列表中的项目以通过索引设置答案

【问题标题】：Items from List to Set by Indexing列表中的项目以通过索引设置
【发布时间】：2013-04-24 20:09:12
【问题描述】：

给定一个列表，

a_list=['chicken','pizza','burger','beer','vodka','potato','fries','mustache']

我正在尝试每六个单词组成一组新的短语...

a_set=(['chicken','pizza','burger','beer','vodka','potato'],['pizza','burger','beer','vodka','potato','fries],['burger','beer','vodka','potato','fries','mustache'])

我正在尝试通过索引来做到这一点..

index1=0
index2=6
a_set=[]
while True:
    a_set.append(a_list[index1:index2])
    index1+=1
    index2+=1
print (a_set)

我似乎无法弄清楚我做错了什么。另外，一旦循环完成并创建了所有六个单词短语并到达列表的末尾，我将如何结束循环，以便它不会从头开始并再次执行所有操作？感谢您的所有帮助。

【问题讨论】：

我没有看到任何集合。我只在您的示例中看到列表。您确定您对Python set type 不感到困惑吗？您将 a_set 设置为 list。

标签： python list python-3.x

【解决方案1】：

您正在寻找 sliding window generator 代替：

from itertools import islice

def window(seq, n=2):
    "Returns a sliding window (of width n) over data from the iterable"
    "   s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...                   "
    it = iter(seq)
    result = tuple(islice(it, n))
    if len(result) == n:
        yield result    
    for elem in it:
        result = result[1:] + (elem,)
        yield result

list(window(a_list, 6))

给出：

>>> list(window(a_list, 6)) 
[('chicken', 'pizza', 'burger', 'beer', 'vodka', 'potato'), ('pizza', 'burger', 'beer', 'vodka', 'potato', 'fries'), ('burger', 'beer', 'vodka', 'potato', 'fries', 'mustache')]

你不是在这里创建python sets，你需要小心你的术语。

具体来说，当第二个索引到达列表末尾时，您不会进行测试：

a_windows = []
index1 = 0
index2 = 6

while index2 <= len(a_list):
    a_windows.append(a_list[index1:index2])
    index1 += 1
    index2 += 1

哪个有效：

>>> a_windows
[['chicken', 'pizza', 'burger', 'beer', 'vodka', 'potato'], ['pizza', 'burger', 'beer', 'vodka', 'potato', 'fries'], ['burger', 'beer', 'vodka', 'potato', 'fries', 'mustache']]

【讨论】：

【解决方案2】：

也许这会有所帮助

def get_set(li, phrase_len):
    l = len(li)
    for i in range(l):
        if phrase_len <= l-i:
            yield(li[i:i+phrase_len])

a_list=['chicken','pizza','burger','beer','vodka','potato','fries','mustache']
print list(get_set(a_list, 6))

在此处运行代码http://codebunk.com/bunk#-IsxjI9Up1AgOfi0nuCd

【讨论】：

【解决方案3】：

如果我理解正确，您需要列表中六项的所有可能组合。 itertools.combinations 应该很容易做到这一点：

>>> import itertools
>>> a_list=['chicken','pizza','burger','beer','vodka','potato','fries','mustache']
>>> a_set=set(itertools.combinations(a_list, 6))
>>> pprint(a_set)
{('burger', 'beer', 'vodka', 'potato', 'fries', 'mustache'),
 ('chicken', 'beer', 'vodka', 'potato', 'fries', 'mustache'),
 ('chicken', 'burger', 'beer', 'potato', 'fries', 'mustache'),
 ('chicken', 'burger', 'beer', 'vodka', 'fries', 'mustache'),
 ('chicken', 'burger', 'beer', 'vodka', 'potato', 'fries'),
 ('chicken', 'burger', 'beer', 'vodka', 'potato', 'mustache'),
 ('chicken', 'burger', 'vodka', 'potato', 'fries', 'mustache'),
 ('chicken', 'pizza', 'beer', 'potato', 'fries', 'mustache'),
 ('chicken', 'pizza', 'beer', 'vodka', 'fries', 'mustache'),
 ('chicken', 'pizza', 'beer', 'vodka', 'potato', 'fries'),
 ('chicken', 'pizza', 'beer', 'vodka', 'potato', 'mustache'),
 ('chicken', 'pizza', 'burger', 'beer', 'fries', 'mustache'),
 ('chicken', 'pizza', 'burger', 'beer', 'potato', 'fries'),
 ('chicken', 'pizza', 'burger', 'beer', 'potato', 'mustache'),
 ('chicken', 'pizza', 'burger', 'beer', 'vodka', 'fries'),
 ('chicken', 'pizza', 'burger', 'beer', 'vodka', 'mustache'),
 ('chicken', 'pizza', 'burger', 'beer', 'vodka', 'potato'),
 ('chicken', 'pizza', 'burger', 'potato', 'fries', 'mustache'),
 ('chicken', 'pizza', 'burger', 'vodka', 'fries', 'mustache'),
 ('chicken', 'pizza', 'burger', 'vodka', 'potato', 'fries'),
 ('chicken', 'pizza', 'burger', 'vodka', 'potato', 'mustache'),
 ('chicken', 'pizza', 'vodka', 'potato', 'fries', 'mustache'),
 ('pizza', 'beer', 'vodka', 'potato', 'fries', 'mustache'),
 ('pizza', 'burger', 'beer', 'potato', 'fries', 'mustache'),
 ('pizza', 'burger', 'beer', 'vodka', 'fries', 'mustache'),
 ('pizza', 'burger', 'beer', 'vodka', 'potato', 'fries'),
 ('pizza', 'burger', 'beer', 'vodka', 'potato', 'mustache'),
 ('pizza', 'burger', 'vodka', 'potato', 'fries', 'mustache')}

【讨论】：