Question

我想在列表中找到连续且相同的元素：

a = [1, 1, 1, 2, 3, 2, 2, 2, 1, 2, 3, 4, 1, 1, 1, 5, 5, 5]

new_list_1 = []
new_list_2 = []

def hello(x):
    for j, i in enumerate(x):
        try:
            if x[j] == x[j + 1] or x[j] == x[j-1]:
                new_list_1.append((i, j))
            else:
                new_list_2.append((i, j))
        except IndexError:
            if x[j]==x[j-1]:
                new_list_1.append((i, j))

print(hello(a))
print(new_list_1)

它正在回归：

[(1, 0), (1, 1), (1, 2), (2, 5), (2, 6), (2, 7), (1, 12), (1, 13), (1, 14), (5, 15), (5, 16), (5, 17)]

但我想要这样的事情：

[[(1, 0), (1, 1), (1, 2)], [(2, 5), (2, 6), (2, 7)], [(1, 12), (1, 13), (1, 14)], [(5, 15), (5, 16), (5, 17)]]

我不想使用任何外部模块，例如 itertools 'chain或groupby。我怎样才能做到这一点？

Answer 1

这是一种基于generator的方法：

def indexed_groups(lst):
    ret_val, crnt = [], None
    for i, x in enumerate(lst):
        if x != crnt:  # for every new item
            if len(ret_val) > 1:  # check if there is a group to yield
                yield ret_val
            ret_val, crnt = [], x  # reset group and current
        ret_val.append((x, i))  # collect (item, index) pairs
    if len(ret_val) > 1:  # check last group
        yield ret_val

>>> list(indexed_groups(a))
[[(1, 0), (1, 1), (1, 2)], 
 [(2, 5), (2, 6), (2, 7)], 
 [(1, 12), (1, 13), (1, 14)], 
 [(5, 15), (5, 16), (5, 17)]]

如何在嵌套列表中获取本地范围的项目？

1 个答案: