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

A

andydtaylor

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
 
M

Mitya Sirenef

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/itertools.html?highlight=itertools#itertools


HTH, -m
 
C

Chris Angelico

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
 
A

andydtaylor

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
 
A

andydtaylor

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
 
D

Dave Angel

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
Please reread Chris Angelico's message. Iterating over the lists
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 ...
 

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