'b' and loop == True:
        q.append(i)
        loop = False
        break

print(q)

But this does not handle all cases and sometimes returns errors. How could I go about doing this without numpy ?

Answer 1

try this

import re
l = ['.','b','w','w','b','.','.','.']
s = ''.join(l)

# print(re.findall('[b]([w]+)[b]',s))
# print(re.findall('[w]([b]+)[w]',s))

if(re.findall('[b]([w]+)[b]',s)):
    witer =re.finditer('w',s)
    wpos = []
    for i in witer:
        wpos.append(i.start())
    print('reverse', wpos)


if(re.findall('[w]([b]+)[w]',s)):
    biter =re.finditer('b',s)
    bpos = []
    for i in biter:
        bpos.append(i.start())
    print('reverse', bpos)

Answer 2

Try this:

ind = [i for i, e in enumerate(l) if e == 'b']
q = []

if len(ind) == 2:
    w = l[ind[0]+1: ind[1]]
    if set(w) == {'w'}:
        for i in range(ind[0]+1, ind[1]):        
            q.append(i)