使用索引Python手动切片列表

Question

Minimal example

I have a list a = [10,20,30,40,50,60,70,80,90,100,110,120,130,140,150,....,]

I want to get a new list new_list = [40,50,60,100,110,120,...] , ie append fourth, fifth and sixth value, skip next three, append next three and so on.

My idea is to create a list called index :

index = [3,4,5,9,10,11,...] new_list = [a[i] for i in index] # This should give me what I want

but how do I create the list index ? I know np.arange has the step option, but that is only for spacing between values.

Answer 1

Here's one way -

[a[i] for i in range(len(a)) if i%6>=3]

Sample run -

In [49]: a = [10,20,30,40,50,60,70,80,90,100,110,120,130,140,150]

In [50]: [a[i] for i in range(len(a)) if i%6>=3]
Out[50]: [40, 50, 60, 100, 110, 120]

Answer 2

Here's an improved and faster version using Python built-in function enumerate building up on Divakar's nice logic .

In [4]: lst = [10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150]
In [6]: [item for idx, item in enumerate(lst) if idx%6 >= 3]
Out[6]: [40, 50, 60, 100, 110, 120]

---

why is this version better & preferable?

In [10]: lst = range(10, 100000, 10)       

In [11]: %timeit [lst[idx] for idx in range(len(lst)) if idx % 6 >= 3]
1.1 ms ± 22.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [12]: %timeit [item for idx, item in enumerate(lst) if idx % 6 >= 3]
788 µs ± 8.67 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

That's more than 300 microseconds gain! Furthermore, enumerate() is more straightforward and intuitive (cf loop like a native )

Answer 3

You can generate the index elements of continuous 3 increment and repeat of 3 elements

a = np.asarray([10,20,30,40,50,60,70,80,90,100,110,120,130,140,150])
b = np.tile(np.arange(1,4),int(len(a)/6)+1) + np.repeat(np.arange(3,int(len(a)/2)+3,3),3)
a.take(b)

Out:

array([ 50,  60,  70,  80,  90, 100, 110, 120, 130])

Explanation

np.tile(np.arange(1,4),int(len(a)/6)+1)
#array([1, 2, 3, 1, 2, 3, 1, 2, 3])

np.repeat(np.arange(3,int(len(a)/2)+3,3),3)
#array([3, 3, 3, 6, 6, 6, 9, 9, 9])