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Optimization of Unique, non-reversible permutations in Python

I'm currently trying to solve the 'dance recital' kattis challenge in Python 3. See here
After taking input on how many performances there are in the dance recital, you must arrange performances in such a way that sequential performances by a dancer are minimized.
I've seen this challenge completed in C++, but my code kept running out of time and I wanted to optimize it.
Question: As of right now, I generate all possible permutations of performances and run comparisons off of that. A faster way to would be to not generate all permutations, as some of them are simply reversed and would result in the exact same output.
import itertools
print(list(itertools.permutations(range(2)))) --> [(0,1),(1,0)] #They're the same, backwards and forwards
print(magic_algorithm(range(2))) --> [(0,1)] #This is what I want
How might I generate such a list of permutations?
I've tried:
-Generating all permutation, running over them again to reversed() duplicates and saving them. This takes too long and the result cannot be hard coded into the solution as the file becomes too big.
-Only generating permutations up to the half-way mark, then stopping, assuming that after that, no unique permutations are generated (not true, as I found out)
-I've checked out questions here, but no one seems to have the same question as me, ditto on the web
Here's my current code:
from itertools import permutations
number_of_routines = int(input()) #first line is number of routines
dance_routine_list = [0]*10
permutation_list = list(permutations(range(number_of_routines))) #generate permutations
for q in range(number_of_routines):
s = input()
for c in s:
v = ord(c) - 65
dance_routine_list[q] |= (1 << v) #each routine ex.'ABC' is A-Z where each char represents a performer in the routine
def calculate():
least_changes_possible = 1e9 #this will become smaller, as optimizations are found
for j in permutation_list:
tmp = 0
for i in range(1,number_of_routines):
tmp += (bin(dance_routine_list[j[i]] & dance_routine_list[j[i - 1]]).count('1')) #each 1 represents a performer who must complete sequential routines
least_changes_possible = min(least_changes_possible, tmp)
return least_changes_possible
print(calculate())
Edit: Took a shower and decided adding a 2-element-comparison look-up table would speed it up, as many of the operations are repeated. Still doesn't fix iterating over the whole permutations, but it should help.
Edit: Found another thread that answered this pretty well. How to generate permutations of a list without "reverse duplicates" in Python using generators
Thank you all!

There are at most 10 possible dance routines, so at most 3.6M permutations, and even bad algorithms like generate 'em all and test will be done very quickly.
If you wanted a fast solution for up to 24 or so routines, then I would do it like this...
Given the the R dance routines, at any point in the recital, in order to decide which routine you can perform next, you need to know:
Which routines you've already performed, because there you can't do those ones next. There are 2R possible sets of already-performed routines; and
Which routine was performed last, because that helps determine the cost of the next one. There are at most R-1 possible values for that.
So there are at less than (R-2)*2R possible recital states...
Imagine a directed graph that connects each possible state to all the possible following states, by an edge for the routine that you would perform to get to that state. Label each edge with the cost of performing that routine.
For example, if you've performed routines 5 and 6, with 5 last, then you would be in state (5,6):5, and there would be an edge to (3,5,6):3 that you could get to after performing routine 3.
Starting at the initial "nothing performed yet" state ():-, use Dijkstra's algorithm to find the least cost path to a state with all routines performed.
Total complexity is O(R2*2R) or so, depending exactly how you implement it.
For R=10, R2*2R is ~100 000, which won't take very long at all. For R=24 it's about 9 billion, which is going to take under half a minute in pretty good C++.

Python pypy: Efficient sum of absolute array/vector difference

I am trying to reduce the computation time of my script,which is run with pypy.
It has to calculate for a large number of lists/vectors/arrays the pairwise sums of absolute differences.
The length of the input vectors is quite small, between 10 and 500.
I tested three different approaches so far:
1) Naive approach, input as lists:
def std_sum(v1, v2):
distance = 0.0
for (a,b) in izip(v1, v2):
distance += math.fabs(a-b)
return distance
2) With lambdas and reduce, input as lists:
lzi = lambda v1, v2: reduce(lambda s, (a,b):s + math.fabs(a-b), izip(v1, v2), 0)
def lmd_sum(v1, v2):
return lzi(v1, v2)
3) Using numpy, input as numpy.arrays:
def np_sum(v1, v2):
return np.sum(np.abs(v1-v2))
On my machine, using pypy and pairs from itertools.combinations_with_replacement
of 500 such lists, the first two approaches are very similar (roughly 5 seconds),
while the numpy approach is significantly slower, taking around 12 seconds.
Is there a faster way to do the calculations? The lists are read and parsed from text
files and an increased preprocessing time would be no problem (such as creating numpy arrays).
The lists contain floating point numbers and are of equal size which is known beforehand.
The script I use for ''benchmarking'' can be found here and some example data here.

