25

我正在使用 python 3.2.3 IDLE，这是我的代码：

originalList = [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, 100]

newList = orginalList[0.05:0.95] #<<<<I have no idea what I'm doing here
print (newList)

我有一个原始数字列表，它们是 1 - 100，我想从原始列表中创建一个新列表，但是新列表只能包含属于原始列表 5%- 95% 子范围的数据

所以新的列表一定是这样的[5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18....95]。我怎么做？我知道我的 newList 代码是错误的

4

7 回答 7

36 
originalList.sort()
newList = originalList[int(len(originalList) * .05) : int(len(originalList) * .95)]
于 2012-11-16T20:18:01.173 回答
4 
size = len(originalList)
newList = originalList[0.05*size - 1:0.95*size + 1]
于 2012-11-16T20:19:30.970 回答
4 
sl = slice(4, 95)
print(originalList[sl])

另见http://docs.python.org/2/library/functions.html#slice

于 2012-11-16T20:20:44.383 回答
1

我还会使用列表推导来创建原始列表......不太容易出错。

originalList = range(1,101)
newList = originalList[(len(originalList)*.05)-1:len(originalList)*.95]
print newList

给出想要的结果...

编辑：根据下面的评论将范围更改为更简洁。

于 2012-11-16T20:33:29.763 回答
0

如果要获取列表的一部分，则语法为

List = [1,2,3,4,5,6,7,8,9,10]
newList = [*start index*:*Index to end AT*]

因此，第一个数字是子列表开始的索引，而第二个数字是子列表停止的索引（不包括该索引）。

希望这可以帮助！

于 2012-11-16T20:18:29.600 回答
0

对于任意长度的列表，您可以执行以下操作：

>>> l = range(200)
>>> percentage = 5
>>> skip = int(len(l) * (float(percentage) / 100) / 2)
>>> len(l[skip:-skip])
190
于 2016-10-16T19:18:11.507 回答
0

您可以使用该fidx模块，它允许百分比作为索引：

import fidx

originalList = fidx([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, 100])
# or better: originalList = fidx.list(range(1,101))

newList = originalList[0.05:0.95]

print (newList)

返回

[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]
于 2020-10-10T22:04:30.650 回答