Question

Access multiple elements of list knowing their index [duplicate] (11 answers)

Closed 3 years ago.

如果你在座标上有一个名单,并希望将指数1、2和5的内容列入新的名单,你将如何这样做?

这是我是如何做的,但我不满意:

>>> a
[10, 11, 12, 13, 14, 15]
>>> [x[1] for x in enumerate(a) if x[0] in [1,2,5]]
[11, 12, 15]

是否有更好的办法?

更笼统地说,鉴于指数的图,你如何利用本图从清单中提取相应内容,甚至重复(例如,图<代码>(1,1,2,1,5)生成<代码>[11,11,12,11,15]。

Answer 1

或许可以这样做:

[a[i] for i in (1,2,5)]
# [11, 12, 15]

Answer 2

我认为,你重新审视了这一问题:

elements = [10, 11, 12, 13, 14, 15]
indices = (1,1,2,1,5)

result_list = [elements[i] for i in indices]

Answer 3

提出

numbers = range(10, 16)
indices = (1, 1, 2, 1, 5)

result = [numbers[i] for i in indices]

Answer 4

使用Numpy直接阵列索引,如MATLAB, Julia,......

a = [10, 11, 12, 13, 14, 15];
s = [1, 2, 5] ;

import numpy as np
list(np.array(a)[s])
# [11, 12, 15]

更糟糕的是,只是停留在Numpy阵列。

a = np.array([10, 11, 12, 13, 14, 15])
a[s]
#array([11, 12, 15])

Answer 5

检查:

 [a[index] for index in (1,2,5,20) if 0 <= index < len(a)]
 # [11, 12, 15]