Is there a faster way to do the calculations? The lists are read and parsed from text files and an increased preprocessing time would be no problem (such as creating numpy arrays). The lists contain floating point numbers and are of equal size which is known beforehand.
PyPy is very good at optimizing list accesses, so you should probably stick to using lists.
One thing that will help PyPy optimize things is to make sure your lists always have only one type of objects. I.e. if you read strings from a file, don't put them in a list, then parse them into floats in-place. Rather, create the list with floats, for example by parsing each string as soon as it is read. Likewise, never try to preallocate a list, especially with [None,]*N, or PyPy will not be able to guess that all the elements have the same type.
Second, iterate the list as few times as possible. Your np_sum function walks both arrays three times (subtract, abs, sum) unless PyPy notices and can optimize it. Both 1. and 2. walk the list once, so they are faster.

Python loop transfer in other programming language

Again I have a question concerning large loops.
Suppose I have a function
limits
def limits(a,b):
*evaluate integral with upper and lower limits a and b*
return float result
A and B are simple np.arrays that store my values a and b. Now I want to calculate the integral 300'000^2/2 times because A and B are of the length of 300'000 each and the integral is symmetrical.
In Python I tried several ways like itertools.combinations_with_replacement to create the combinations of A and B and then put them into the integral but that takes huge amount of time and the memory is totally overloaded.
Is there any way, for example transferring the loop in another language, to speed this up?
I would like to run the loop
for i in range(len(A)):
for j in range(len(B)):
np.histogram(limits(A[i],B[j]))
I think histrogramming the return of limits is desirable in order not to store additional arrays that grow squarely.
From what I read python is not really the best choice for this iterative ansatzes.
So would it be reasonable to evaluate this loop in another language within Python, if yes, How to do it. I know there are ways to transfer code, but I have never done it so far.
Thanks for your help.

If you're worried about memory footprint, all you need to do is bin the results as you go in the for loop.
num_bins = 100
bin_upper_limits = np.linspace(-456, 456, num=num_bins-1)
# (last bin has no upper limit, it goes from 456 to infinity)
bin_count = np.zeros(num_bins)
for a in A:
for b in B:
if b<a:
# you said the integral is symmetric, so we can skip these, right?
continue
new_result = limits(a,b)
which_bin = np.digitize([new_result], bin_upper_limits)
bin_count[which_bin] += 1
So nothing large is saved in memory.
As for speed, I imagine that the overwhelming majority of time is spent evaluating limits(a,b). The looping and binning is plenty fast in this case, even in python. To convince yourself of this, try replacing the line new_result = limits(a,b) with new_result = 234. You'll find that the loop runs very fast. (A few minutes on my computer, much much less than the 4 hour figure you quote.) Python does not loop very fast compared to C, but it doesn't matter in this case.
Whatever you do to speed up the limits() call (including implementing it in another language) will speed up the program.
If you change the algorithm, there is vast room for improvement. Let's take an example of what it seems you're doing. Let's say A and B are 0,1,2,3. You're integrating a function over the ranges 0-->0, 0-->1, 1-->1, 1-->2, 0-->2, etc. etc. You're re-doing the same work over and over. If you have integrated 0-->1 and 1-->2, then you can add up those two results to get the integral 0-->2. You don't have to use a fancy integration algorithm, you just have to add two numbers you already know.
Therefore it seems to me that you can compute integrals in all the smallest ranges (0-->1, 1-->2, 2-->3), store the results in an array, and add subsets of the results to get the integral over whatever range you want. If you want this program to run in a few minutes instead of 4 hours, I suggest thinking through an alternative algorithm along those lines.
(Sorry if I'm misunderstanding the problem you're trying to solve.)

