python list分成多个_根据Python中的一组索引将列表分成多个部分

What is the best way to split a list into parts based on an arbitrary number of indexes? E.g. given the code below

indexes = [5, 12, 17]

list = range(20)

return something like this

part1 = list[:5]

part2 = list[5:12]

part3 = list[12:17]

part4 = list[17:]

If there are no indexes it should return the entire list.

解决方案

This is the simplest and most pythonic solution I can think of:

def partition(alist, indices):

return [alist[i:j] for i, j in zip([0]+indices, indices+[None])]

if the inputs are very large, then the iterators solution should be more convenient:

from itertools import izip, chain

def partition(alist, indices):

pairs = izip(chain([0], indices), chain(indices, [None]))

return (alist[i:j] for i, j in pairs)

and of course, the very, very lazy guy solution (if you don't mind to get arrays instead of lists, but anyway you can always revert them to lists):

import numpy

partition = numpy.split