Hi all,
I have a list whose length is a multiple of 3. I want to get a list of
tuples each of which has 3 consecutive elements from the original list,
thus packing the list into smaller packets. Like this:
l = [1,2,3,4,5,6]
tups = [(1,2,3), (4,5,6)]
or
l = [1,2,3,4,5,6,7,8,9]
tups = [(1,2,3), (4,5,6), (7,8,9)]
if i can dictionaries it would be even better:
l = [1,2,3,4,5,6]
tups = [
{'first':1,'second':2,'third':3},
{'first':4,'second':5,'third':6}
]
and so on.
Any ideas?
Many thanks in advance!
Cheers,
 Nickolay 8 2545
"Nickolay Kolev" <nm*****@unibonn.de> wrote in message
news:cm**********@f1node01.rhrz.unibonn.de... Hi all,
I have a list whose length is a multiple of 3. I want to get a list of tuples each of which has 3 consecutive elements from the original list, thus packing the list into smaller packets. Like this:
l = [1,2,3,4,5,6,7,8,9]
tups = [(1,2,3), (4,5,6), (7,8,9)]
How's this? : alist
[1, 2, 3, 4, 5, 6, 7, 8, 9] tups = [tuple(alist[i:i+3]) for i in xrange(0, len(alist), 3)] tups
[(1, 2, 3), (4, 5, 6), (7, 8, 9)] if i can dictionaries it would be even better:
l = [1,2,3,4,5,6]
tups = [ {'first':1,'second':2,'third':3}, {'first':4,'second':5,'third':6} ]
Well, now that you've been introduced to the concept of list indexes and
list comprehensions, I think that extending it to dictionaries is fairly
simple.

I don't actually read my hotmail account, but you can replace hotmail with
excite if you really want to reach me.
Nickolay Kolev wrote: Hi all,
I have a list whose length is a multiple of 3. I want to get a list of tuples each of which has 3 consecutive elements from the original list, thus packing the list into smaller packets. Like this:
I found it in the cookbook: http://aspn.activestate.com/ASPN/Coo.../Recipe/303060
It looks quite cryptic, could anyone give me some pointers at what
exactly is going on in the function?
Cheers,
 Nickolay
Nickolay Kolev <nmkolev <at> unibonn.de> writes: l = [1,2,3,4,5,6]
tups = [(1,2,3), (4,5,6)]
My favorite idiom: l = [1,2,3,4,5,6] zip(*(iter(l),)*3)
[(1, 2, 3), (4, 5, 6)] l = [1,2,3,4,5,6,7,8,9] zip(*(iter(l),)*3)
[(1, 2, 3), (4, 5, 6), (7, 8, 9)]
If you want dicts, try:
l = [1,2,3,4,5,6] tuples = zip(*(iter(l),)*3) labels = 'first second third'.split() [dict(zip(labels, t)) for t in tuples]
[{'second': 2, 'third': 3, 'first': 1}, {'second': 5, 'third': 6, 'first': 4}]
Basically, you can make a dict with whatever labels you want by zipping your
labels with the appropriate tuples and calling the dict builtin.
Steve
I kind of like:
[dict(first=lst[k], second=lst[k+1], third=lst[k+2])
for k in range(0, len(lst), 3)]
Scott David Daniels Sc***********@Acm.Org
* Nickolay Kolev (20041109 22:29 +0100) I have a list whose length is a multiple of 3. I want to get a list of tuples each of which has 3 consecutive elements from the original list, thus packing the list into smaller packets. Like this:
l = [1,2,3,4,5,6]
tups = [(1,2,3), (4,5,6)]
or
l = [1,2,3,4,5,6,7,8,9]
tups = [(1,2,3), (4,5,6), (7,8,9)]
Fragment[1] from a general purpose partitioning utility:
def part(seq, slice):
""" partition seq """
return [seq[slice * index:slice * (index + 1)]
for index in range(len(seq) / slice + bool(len(seq) % slice))]
if i can dictionaries it would be even better:
l = [1,2,3,4,5,6]
tups = [ {'first':1,'second':2,'third':3}, {'first':4,'second':5,'third':6} ]
You're losing order and therfore you're attaching 'first', 'second'
and so on because you obviously need the original order. So don't use
a dictionary.
Thorsten
[1] Just for the record:
def part(seq, slice = None, modus = None):
"""
partition seq
syntax:
part(seq, boolean_function, modus = 'bool')
> [[first_true_items], [first_false_item, remaining_items]]
part('str', 'separator', modus = 'sep') or
part('str', ['separators'], modus = 'sep')
part(list, item, modus = 'sep')
part(n, modus = 'set')
> len([all_possible_partitions_of_[0, 1, ..., n]])
part(list, modus = 'set'
> [all_possible_partitions_of_list]
part(seq, int, modus = 'size')
> [seq0, seq1, ..., seqn]  where len(seq(i)) = int
part(seq, [n0, n1, ..., n(i)], modus = 'size'
> [seq0, seq1, ..., seq(i)]  where len(seq(i)) = n(i)
"""
Nickolay Kolev wrote: I have a list whose length is a multiple of 3. I want to get a list**of tuples each of which has 3 consecutive elements from the original list, thus packing the list into smaller packets. Like this:
... dictionaries ... would be even better:
l = [1,2,3,4,5,6]
tups = [ {'first':1,'second':2,'third':3}, {'first':4,'second':5,'third':6} ]
itertools to the rescue: from itertools import * names = "first second third".split() items = range(6) map(dict, takewhile(bool, starmap(zip, repeat((names, iter(items))))))
[{'second': 1, 'third': 2, 'first': 0}, {'second': 4, 'third': 5, 'first':
3}]
Fun, but not recommended. There must be a simpler expression where you need
not mention len(names) explicitly  but right now I can't think of one.
Note how the nonzero len(items) % len(names) case is handled:
map(dict, takewhile(bool, starmap(zip, repeat((names,
iter(range(5)))))))
[{'second': 1, 'third': 2, 'first': 0}, {'second': 4, 'first': 3}]
However, in real life I would rather go with Steven Bethard's twostep
approach...
Peter
"Nickolay Kolev" <nm*****@unibonn.de> wrote in message
news:cm**********@f1node01.rhrz.unibonn.de... Nickolay Kolev wrote:
I found it in the cookbook: http://aspn.activestate.com/ASPN/Coo.../Recipe/303060
It looks quite cryptic, could anyone give me some pointers at what exactly is going on in the function?
The function contains only one significant line, which is:
return zip(*[lst[i::n] for i in range(n)])
To figure out what it is doing, take it one piece at a time. Start with a
list: lst
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
Pick how many elements you want in each group:
n=3
Now see what the list comprehension does:
[lst[i::n] for i in range(n)]
[[1, 4, 7, 10], [2, 5, 8, 11], [3, 6, 9, 12]]
So this breaks up the list into three separate lists by taking each n'th
element from the original list, then combines them. The "n" in lst[i::n]
determines how many elements will be skipped in creating each sublist.
Unfortunately, this syntax isn't mentioned in the tutorial or the library
reference; you have to dig to find an explanation of it in http://www.python.org/doc/2.3.4/what...onslices.html
Let's store the result of the last step so we can see what the following
step does:
m = [lst[i::n] for i in range(n)] zip(*m)
[(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12)]
The magic function zip() takes a list of sequences as its argument, and
returns a new list that rearranges them. Imagine the first list (m) as a
twodimensional table in which each subsequence is a row; zip() goes
through the table and returns the columns as the subsequences. See http://www.python.org/doc/2.3.4/lib/builtinfuncs.html (all the way at the
end).