Iterate two or more lists / numpy arrays... and compare each item with each other and avoid loops in python

I am new to python and my problem is the following:
I have defined a function func(a,b) that return a value, given two input values.
Now I have my data stored in lists or numpy arrays A,Band would like to use func for every combination. (A and B have over one million entries)
ATM i use this snippet:
for p in A:
for k in B:
value = func(p,k)
This takes really really a lot of time.
So i was thinking that maybe something like this:
C=(map(func,zip(A,B)))
But this method only works pairwise... Any ideas?
Thanks for help

First issue
You need to calculate the output of f for many pairs of values. The "standard" way to speed up this kind of loops (calculations) is to make your function f accept (NumPy) arrays as input, and do the calculation on the whole array at once (ie, no looping as seen from Python). Check any NumPy tutorial to get an introduction.
Second issue
If A and B have over a million entries each, there are one trillion combinations. For 64 bits numbers, that means you'll need 7.3 TiB of space just to store the result of your calculation. Do you have enough hard drive to just store the result?
Third issue
If A and B where much smaller, in your particular case you'd be able to do this:
values = f(*meshgrid(A, B))
meshgrid returns the cartesian product of A and B, so it's simply a way to generate the points that have to be evaluated.
Summary
You need to use NumPy effectively to avoid Python loops. (Or if all else fails or they can't easily be vectorized, write those loops in a compiled language, for instance by using Cython)
Working with terabytes of data is hard. Do you really need that much data?
Any solution that calls a function f 1e12 times in a loop is bound to be slow, specially in CPython (which is the default Python implementation. If you're not really sure and you're using NumPy, you're using it too).

suppose, itertools.product does what you need:
from itertools import product
pro = product(A,B)
C = map(lambda x: func(*x), pro)
so far as it is generator it doesn't require additional memory

One million times one million is one trillion. Calling f one trillion times will take a while.
Unless you have a way of reducing the number of values to compute, you can't do better than the above.

If you use NumPy, you should definitely look the np.vectorize function which is designed for this kind of problems...

Python: create a polynomial of degree n

I have a feature set
[x1,x2....xm]
Now I want to create polynomial feature set
What that means is that if degree is two, then I have the feature set
[x1.... xm,x1^2,x2^2...xm^2, x1x2, x1x3....x1,xm......xm-1x1....xm-1xm]
So it contains terms of only of order 2..
same is if order is three.. then you will have cubic terms as well..
How to do this?
Edit 1: I am working on a machine learning project where I have close to 7 features... and a non-linear regression on this linear features are giving ok result...Hence I thought that to get more number in features I can map these features to a higher dimension..
So one way is to consider polynomial order of the feature vector...
Also generating x1*x1 is easy.. :) but getting the rest of the combinations are a bit tricky..
Can combinations give me x1x2x3 result if the order is 3?

Use
itertools.combinations(list, r)
where list is the feature set, and r is the order of desired polynomial features. Then multiply elements of the sublists given by the above. That should give you {x1*x2, x1*x3, ...}. You'll need to construct other ones, then union all parts.
[Edit]
Better: itertools.combinations_with_replacement(list, r) will nicely give sorted length-r tuples with repeated elements allowed.

You could use itertools.product to create all the possible sets of n values that are chosen from the original set; but keep in mind that this will generate (x2, x1) as well as (x1, x2).
Similarly, itertools.combinations will produce sets without repetition or re-ordering, but that means you won't get (x1, x1) for example.
What exactly are you trying to do? What do you need these result values for? Are you sure you do want those x1^2 type terms (what does it mean to have the same feature more than once)? What exactly is a "feature" in this context anyway?

Using Karl's answer as inspiration, try using product and then taking advantage of the set object. Something like,
set([set(comb) for comb in itertools.product(range(5),range(5)])
This will get rid of recurring pairs. Then you can turn the set back into a list and sort it or iterate over it as you please.
EDIT:
this will actually kill the x_m^2 terms, so build sorted tuples instead of sets. this will allow the terms to be hashable and nonrepeating.
set([tuple(sorted(comb)) for comb in itertools.product(range(5),range(5))])