Velocity Reviews > Best way to do this? List loop (matrix?) iteration

# Best way to do this? List loop (matrix?) iteration

andydtaylor@gmail.com
Guest
Posts: n/a

 01-09-2013
Hi!

I might be missing the obvious, or I may have found something more complicated than the VBA I am used to. Could it be I need to use a maths library?

For a given list of k items I'd like to turn it into an k*k matrix of item pairs.

List_sample = ['a', 'b', 'c']

Output:

aa ab ac
ba bb bc
ca cb cc

I'd like to have 2 hooks into this process
1. I want the opportunity to use a value pair each time they are generated (because I need to send these to an api and get a number back to put into a temporary list or dictionary - still tbd).
2. I'd also like to know each time a row is completed so I can bank that temporary list to a database table. Else build one big list and do it at the end, I'm still figuring this out.

#Code I've tried:

stn_count = len(stn_list_short)
for rowcount in range (0, stn_count):
for colcount in range (0, stn_count):
print stn_list_long[rowcount] stn_list_long[colcount]

I've found itertools, tee, and product and felt I was getting warmer. I'm still looking, but any pointers would be appreciated!

Thanks,

Andy

Mitya Sirenef
Guest
Posts: n/a

 01-09-2013
On Tue 08 Jan 2013 07:19:59 PM EST, http://www.velocityreviews.com/forums/(E-Mail Removed) wrote:
> Hi!
>
> I might be missing the obvious, or I may have found something more complicated than the VBA I am used to. Could it be I need to use a maths library?
>
> For a given list of k items I'd like to turn it into an k*k matrix of item pairs.
>
> List_sample = ['a', 'b', 'c']
>
> Output:
>
> aa ab ac
> ba bb bc
> ca cb cc
>
> I'd like to have 2 hooks into this process
> 1. I want the opportunity to use a value pair each time they are generated (because I need to send these to an api and get a number back to put into a temporary list or dictionary - still tbd).
> 2. I'd also like to know each time a row is completed so I can bank that temporary list to a database table. Else build one big list and do it at the end, I'm still figuring this out.
>
> #Code I've tried:
>
> stn_count = len(stn_list_short)
> for rowcount in range (0, stn_count):
> for colcount in range (0, stn_count):
> print stn_list_long[rowcount] stn_list_long[colcount]
>
> I've found itertools, tee, and product and felt I was getting warmer. I'm still looking, but any pointers would be appreciated!
>
> Thanks,
>
> Andy
>
>

You can use itertools.product("abc", repeat=2) together with itertools
recipe grouper() from the same page:

http://docs.python.org/3.3/library/i...ools#itertools

HTH, -m

--
Lark's Tongue Guide to Python: http://lightbird.net/larks/

Chris Angelico
Guest
Posts: n/a

 01-09-2013
On Wed, Jan 9, 2013 at 11:19 AM, <(E-Mail Removed)> wrote:
> stn_count = len(stn_list_short)
> for rowcount in range (0, stn_count):
> for colcount in range (0, stn_count):
> print stn_list_long[rowcount] stn_list_long[colcount]

First off, you can iterate over the list directly:

for row in stn_list_short:
for col in stn_list_short:
print row + col

(I'm not sure what your code was doing with the print line, because it
ought to have failed. Explicit concatenation will work.)

Secondly, you can make a list of all of those pairs with a compact
notation called a comprehension:

pairs = [row + col for row in stn_list_short for col in stn_list_short]

That covers your requirement #2, giving you a full list of all of
them. How big is k going to be? Storing the whole list in memory will
get a little awkward if you have very large k; on this system, I
started seeing performance issues with a thousand elements in the
list, but you could probably go to ten thousand (ie a hundred million
pairs) if you have a decent bit of RAM.

ChrisA

andydtaylor@gmail.com
Guest
Posts: n/a

 01-09-2013
Thanks for your help - this is what I did - though it's probably obvious to most people reading this.

for rowcount in range (0, stn_count):
row_durations.append(stn_list_short[rowcount])
for colcount in range (0, stn_count):
# 3. Determine Station pairs for API query
query_origin_stop = stn_list_long[rowcount]
query_destination_stop = stn_list_long[colcount]
# 4. Paths for determining duration. "station x = station x" has journey time 0
# 4a. Stations are SAME

.....etc. and this part works! I am now stuck on something else though, but I'll start a new topic for that.

Thanks

Andy

andydtaylor@gmail.com
Guest
Posts: n/a

 01-09-2013
Thanks for your help - this is what I did - though it's probably obvious to most people reading this.

for rowcount in range (0, stn_count):
row_durations.append(stn_list_short[rowcount])
for colcount in range (0, stn_count):
# 3. Determine Station pairs for API query
query_origin_stop = stn_list_long[rowcount]
query_destination_stop = stn_list_long[colcount]
# 4. Paths for determining duration. "station x = station x" has journey time 0
# 4a. Stations are SAME

.....etc. and this part works! I am now stuck on something else though, but I'll start a new topic for that.

Thanks

Andy

Dave Angel
Guest
Posts: n/a

 01-09-2013
On 01/09/2013 06:24 PM, (E-Mail Removed) wrote:
> Thanks for your help - this is what I did - though it's probably obvious to most people reading this.
>
> for rowcount in range (0, stn_count):
> row_durations.append(stn_list_short[rowcount])
> for colcount in range (0, stn_count):
> # 3. Determine Station pairs for API query
> query_origin_stop = stn_list_long[rowcount]
> query_destination_stop = stn_list_long[colcount]
> # 4. Paths for determining duration. "station x = station x" has journey time 0
> # 4a. Stations are SAME
>

themselves, instead of making a range, is shorter, easier to read, and
usually quicker.

> ....etc. and this part works! I am now stuck on something else though, but I'll start a new topic for that.
>
> Thanks
>
> Andy

Untested:

for rowshort, query_origin_stop in zip(stn_list_short, stn_list_long):
row_durations.append(rowshort)
for query_destination_stop in stn_list_long:
#4 . Determine ...

--

DaveA