重复列表Python中的元素

Question

假设我有一个字符串列表:

a = ['a', 'a', 'b', 'c', 'c', 'c', 'd']

我想列出连续出现至少两次的项目列表:

result = ['a', 'c']

我知道我必须使用for循环,但我无法弄清楚如何定位连续重复的项目.我怎么能这样做？

编辑:如果相同的项目在一次重复两次怎么办？然后设置功能将无效

a = ['a', 'b', 'a', 'a', 'c', 'a', 'a', 'a', 'd', 'd']
result = ['a', 'a', 'd']

Answer 1

试试itertools.groupby()这里:

>>> from itertools import groupby,islice
>>> a = ['a', 'a', 'b', 'c', 'c', 'c', 'b']

>>> [list(g) for k,g in groupby(a)]
[['a', 'a'], ['b'], ['c', 'c', 'c'], ['b']] 

>>> [k for k,g in groupby(a) if len(list(g))>=2]
['a', 'c']

使用islice():

>>> [k for k,g in groupby(a) if len(list(islice(g,0,2)))==2]
>>> ['a', 'c']

使用zip()和izip():

In [198]: set(x[0] for x in izip(a,a[1:]) if x[0]==x[1])
Out[198]: set(['a', 'c'])

In [199]: set(x[0] for x in zip(a,a[1:]) if x[0]==x[1])
Out[199]: set(['a', 'c'])

timeit 结果:

from itertools import *

a='aaaabbbccccddddefgggghhhhhiiiiiijjjkkklllmnooooooppppppppqqqqqqsssstuuvv'

def grp_isl():
    [k for k,g in groupby(a) if len(list(islice(g,0,2)))==2]

def grpby():
    [k for k,g in groupby(a) if len(list(g))>=2]

def chn():
    set(x[1] for x in chain(izip(*([iter(a)] * 2)), izip(*([iter(a[1:])] * 2))) if x[0] == x[1])

def dread():
    set(a[i] for i in range(1, len(a)) if a[i] == a[i-1])

def xdread():
    set(a[i] for i in xrange(1, len(a)) if a[i] == a[i-1])

def inrow():
    inRow = []
    last = None
    for x in a:
        if last == x and (len(inRow) == 0 or inRow[-1] != x):
            inRow.append(last)
        last = x

def zipp():
    set(x[0] for x in zip(a,a[1:]) if x[0]==x[1])

def izipp():
    set(x[0] for x in izip(a,a[1:]) if x[0]==x[1])

if __name__=="__main__":
    import timeit
    print "islice",timeit.timeit("grp_isl()", setup="from __main__ import grp_isl")
    print "grpby",timeit.timeit("grpby()", setup="from __main__ import grpby")
    print "dread",timeit.timeit("dread()", setup="from __main__ import dread")
    print "xdread",timeit.timeit("xdread()", setup="from __main__ import xdread")
    print "chain",timeit.timeit("chn()", setup="from __main__ import chn")
    print "inrow",timeit.timeit("inrow()", setup="from __main__ import inrow")
    print "zip",timeit.timeit("zipp()", setup="from __main__ import zipp")
    print "izip",timeit.timeit("izipp()", setup="from __main__ import izipp")

输出:

islice 39.9123107277
grpby 30.1204478987
dread 17.8041124706
xdread 15.3691785568
chain 17.4777339702
inrow 11.8577565327           
zip 16.6348844045
izip 15.1468557105

结论:

与其他替代方案相比,Poke的解决方案是最快的解决方案.

Answer 2

这听起来像是家庭作业,所以我只想概述一下我会做什么:

1. 迭代a,但保持每个元素的索引在变量中.enumerate()会有用的.
2. 在for循环内部,while从当前项的索引开始循环.
3. 只要下一个元素与上一个元素(或原始元素)相同,就重复循环.break在这里很有用.
4. 计算循环重复的次数(为此需要一些计数器变量).
5. result如果您的计数器变量为>=2,则将项目附加到您的项目.