# A way to re-organize a list

beginner zyzhu2000 at gmail.com
Wed Aug 1 03:33:49 CEST 2007

```On Jul 19, 10:05 am, beginner <zyzhu2... at gmail.com> wrote:
> Hi Everyone,
>
> I have a simple list reconstruction problem, but I don't really know
> how to do it.
>
> I have a list that looks like this:
>
> l=[ ("A", "a", 1), ("A", "a", 2), ("A", "a", 3), ("A", "b", 1), ("A",
> "b", 2), ("B", "a", 1), ("B", "b", 1)]
>
> What I want to do is to reorganize it in groups, first by the middle
> element of the tuple, and then by the first element. I'd like the
> output look like this:
>
> out=[
>    [    #group by first element "A"
>           [("A", "a", 1), ("A", "a", 2), ("A", "a", 3)], #group by
> second element "a"
>           [ ("A", "b", 1), ("A", "b", 2)], #group by second element
> "b"
>    ],
>    [   #group by first element "B"
>           [("B", "a", 1)],
>           [("B", "b", 1)]
>    ]
> ]
>
> All the solutions I came up with are difficult to read and even harder
> to go back and change, and I just feel are too complicated for such a
> simple problem. I am wondering if anyone here has some insight.
>
> If there is a 'functional' way to do this, it would be even greater.
>
> Thanks,
> Geoffrey

I guess I still don't quite get functional programming. Here is my
imperitive way to do it in O(n).

"""Put a sorted hirerchical structure into nested groups"""

def group_items(source_list, f_list, target=[]):
"""Group_items: Group a list based on the grouping functions.

source_list is a list or iterator that produces a stream of source
records.

f_list is a list of functions. Each function takes the form
f(a,b), where
a and b are two records. If the two records should be in the same
group, it returns
True, otherwise it returns False.

If target is not provided, a new list will be created. Otherwise,
the records are
appended to the provided list.
"""

last_item=None

for new_item in source_list:
level=len(f_list)
t=target
for fn in f_list:
if  t==[] or last_item==None or not fn(last_item,
new_item):
ng=new_item
for i in range(level): ng=[ng]
t.append(ng)
break
else:
t=t[-1]
level -= 1
else:
t.append(new_item)
last_item=new_item

return target

if __name__=="__main__":
import pprint
def f(a,b):
return a[0]==b[0]
def g(a,b):
return a[1]==b[1]

mydata=[[1,2,3,4],[1,2,4,5],[1,2,"A","B"],[2,2,"A","C"]]
t=group_items(mydata, [f,g])
pprint.pprint(t)

```