Sorting and list comprehension

[Alex Martelli]

Tran Tuan Anh <anhtt at hotmail.com> wrote:

> Hi all,
> 
> I would like to do the followings, but I am quite new to Python hence
> have not figured it out how...
> 
> salary = dictionary
> salary["Bob"] = 11
> salary["Marry"] = 4
> salary["me"]= 45
> 
> How can I have a sort_value() function acts like this:
> result = sortvalue(salary)
> result = ["Marry","Bob","me"]

You could upgrade to Python 2.4 and define

def sortvalue(d):
    return sorted(d, key=d.get)

(or code it inline).  If you're stuck with Python 2.3,

def sortvalue(d):
    aux = [ (v,k) for k, v in d.iteritems() ]
    aux.sort()
    return [ k for v, k in aux ]

is probably the best you can do.


> Regarding the list construction, how can I do this:
> 
> list = [item for item in list1 if for all item2 in list2 NOT
> bad_condition(item, item2]

One possibility (I think it's not that good...):

from itertools import ifilter, imap

[x for x in list1
   if 23 not in imap(lambda x: 23,
                ifilter(lambda y: not badcondition(x, y), list2))
]

I think it's too fancy and you should define an auxiliary function:

def itemisok(item, list2, bad_condition):
    for y in list2:
        if bad_condition(item, y): return False
    return True

so that you can code

[x for x in list1 if itemisok(x, list2, bad_condition)]

or thereabouts...


Alex