使用不一致的间隔切片Python列表。

Question

我有一家公司的股票价格清单。现在我想将列表分割成多个区间。 我们将像这样存储价格：前2个元素，然后是接下来的3个元素，然后是2个元素，以此类推。

meta_stocks = [10, 9, 11, 15, 19, 22, 25, 11, 15, 17]

输出

meta_stocks = [[10, 9],[11, 15, 19],[22, 25],[ 11, 15, 17]]

我可以将列表每5个项目拆分一次，但无法进一步拆分。

>>> [meta_stocks[i:i+interval2] for i in range(0, len(meta_stocks), interval2)]
>>> [[10, 9, 11, 15, 19], [22, 25, 11, 15, 17]]

- Tarun K

---

你可以将列表分成每个包含5个元素的小列表，然后按照你想要的方式将它们拆分为3+2或其他组合。 - MSH

这就是我一直在尝试的，但是无法弄清楚。 - Tarun K

Answer 1

你可以使用 itertools.cycle 和列表推导式完成以下操作：

meta_stocks = [10, 9, 11, 15, 19, 22, 25, 11, 15, 17]

from itertools import cycle

start = 0
l = [2,3]
c = cycle(l)

[meta_stocks[start:(start:=start+next(c))]
 for i in range(len(l)*len(meta_stocks)//sum(l))]

输出:

[[10, 9], [11, 15, 19], [22, 25], [11, 15, 17]]

- mozway

Answer 2

您可以这样做，无需使用其他导入：

meta_stocks = [10, 9, 11, 15, 19, 22, 25, 11, 15, 17]
meta_stocks_out = []
offset = 0
interval = 2
while offset + interval <= len(meta_stocks):
    meta_stocks_out.append(meta_stocks[offset:offset+interval])
    offset += interval
    interval = 2 if interval == 3 else 3
print(meta_stocks_out)

- user2668284

Answer 3

一些itertools来拯救:

from itertools import islice, cycle, takewhile

i = iter(meta_stocks)
intervals = [2, 3]

[*takewhile(lambda _: _, ([*islice(i, n)] for n in cycle(intervals)))]
# [[10, 9], [11, 15, 19], [22, 25], [11, 15, 17]]

- user2390182

Answer 4

使用itertools.islice

from itertools import islice

meta_stocks = [10, 9, 11, 15, 19, 22, 25, 11, 15, 17]
chunks = [2, 3, 2, 3                 # length of each sublist

it = iter(meta_stocks)               
result = [list(islice(it, i)) for i in chunks]

print(result)
#Output: [[10, 9], [11, 15, 19], [22, 25], [11, 15, 17]]

- DarrylG

Answer 5

你可以尝试这个方法。首先，我们获取一个列表，其中每个元素都是包含两个子列表的列表，分别包含2个和3个元素。然后，我们将列表的第一层展开。

import pprint
meta_stocks = [10, 9, 11, 15, 19, 22, 25, 11, 15, 17]
new_list = [item for sublist in [[meta_stocks[i:i+2], meta_stocks[i+2:i+5]] for i in range(0, len(meta_stocks), 5)] for item in sublist]
pprint.pprint(new_list)

以及输出

[[10, 9], [11, 15, 19], [22, 25], [11, 15, 17]]

- Marcel Preda

Answer 6

尝试：

meta_stocks = [10, 9, 11, 15, 19, 22, 25, 11, 15, 17]

res = [x for i in range(0, len(meta_stocks), 5) for x in [meta_stocks[i:i+2], meta_stocks[i+2:i+5]]]
print(res)

输出

[[10, 9], [11, 15, 19], [22, 25], [11, 15, 17]]

使用 itertools.accumulate 的一般方法：

from itertools import accumulate

meta_stocks = [10, 9, 11, 15, 19, 22, 25, 11, 15, 17]

step = sum([2, 3])
steps = list(accumulate([0, 2, 3]))

res = [x for i in range(0, len(meta_stocks), 5) for x in [meta_stocks[i+s:i+e] for s, e in zip(steps, steps[1:])]]
print(res)

                                                                                                       

- Dani Mesejo

Answer 7

假设你已经把区间表示成一个整数列表

meta_stocks = [10, 9, 11, 15, 19, 22, 25, 11, 15, 17]
intervals = [2,3,2,3]

cintervals = [sum(intervals[:i]) for i in range(0,len(intervals)+1)]

sliced = [meta_stocks[cintervals[i-1]:cintervals[i]] for i in range(1, len(cintervals))]

- nikeros