OGlist = [A, B, C, D, E, 1, 2, 3, 4, 5, F, G, H, I, J, 6, 7, 8, 9, 10]
_list = []
_list2 = []所以,我有一个名单OGlist..。我希望将OGlist的前5个元素转换为_list,将OGlist的第二个5个元素转化为_list2,将第三个5个元素转化为_list,将第4个5个元素转化为_list2等等。
我怎样才能做到这一点?
我试过这样做:
for x in range(1, len(OGlist) + 1):
_list.append(OGlist[x-1])
if x%5 == 0:
y = x
while True:
_list2.append(OGlist[x])
x += 1
if x == y + 5:
break我应该使用什么条件和逻辑来获得所需的输出?
期望产出:
_list = [A, B, C, D, E, F, G, H, I, J]
_list2 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]发布于 2020-12-25 07:53:48
我会这样对待:
from itertools import chain
_list = list(chain.from_iterable(OGlist[i:i+5] for i in range(0, len(OGlist), 10)))
_list2 = list(chain.from_iterable(OGlist[i:i+5] for i in range(5, len(OGlist), 10)))OGlist[i:i+5] for i in range(0, len(OGlist), 10)返回类似于[[0, 1, 2, 3, 4], [10, 11, 12, 13, 14], ...]的序列,chain.from_iterable将该列表连接到一个列表中。
发布于 2020-12-25 07:49:59
如果您正在寻找一个线,这应该可以做到。
OGlist = [x, y, foo, bar, etc]
list_of_lists = [OGlist[i:i+5] for i in range(len(OGlist)-5) if i%5==0]
##GETTING LISTS AT EVEN INDEXES
_list = [l_ for i,l_ in enumerate(list_of_lists) if i%2 == 0]
##GETTING LISTS AT ODD INDEXES
_list2 = [l_ for i,l_ in enumerate(list_of_lists) if i%2 == 1]
##FLATTENING
from itertools import chain
_list = list(chain.from_iterable(_list))
_list2 = list(chain.from_iterable(_list2))它遍历列表中的每5个元素(使用OGlisti:i+5) (由i%5==0强制执行),直到达到少于5个元素(由len(OGlist) -5强制执行)为止。
测试:
╰─$ python 25ms
Python 3.8.3 (default, Jul 2 2020, 11:26:31)
[Clang 10.0.0 ] :: Anaconda, Inc. on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> OGlist = list(range(100))
>>> OGlist
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99]
>>> list_of_lists = [OGlist[i:i+5] for i in range(len(OGlist)-5) if i%5==0]
>>> list_of_lists
[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13, 14], [15, 16, 17, 18, 19], [20, 21, 22, 23, 24], [25, 26, 27, 28, 29], [30, 31, 32, 33, 34], [35, 36, 37, 38, 39], [40, 41, 42, 43, 44], [45, 46, 47, 48, 49], [50, 51, 52, 53, 54], [55, 56, 57, 58, 59], [60, 61, 62, 63, 64], [65, 66, 67, 68, 69], [70, 71, 72, 73, 74], [75, 76, 77, 78, 79], [80, 81, 82, 83, 84], [85, 86, 87, 88, 89], [90, 91, 92, 93, 94]]
>>> _list = [val for val in [l_ for i,l_ in enumerate(list_of_lists) if i%2 == 0]]
>>> _list2 = [val for val in [l_ for i,l_ in enumerate(list_of_lists) if i%2 == 1]]
>>> _list
[[0, 1, 2, 3, 4], [10, 11, 12, 13, 14], [20, 21, 22, 23, 24], [30, 31, 32, 33, 34], [40, 41, 42, 43, 44], [50, 51, 52, 53, 54], [60, 61, 62, 63, 64], [70, 71, 72, 73, 74], [80, 81, 82, 83, 84], [90, 91, 92, 93, 94]]
>>> _list2
[[5, 6, 7, 8, 9], [15, 16, 17, 18, 19], [25, 26, 27, 28, 29], [35, 36, 37, 38, 39], [45, 46, 47, 48, 49], [55, 56, 57, 58, 59], [65, 66, 67, 68, 69], [75, 76, 77, 78, 79], [85, 86, 87, 88, 89]]
>>> from itertools import chain
>>> _list = list(chain.from_iterable(_list))
>>> _list2 = list(chain.from_iterable(_list2))
>>> _list
[0, 1, 2, 3, 4, 10, 11, 12, 13, 14, 20, 21, 22, 23, 24, 30, 31, 32, 33, 34, 40, 41, 42, 43, 44, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 70, 71, 72, 73, 74, 80, 81, 82, 83, 84, 90, 91, 92, 93, 94]
>>> _list2
[5, 6, 7, 8, 9, 15, 16, 17, 18, 19, 25, 26, 27, 28, 29, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 55, 56, 57, 58, 59, 65, 66, 67, 68, 69, 75, 76, 77, 78, 79, 85, 86, 87, 88, 89]
>>>发布于 2020-12-25 08:21:16
这是我的密码
import numpy as np
size_array = 500 # example
big_list = [x for x in range(size_array)]
half_list1 = []
half_list2 = []
idx = [x for x in range(0,int(size_array/5), 2)]
idx = np.repeat(idx, 5)*5 + np.tile([1, 2, 3, 4, 5], int(size_array/10))
half_list1 = [big_list[i] for i in idx]
half_list2 = [big_list[i+5] for i in idx]https://stackoverflow.com/questions/65446180
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