将列表拆分为嵌套列表

Question

我有一个嵌套列表，我正试图将其拆分为嵌套在原始列表中的较小的列表。我要用一个特定的元素来划分列表。我已经知道了这个部分，但是当我试图迭代原始列表时，却只返回了最后一个列表。

animals  = [["animal", "cat", "dog", "bird", "animal", "cat", "snake", "bird"], 
     ["animal", "cat", "iguana", "bird", "animal", "lizard", "dog", "bird"]]

for n in animals:
    it  = iter(n)
    combined = [[next(it)]]
    for ele in it:
        if ele != "animal":
            combined[-1].append(ele)
        else:
            combined.append([ele])
    
print combined

输出：

[例][“动物”，“猫”，“蜥蜴”，“鸟”，“动物”，“蜥蜴”，“狗”，“鸟”)

期望产出：

[“动物”、“猫”、“狗”、“鸟”、“动物”、“猫”、“蛇”、“鸟”、“动物”、“猫”、“蜥蜴”、“鸟”、“动物”、“蜥蜴”、“狗”、“鸟”)

Answer 1

试试这个丙酮溶液：

str_ = ' '.join([a for animal in animals for a in animal])
list_of_animals = [['animal']+st.split() for st in str_.split("animal") if len(st)>1]

它提供了所需的输出。

Answer 2

animals  = [["animal", "cat", "dog", "bird", "animal", "cat", "snake", "bird"], ["animal", "cat", "iguana", "bird", "animal", "lizard", "dog", "bird"]]

result = []
for row in animals:
    result_row = []
    for element in row:
        if element == "animal":
            if result_row:
                result.append(result_row)
            result_row = []
        result_row.append(element)
    result.append(result_row)
print(result)

Answer 3

由于animals的结构可能是不规则的，这可能是一个解决方案：

from itertools import groupby

combined = [
    ['animal'] + list(group)
    for sublist in animals
    for key, group in groupby(sublist, key=lambda s: s == 'animal') if not key
]

输出：

[['animal', 'cat', 'dog', 'bird'],
 ['animal', 'cat', 'snake', 'bird'],
 ['animal', 'cat', 'iguana', 'bird'],
 ['animal', 'lizard', 'dog', 'bird']]