通过元组分隔符拆分列表 [英] Split list by tuple separator

问题描述

我有清单:

print (L)
[('I', 'WW'), ('am', 'XX'), ('newbie', 'YY'), ('.', 'ZZ'), 
 ('You', 'WW'), ('are', 'XX'), ('cool', 'YY'), ('.', 'ZZ')]

我想用分隔符('.', 'ZZ')将列表拆分为子列表:

I want split list to sublists with separator ('.', 'ZZ'):

print (new_L)
[[('I', 'WW'), ('am', 'XX'), ('newbie', 'YY'), ('.', 'ZZ')], 
 [('You', 'WW'), ('are', 'XX'), ('cool', 'YY'), ('.', 'ZZ')]]

我对另一种可能的解决方案很感兴趣，性能很重要.

I am interested about another possible solutions, performance is important.

推荐答案

for循环方法将更快，这仅需一次通过:

The for-loop approach will be faster, this requires only one-pass:

>>> def juan(L, sep):
...     L2 = []
...     sub = []
...     for x in L:
...         sub.append(x)
...         if x == sep:
...             L2.append(sub)
...             sub = []
...     if sub:
...         L2.append(sub)
...     return L2
...
>>> juan(L, sep)
[[('I', 'WW'), ('am', 'XX'), ('newbie', 'YY'), ('.', 'ZZ')], [('You', 'WW'), ('are', 'XX'), ('cool', 'YY'), ('.', 'ZZ')]]

一些比较:

>>> def jezrael(L, sub):
...     return [list(g) + [sep] for k, g in groupby(L, lambda x: x==sep) if not k]
...
>>> def coldspeed(L, sep):
...     L2 = []
...     for i in reversed(L):
...         if i == sep:
...             L2.append([])
...         L2[-1].append(i)
...     return [x[::-1] for x in reversed(L2)]
...
>>> def pm2ring(L, sep):
...     seplist = [sep]
...     return [list(g) + seplist for k, g in groupby(L, sep.__eq__) if not k]
...
>>> setup = "from __main__ import L, sep, juan, coldspeed, pm2ring, jezrael"

更多时间

>>> def buzzycoder(L, sep):
...     a = []
...     length = len(L)
...     start = 0
...     end = L.index(sep)
...     if start < length: a.append(L[start:end+1])
...     start = end + 1
...     while start < length:
...         end = L.index(sep, start) + 1
...         a.append(L[start:end])
...         start = end
...     return a
...

>>> def splitList(l, s):
...     ''' l is list, s is separator, simular to split, but keep separator'''
...     i = 0
...     for _ in range(l.count(s)): # break using slices
...         e = l.index(s,i)
...         yield l[i:e+1] # sublist generator value
...         i = e+1
...     if e+1 < len(l): yield l[e+1:] # pick up
...

>>> def bharath(x,sep):
...     n = [0] + [i+1 for i,j in enumerate(x) if j == sep]
...     m= list()
...     for first, last in zip(n, n[1:]):
...         m.append(x[first:last])
...     return m
...

结果:

>>> timeit.timeit("jezrael(L, sep)", setup)
4.1499102029483765
>>> timeit.timeit("pm2ring(L, sep)", setup)
3.3499899921007454
>>> timeit.timeit("coldspeed(L, sep)", setup)
2.868469718960114
>>> timeit.timeit("juan(L, sep)", setup)
1.5428746730322018
>>> timeit.timeit("buzzycoder(L, sep)", setup)
1.5942967369919643
>>> timeit.timeit("list(splitList(L, sep))", setup)
2.7872562300181016
>>> timeit.timeit("bharath(L, sep)", setup)
2.9842335029970855

列表更大:

>>> L = L*100000
>>> timeit.timeit("jezrael(L, sep)", setup, number=10)
3.3555950550362468
>>> timeit.timeit("pm2ring(L, sep)", setup, number=10)
2.337177241919562
>>> timeit.timeit("coldspeed(L, sep)", setup, number=10)
2.2037084710318595
>>> timeit.timeit("juan(L, sep)", setup, number=10)
1.3625159269431606
>>> timeit.timeit("buzzycoder(L, sep)", setup, number=10)
1.4375156159512699
>>> timeit.timeit("list(splitList(L, sep))", setup, number=10)
1.6824725979240611
>>> timeit.timeit("bharath(L, sep)", setup, number=10)
1.5603888860205188

注意事项

鉴于L中sep的比例，结果无法解决性能问题，这对于其中某些解决方案的时序会产生很大影响.

Caveat

The results do not address performance given the proportion of sep in L, which will affect timings a lot for some of these solutions.

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