I don't actually read my hotmail account, but you can replace hotmail with
excite if you really want to reach me.
On Wed, 10 Nov 2004 19:00:52 +0800, Deepak Sarda <de**********@gmail.com> wrote: >> l = [1,2,3,4,5,6] >> zip(*(iter(l),)*3) [(1, 2, 3), (4, 5, 6)]>> l = [1,2,3,4,5,6,7,8,9] >> zip(*(iter(l),)*3) [(1, 2, 3), (4, 5, 6), (7, 8, 9)]
Can someone explain how this works?!
[snip] So zip() basically gets zip(iter(l), iter(l), iter(l)) , right?
Close, but note that zip(iter(l), iter(l), iter(l)) creates three
iterators to the list l, while zip(*(iter(l),)*3) uses the same
iterator at each position in the tuple. l = [1,2,3,4,5,6] zip(iter(l), iter(l), iter(l))
[(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6, 6)] itr = iter(l) zip(itr, itr, itr)
[(1, 2, 3), (4, 5, 6)]
Here's basically what zip ends up doing when you give it 3 names bound
to the same iterator:
itr = iter(l) tuple1 = (itr.next(), itr.next(), itr.next()) tuple1
(1, 2, 3) tuple2 = (itr.next(), itr.next(), itr.next()) tuple2
(4, 5, 6) result = [tuple1, tuple2] result
[(1, 2, 3), (4, 5, 6)]
Note that when they're not the same iterator, zip does something like:
itr1, itr2, itr3 = iter(l), iter(l), iter(l) tuple1 = (itr1.next(), itr2.next(), itr3.next()) tuple1
(1, 1, 1) tuple2 = (itr1.next(), itr2.next(), itr3.next()) tuple2
(2, 2, 2)
....
So you just get 3 copies of the elements at each index in your list.
Hope that helped!
Steve

You can wordify anything if you just verb it.
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