Python，有规律的空白列表

Question

I want to create a python list with regular gaps. I have a general idea of how I can do it, but is there a short way to do it or an inbuilt function to do this.

Suppose I want to create a list from 1 to 200, with gaps of 50. The list would be [1,..,50,101,..,150], ie the sequence from 51 to 100 and 151 to 200 is not in the list.

def create_list_step(start,end,step):
     result = []
     current = start
     while current < end:
          smallinterval = current + step
          result = result + list(range(current, smallinterval))
          current = smallinterval + step
    return result

Answer 1

我不认为有一个内置函数，但这可以使用range的串联轻松完成：

final = list(range(1, 51)) + list(range(101, 151))

Answer 2

One way is to define a function which skips the gaps for you.

from itertools import chain

def gappy(start, rng, n):
    return chain(*(list(range(2*i*rng+1, (2*i+1)*rng+1)) for i in range(n)))

list(gappy(1, 50, 3))

# [1, 2, 3, ..., 48, 49, 50, 101, 102, 103, ...,
#  148, 149, 150, 201, 202, 203, ..., 248, 249, 250]

Answer 3

You could use list comprehensions:

start = 5
stop = 17
gap = 3

a = [x for x in range(start, stop + 1) if ((x - start) // gap) % 2 == 0]

Answer 4

This is one-liner that should do it:

[i for i in range(200) if (i-1)/50%2==0] #Python 2.x
[i for i in range(200) if (i-1)//50%2==0] #Python 3.x