related lists mean value

[dimitri pater - serpia]

thanks Chris and MRAB!
Looks good, I'll try it out

On Tue, Mar 9, 2010 at 12:22 AM, Chris Rebert <clp2 at rebertia.com> wrote:
> On Mon, Mar 8, 2010 at 2:34 PM, dimitri pater - serpia
> <dimitri.pater at gmail.com> wrote:
>> Hi,
>>
>> I have two related lists:
>> x = [1 ,2, 8, 5, 0, 7]
>> y = ['a', 'a', 'b', 'c', 'c', 'c' ]
>>
>> what I need is a list representing the mean value of 'a', 'b' and 'c'
>> while maintaining the number of items (len):
>> w = [1.5, 1.5, 8, 4, 4, 4]
>>
>> I have looked at iter(tools) and next(), but that did not help me. I'm
>> a bit stuck here, so your help is appreciated!
>
> from __future__ import division
>
> def group(keys, values):
>    #requires None not in keys
>    groups = []
>    cur_key = None
>    cur_vals = None
>    for key, val in zip(keys, values):
>        if key != cur_key:
>            if cur_key is not None:
>                groups.append((cur_key, cur_vals))
>            cur_vals = [val]
>            cur_key = key
>        else:
>            cur_vals.append(val)
>    groups.append((cur_key, cur_vals))
>    return groups
>
> def average(lst):
>    return sum(lst) / len(lst)
>
> def process(x, y):
>    result = []
>    for key, vals in group(y, x):
>        avg = average(vals)
>        for i in xrange(len(vals)):
>            result.append(avg)
>    return result
>
> x = [1 ,2, 8, 5, 0, 7]
> y = ['a', 'a', 'b', 'c', 'c', 'c' ]
>
> print process(x, y)
> #=> [1.5, 1.5, 8.0, 4.0, 4.0, 4.0]
>
> It could be tweaked to use itertools.groupby(), but it would probably
> be less efficient/clear.
>
> Cheers,
> Chris
> --
> http://blog.rebertia.com
>



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