在Python中对元素进行分组的list方法

谢谢你的回答，但我编辑了我的问题，因为它不是所有人都清楚。 我有以下元组列表：

[("ok",1),("yes",1),("no",0),("why",1),("some",1),("eat",0),("give",0),("about",0),("tell",1),("ask",0),("be",0)]

我希望：

[("ok yes","no"),("why some","eat give about"),("tell","ask be")]

谢谢大家!

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因此，我想重新组合所有1，当出现0时，我在列表中添加该值，并为下一个值创建一个新元素。

根据我的理解，以下代码应该适用于上述问题

list_tuples = [("ok",1),("yes",1),("no",0),("why",1),("some",1),("eat",0)]

tups=[]
updated_list=[]
for elem in list_tuples:
        if elem[1] == 0:
                updated_list.append(tuple([' '.join(tups), elem[0]]))
                tups=[]
        else:
                tups.append(elem[0])

print updated_list

---

一种可能的解决方案是使用

itertools.groupby

：

from operator import itemgetter
from itertools import groupby

lst = [("ok",1), ("yes",1), ("no",0), ("why",1), ("some",1), ("eat",0)]

def generate(lst):
    rv = []
    for v, g in groupby(lst, itemgetter(1)):
        if v:
            rv.append(' '.join(map(itemgetter(0), g)))
        else:
            for i in g:
                rv.append(i[0])
                yield tuple(rv)
                rv = []
    # yield last item if==1:
    if v:
        yield tuple(rv)

print([*generate(lst)])

from itertools import groupby
d = [("ok",1),("yes",1),("no",0),("why",1),("some",1),("eat",0),("give",0),("about",0),("tell",1),("ask",0),("be",0)]
new_d = [' '.join(j for j, _ in b) for _, b in groupby(d, key=lambda x:x[-1])]
result = [(new_d[i], new_d[i+1]) for i in range(0, len(new_d), 2)]

印刷品：

[('ok yes', 'no'), ('why some', 'eat')]

---

您可以使用

itertools.groupby

：

from operator import itemgetter
from itertools import groupby

lst = [("ok",1), ("yes",1), ("no",0), ("why",1), ("some",1), ("eat",0)]

def generate(lst):
    rv = []
    for v, g in groupby(lst, itemgetter(1)):
        if v:
            rv.append(' '.join(map(itemgetter(0), g)))
        else:
            for i in g:
                rv.append(i[0])
                yield tuple(rv)
                rv = []
    # yield last item if==1:
    if v:
        yield tuple(rv)

print([*generate(lst)])

from itertools import groupby
d = [("ok",1),("yes",1),("no",0),("why",1),("some",1),("eat",0),("give",0),("about",0),("tell",1),("ask",0),("be",0)]
new_d = [' '.join(j for j, _ in b) for _, b in groupby(d, key=lambda x:x[-1])]
result = [(new_d[i], new_d[i+1]) for i in range(0, len(new_d), 2)]

输出：

[('ok yes', 'no'), ('why some', 'eat give about'), ('tell', 'ask be')]

---

我认为你应该先尝试一个解决方案，然后指出你的困境。