list: from 2 to 3 dimensions..looking for a nice way

[sven]

I've got a nested list->
a = [[1,'house'],[2,'house'],[3,'garden']]
and I want to get one level deeper with the lists having the same value in
index value 1

b =[[[1, 'house'], [2, 'house']], [[3, 'garten']]]
I' achieving this with an ugly syntax:

a = [[1,'house'],[2,'house'],[3,'garden']]
b = [[]]
b[0].append(a.pop(0))

for id in range(len(a)):
if(b[-1][0][1]==a[id][1]):
b[-1].append(a[id])
else:
b.append([a[id]])

What is the pythonic way to do this?
Thanks for any insight

[Peter Otten]

sven wrote:

I've got a nested list->
a = [[1,'house'],[2,'house'],[3,'garden']]
and I want to get one level deeper with the lists having the same value
in index value 1

b =[[[1, 'house'], [2, 'house']], [[3, 'garten']]]

How about:

a = [[1,'house'],[2,'house'],[3,'garden']]

d = {}
for n, s in a:
d.setdefault(s,[]).append(n)

b = [[[n, s] for n in nlist] for s, nlist in d.iteritems()]
print b

Peter

[Peter Otten]

Peter Otten wrote:

sven wrote:

I've got a nested list->
a = [[1,'house'],[2,'house'],[3,'garden']]
and I want to get one level deeper with the lists having the same value
in index value 1

b =[[[1, 'house'], [2, 'house']], [[3, 'garten']]]

How about:

a = [[1,'house'],[2,'house'],[3,'garden']]

d = {}
for n, s in a:
d.setdefault(s,[]).append(n)

b = [[[n, s] for n in nlist] for s, nlist in d.iteritems()]
print b

Or rather

a = [[1,'house'],[2,'house'],[3,'garden']]

d = {}
for ns in a:
d.setdefault(ns[1], []).append(n)
b = d.items()

Peter

[Peter Otten]

Peter Otten wrote:

Peter Otten wrote:

sven wrote:

I've got a nested list->
a = [[1,'house'],[2,'house'],[3,'garden']]
and I want to get one level deeper with the lists having the same value
in index value 1

b =[[[1, 'house'], [2, 'house']], [[3, 'garten']]]

[...]
Or rather

a = [[1,'house'],[2,'house'],[3,'garden']]

d = {}
for ns in a:
d.setdefault(ns[1], []).append(n)
b = d.items()

So many errors in so few lines :-(
Hope I get it right this time:

a = [[1,'house'],[2,'house'],[3,'garden']]

d = {}
for ns in a:
d.setdefault(ns[1], []).append(ns)
b = d.values()

Peter

[Mike C. Fletcher]

You want to collect those items in a with equal values for item[1], so:

c = {}
for item in a: .... c.setdefault(item[1],[]).append( item )
.... c.values()

[[[1, 'house'], [2, 'house']], [[3, 'garden']]]

HTH,
Mike

sven wrote:
I've got a nested list->
a = [[1,'house'],[2,'house'],[3,'garden']]
and I want to get one level deeper with the lists having the same value in
index value 1

b =[[[1, 'house'], [2, 'house']], [[3, 'garten']]]
I' achieving this with an ugly syntax:

a = [[1,'house'],[2,'house'],[3,'garden']]
b = [[]]
b[0].append(a.pop(0))

for id in range(len(a)):
if(b[-1][0][1]==a[id][1]):
b[-1].append(a[id])
else:
b.append([a[id]])

What is the pythonic way to do this?
Thanks for any insight

--
_______________________________________
Mike C. Fletcher
Designer, VR Plumber, Coder
http://members.rogers.com/mcfletch/