Need a strange sort method...

[SpreadTooThin]

I have a list and I need to do a custom sort on it...

for example:
a = [1,2,3,4,5,6,7,8 ,9,10] #Although not necessarily in order

def cmp(i,j): #to be defined in this thread.

a.sort(cmp)

print a
[1,4,7,10, 2,5,8, 3,6,9]

So withouth making this into an IQ test.
Its more like
1 4 7 10
2 5 8
3 6 9

Read that top to bottom from column 1 to column 4.
When you get to the bottom of a column move to the next column.
Get it?

[Paul Rubin]

"SpreadTooT hin" <bj********@gma il.comwrites:

a = [1,2,3,4,5,6,7,8 ,9,10] #Although not necessarily in order

def cmp(i,j): #to be defined in this thread.

a.sort(cmp)

print a
[1,4,7,10, 2,5,8, 3,6,9]

def k(n): return (n-1) % 3, (n-1) // 3
a.sort(key=k)

[Simon Brunning]

On 16 Oct 2006 11:13:08 -0700, SpreadTooThin <bj********@gma il.comwrote:

I have a list and I need to do a custom sort on it...

for example:
a = [1,2,3,4,5,6,7,8 ,9,10] #Although not necessarily in order

def cmp(i,j): #to be defined in this thread.

a.sort(cmp)

print a
[1,4,7,10, 2,5,8, 3,6,9]

So withouth making this into an IQ test.
Its more like
1 4 7 10
2 5 8
3 6 9

>>a = [1,2,3,4,5,6,7,8 ,9,10]
a.sort(key=la mbda item: (((item-1) %3), item))
a

[1, 4, 7, 10, 2, 5, 8, 3, 6, 9]

--
Cheers,
Simon B
si***@brunningo nline.net
http://www.brunningonline.net/simon/blog/

[Fredrik Lundh]

SpreadTooThin wrote:

I have a list and I need to do a custom sort on it...

Its more like
1 4 7 10
2 5 8
3 6 9

that's trivial to do with slicing, of course. what makes you think you
need to do this by calling the "sort" method ?

</F>

[Simon Brunning]

On 10/16/06, Simon Brunning <si***@brunning online.netwrote :

>a = [1,2,3,4,5,6,7,8 ,9,10]
a.sort(key=lam bda item: (((item-1) %3), item))
a

[1, 4, 7, 10, 2, 5, 8, 3, 6, 9]

Re-reading the OP's post, perhaps sorting isn't what's required:

>>a[::3] + a[1::3] + a[2::3]

[1, 4, 7, 10, 2, 5, 8, 3, 6, 9]

--
Cheers,
Simon B
si***@brunningo nline.net
http://www.brunningonline.net/simon/blog/

[Gerard Flanagan]

SpreadTooThin wrote:

I have a list and I need to do a custom sort on it...

for example:
a = [1,2,3,4,5,6,7,8 ,9,10] #Although not necessarily in order

def cmp(i,j): #to be defined in this thread.

a.sort(cmp)

print a
[1,4,7,10, 2,5,8, 3,6,9]

So withouth making this into an IQ test.
Its more like
1 4 7 10
2 5 8
3 6 9

Read that top to bottom from column 1 to column 4.
When you get to the bottom of a column move to the next column.
Get it?

maybe the columnise function here would help:

http://www.gflanagan.net/site/python/utils/sequtils/

from math import sqrt

for i in range(2,12):
seq = range(1,i)
numcols = int(sqrt(len(se q)))
print columnise(seq, numcols)

[(1,)]
[(1, 2)]
[(1, 2, 3)]
[(1, 3), (2, 4)]
[(1, 3, 5), (2, 4, None)]
[(1, 3, 5), (2, 4, 6)]
[(1, 3, 5, 7), (2, 4, 6, None)]
[(1, 3, 5, 7), (2, 4, 6, 8)]
[(1, 4, 7), (2, 5, 8), (3, 6, 9)]
[(1, 4, 7, 10), (2, 5, 8, None), (3, 6, 9, None)]

--------------------------

def chunk( seq, size, pad=None ):
'''
Slice a list into consecutive disjoint 'chunks' of
length equal to size. The last chunk is padded if necessary.

