[portland] Still Having List/Tuple Problems

[kirby urner]

On Wed, Apr 16, 2008 at 4:24 PM, Rich Shepard <rshepard at appl-ecosys.com> wrote:
> On Wed, 16 Apr 2008, kirby urner wrote:
>
>  >>>> thekeys = set([i for (i,j) in mydata])
>
>    I thought this more elegant, but when I could not use the 'j' values
>  associated with each 'i' value I did some reading on sets. They are
>  iterable, but not indexable. And I need to access each 'j' associated with
>  each 'i'.
>

Yeah, was just using the set to garner / harvest unique keys
(throwing away j values, keeping only i values), then in the
next loop building a data structure with filtration.

Here'd be a complete program, data set of my own invention:

IDLE 1.2.1
>>> thedata = [('salmon',1),('tuna',1),('salmon',3),('rockfish',4),('salmon',2),
('rockfish',2),('tuna',9),('barracuda',4),('trout',1),('barracuda',18)]


def groupstuff(anydata):
	thekeys = set([i for (i,j) in anydata])  # get unique i values
	grouped = {}  # return a dictionary listing j values, indexed by i values
	for key in thekeys:
		grouped[key] = [ j for (i,j) in anydata if i == key ]  # scoop out j values
	return grouped

>>> groupstuff(thedata)
{'salmon': [1, 3, 2], 'tuna': [1, 9], 'barracuda': [4, 18],
'rockfish': [4, 2], 'trout': [1]}

def graphit(groups):
    for key in groups.keys():  # loop through indexes
	print key
	for graph in sorted(groups[key]):  # assume ordering of j-values matters
	    print graph
			
>>> graphit(groupstuff(thedata))
salmon
1
2
3
tuna
1
9
barracuda
4
18
rockfish
2
4
trout
1

>    But, I learned quite a bit. I think the groupby() approach is most
>  suitable for my need.
>
>  Thanks all,
>
>
>
>  Rich