Python - 根据计算的百分比返回列表的区域

Question

I have a fairly long list in python which I want to return parts of it in a single, shorter list based on 3 percentages. If the first percentage was 30%, it would need to return the first 30% of the values and if the 3rd percentage was 50% it would need to return the last half of the values of my long list.

This is what I have so far, it has issues with rounding and is an ugly solution

  class OM():
    def __init__(self,name):
      self.name = name
      self.total = 120
      self.a = 21
      self.b = 34
      self.c = 65

    def hRange(self,action):
      if self.total > 0:
        a_perc = int(self.a / float(self.total) *169)
        b_perc = int(self.b / float(self.total) *169)
        c_perc = int(self.c / float(self.total) *169)
        if action=='a': return lst[:aperc]
        elif action=='b': return lst[a_perc:a_perc+b_perc]
        elif action=='c': return lst[-c_perc:]

      else:
        raise Exception

I realise this isn't well coded at all, (hard coded the lst length as 169, doesn't catch different actions etc etc) I just wanted to help explain what I was trying to do.

In my actual implementation the values total,a,b,c would be initialised at 0 and another method would update them, as they're just tallys. I just set them as some random values here so the code returns percentages.

I would be hugely appreciative if anyone could give me any advice on how to go about doing this a better way.

Answer 1

If I understand correctly, you want to keep three lists, based on a certain percentage at the start and end plus the "in between":

def split_list(mylist, a, b):
    l = len(mylist)
    return ( mylist[:int(a*l/100)], 
             mylist[int(a*l/100):-int(b*l/100)],
             mylist[-int(b*l/100):] )

where a and b are the percentages (given from 0 to 100). It's nearly the same implementation as yours, but you only need to use two percentages (if the middle list is the remaining part when both ends are trimmed off). I will leave it up to you to incorporate this in your code...

Answer 2

Got carried away with a general solution... I belive this should handle all cases:

def split_list(mylist, *args):
    ilist = map(lambda p : int(p * len(mylist) / 100.0), args) + [len(mylist)]
    return reduce(lambda l, v : [l[0] + [mylist[l[1]:v]], v], ilist, [[],0])[0]

The function takes a list and a list of percentage split points (in order) and returns you the lists as you need them. Handles cases where there are more percentages of splits than items in the list.