>>list(chunk(ra nge(1,10),3))

[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

>>list(chunk(ra nge(1,9),3))

[[1, 2, 3], [4, 5, 6], [7, 8, None]]

>>list(chunk(ra nge(1,8),3))

[[1, 2, 3], [4, 5, 6], [7, None, None]]

>>list(chunk(ra nge(1,10),1))

[[1], [2], [3], [4], [5], [6], [7], [8], [9]]

>>list(chunk(ra nge(1,10),9))

[[1, 2, 3, 4, 5, 6, 7, 8, 9]]

>>for X in chunk([],3): print X

'''
n = len(seq)
mod = n % size
for i in xrange(0, n-mod, size):
yield seq[i:i+size]
if mod:
padding = [pad] * (size-mod)
yield seq[-mod:] + padding

def columnise( seq, numcols, pad=None ):
'''

>>columnise(ran ge(1,10),3)

[(1, 4, 7), (2, 5, 8), (3, 6, 9)]

>>columnise(ran ge(1,9),3)

[(1, 4, 7), (2, 5, 8), (3, 6, None)]

>>columnise(ran ge(1,8),3)

[(1, 4, 7), (2, 5, None), (3, 6, None)]
'''
return zip( *chunk(seq, numcols, pad) )
-------------------------------

Gerard

[Saizan]

SpreadTooThin wrote:

I have a list and I need to do a custom sort on it...

for example:
a = [1,2,3,4,5,6,7,8 ,9,10] #Although not necessarily in order

def cmp(i,j): #to be defined in this thread.

a.sort(cmp)

print a
[1,4,7,10, 2,5,8, 3,6,9]

So withouth making this into an IQ test.
Its more like
1 4 7 10
2 5 8
3 6 9

Read that top to bottom from column 1 to column 4.
When you get to the bottom of a column move to the next column.
Get it?

It's a little vague, but i'm supposing that if you have an 11 in a the
order will be:
[1,4,7,10, 2,5,8,11, 3, 6,9]
If this holds then your order is based on x%3. You place first all x
for which x%3 == 1, then x%3 == 2, and last x%3 == 0. And among these
three group you use "natural" order.
so:
def yourcmp(i,j):
ri = i%3
rj = j%3
if ri == rj: return i.__cmp__(j)
elif ri == 0: return 1
elif rj == 0: return -1
else: return ri.__cmp__(rj)
This works with your example, and with my assumption, feel free to
"optimize" the if/elif block.

However you shuold pay attention to the 0 behavior.
a = range(11)
a.sort(yourcmp)
print a
[1, 4, 7, 10, 2, 5, 8, 0, 3, 6, 9]

[Tim Chase]

for example:

a = [1,2,3,4,5,6,7,8 ,9,10] #Although not necessarily in order

def cmp(i,j): #to be defined in this thread.

Well, if you're willing to give up doing it in a cmp() method,
you can do it as such:

>>a.sort()
chunk_size = 3
[a[i::chunk_size] for i in range(chunk_siz e)]

[[1, 4, 7, 10], [2, 5, 8], [3, 6, 9]]

If you need it in a flat list, rather than as a list of
chunk_size lists (which are handy for iterating over in many
cases), there are ways of obtaining it, such as the hackish

>>sum([a[i::chunk_size] for i in range(chunk_siz e)], [])

[1, 4, 7, 10, 2, 5, 8, 3, 6, 9]

There are likely good recipes for flattening a list. I just
happen not to have any at my fingertips.

I'm not sure it's possible to do in a cmp() method, given that it
requires apriori knowledge of the dataset (are the numbers
contiguous?). Unless, of course, you have such a list...

However, as a benefit, this method should work no matter what the
list contains, as long as they're comparable to each other for an
initial sorting:

>>a = [chr(ord('a') + i) for i in range(10)]
# a.sort() if it were needed
a

['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']

>>sum([a[i::chunk_size] for i in range(chunk_siz e)], [])

['a', 'd', 'g', 'j', 'b', 'e', 'h', 'c', 'f', 'i']
-tkc

[SpreadTooThin]

Fredrik Lundh wrote:

SpreadTooThin wrote:

I have a list and I need to do a custom sort on it...

Its more like
1 4 7 10
2 5 8
3 6 9

that's trivial to do with slicing, of course. what makes you think you
need to do this by calling the "sort" method ?

</F>

You are of course correct.. There might be a way to do this with
slicing
and i % 3

[Neil Cerutti]

On 2006-10-16, Tim Chase <py*********@ti m.thechases.com wrote:

If you need it in a flat list, rather than as a list of
chunk_size lists (which are handy for iterating over in many
cases), there are ways of obtaining it, such as the hackish

>sum([a[i::chunk_size] for i in range(chunk_siz e)], [])

[1, 4, 7, 10, 2, 5, 8, 3, 6, 9]

There are likely good recipes for flattening a list. I just
happen not to have any at my fingertips.

Actually, there isn't a good recipe in Python for flattening a
list. They all come out tasting like Circus Peanuts (Turkish
Delight for you non-Yanks).

--
Neil Cerutti