lazy mapping between integers and arrays permutations without generating all the permutations-CodePudding

I want to generate a mapping between a set of integer arrays and integer values in python, e.g for n=2 (size of the array) and m=2 (array item range), I want to have:

{
   0: [0, 0, 0],
   1: [0, 0, 1],
   2: [0, 1, 1],
   3: [1, 0, 0],
   ...
   7: [1, 1, 1]

}

One way to do this is to use the itertools.product and do the following:

from itertools import product
import numpy as np
n = 2
m = 3
a = list(product(np.arange(n), repeat=3))
a = list(map(lambda l: list(l), a))
b = dict(zip(np.arange(len(a)), a))

and b will be:

{0: [0, 0, 0], 1: [0, 0, 1], 2: [0, 1, 0], 3: [0, 1, 1], 4: [1, 0, 0], 5: [1, 0, 1], 6: [1, 1, 0], 7: [1, 1, 1]}

and I can have the mentioned mapping e.g.

>>> b[1]
[0, 0, 1]

However, in my case the n=8 and m=16 and storing such a dictionary (8**16 entries!) will take more than 100 Gb. I want to know if there is a lazy way of generating each index of the b everytime I want to access that without generating the entire dictionary?

CodePudding user response：

You can make a function instead of full map. Your output is actually array of size m representing value of input a in n-th scale of notation. If you want to make a map just save the result of the function in map if key doesn't exist.

def f(a, n, m):
    res = []
    while a != 0:
        res = [a % n]   res
        a //= n
        m -= 1
    while m != 0:
        res = [0]   res
        m -= 1
    return res

CodePudding user response：

You can generate the xth element by taking the successive x//n**(i-1)%n for i that varies between m-1 and 0.

n = 3
m = 4

x = 6
[x//n**(i-1)%n for i in range(m, 0, -1)]

output: [0, 0, 2, 0]

as a function

def f(x, n, m):
    return [x//n**(i-1)%n for i in range(m, 0, -1)]

f(1, n=2, m=3)
# [0, 0, 1]

f(1234567890, n=8, m=16)
# [0, 0, 0, 0, 0, 1, 1, 1, 4, 5, 4, 0, 1, 3, 2, 2]

CodePudding user response：

In your example, you are looking more as mapping the product than permutations.

The specific example given can be converted to a function that will give you the bits of an integer (without having to store anything):

def bits(n,w=16): return [*map(int,f'{n:0{w}b}')]

bits(5,3)  # [1, 0, 1]
bits(13,8) # [0, 0, 0, 0, 1, 1, 0, 1]

For a more generalized virtual lists of 'products', you can make a function to get the Nth product from a list of sequences which will cover your example and more:

def prodAtIndex(i,L):
    result = []                     # product fed in reversed order           
    for v in reversed(L):           # from least significant to most
        i,p = divmod(i,len(v))      # get digit in current 'base'
        result.insert(0,v[p])       # build product
    return result if i==0 else None # i will end at zero if in range

prodAtIndex(16,[(0,1,2)]*3)      # [1,2,1]
prodAtIndex(13,[(0,1)]*8)        # [0, 0, 0, 0, 1, 1, 0, 1]
prodAtIndex(10,["ABCD",(1,2,3)]) # ['D', 2]

A class can be made to manipulate virtual lists of products more seamlessly:

class Products:

    def __init__(self,*values):
        self.values = values
        
    def __len__(self):
        result = 1
        for s in map(len,self.values): result *= s
        return result

    def __getitem__(self,index):
        if isinstance(index,slice):
            return (self[i] for i in range(len(self))[index])
        if index<0: index  = len(self)
        result = []                        # product fed in reversed order
        for v in reversed(self.values):    # from least significant to most
            index,p = divmod(index,len(v)) # get digit in current 'base'
            result.insert(0,v[p])          # build product
        if index: raise IndexError         # will be zero if in range
        return tuple(result)

    def index(self,P):
        index = 0
        for v,p in zip(self.values,P):
            index *= len(v)
            index  = v.index(p)
        return index

Usage:

 p = Products("ABCD",(1,2,3,4,5))

 len(p)             # 20

 p[4]               # ('A', 5)
 p[-4]              # ('D', 2) 

 p.index(['A',5])   # 4 

 for prod in p[3:5]: print(prod)

 ('A', 4)
 ('A', 5)


 # For your example...

 m  = 3
 n  = 2
 b3 = Products(*[range(n)]*m)

 b3[5] # [1, 0, 1]

Permutations?

If you actually want to manipulate permutations as if you had a list but without generating the actual list, you can make a function that will give you then Nth permutation in a "virtual" list:

from math import factorial
def permCount(A,k): return factorial(len(A))//factorial(len(A)-k)
def permAtIndex(A,k,index):
    A      = list(A)
    base   = permCount(A,k)
    result = []
    for _ in range(k):                    # for required size
        base  //= len(A)                  # chunk size for position
        i,index = divmod(index,base or 1) # i is value index at position
        result.append(A.pop(i))           # build permutation
    return result

This will give you the size of the permutation "list" and allow you to get a permutation at a given index:

print(permCount([1,2,3,4],4))       # 24
print(permAtIndex([1,2,3,4],4,22))  # [4, 3, 1, 2]

print(permCount(range(15),10))              # 10897286400
print(permAtIndex(range(15),10,5897286400)) # [8,1,10,5,12,0,13,7,14,9]

You can also create a reverse function to give you the index of a given permutation:

def indexOfPerm(A,k,P):
    A      = list(A)
    base   = permCount(A,k)
    result = 0
    for n in P:            
        base //= len(A)  # chunk size for position
        i = A.index(n)   # index in remaining items
        result  = base*i # position's component of index
        A.pop(i)         # next position on rest of items
    return result

print(indexOfPerm([1,2,3,4],4,[4, 3, 1, 2])) # 22
print(indexOfPerm(range(15),10,[8,1,10,5,12,0,13,7,14,9])) # 5897286400

All this can be turned into a class that allows manipulation of virtual lists of permutations:

from math import factorial as fact
class Permutations:

    def __init__(self,values,size):
        self.values = values
        self.size   = size
        
    def __len__(self):
        return fact(len(self.values))//fact(len(self.values)-self.size)

    def __getitem__(self,index):
        if isinstance(index,slice):
            return (self[i] for i in range(len(self))[index])
        A      = list(self.values)
        base   = len(self)
        value  = []
        if index<0: index  = len(self)
        for _ in range(self.size):            # for required size
            base  //= len(A)                  # chunk size for position
            i,index = divmod(index,base or 1) # i is value index at position
            value.append(A.pop(i))            # build permutation
        return tuple(value)

    def index(self,P):
        A      = list(self.values)
        base   = len(self)
        index  = 0
        for n in P:            
            base //= len(A)  # chunk size for position
            i = A.index(n)   # index in remaining items
            index  = base*i  # position's component of index
            A.pop(i)         # next position on rest of items
        return index

Usage:

p = Permutations(range(15),10)

print(len(p))                  # 10897286400

print(p[5897286400])           # (8, 1, 10, 5, 12, 0, 13, 7, 14, 9)
print(p[-100])                 # (14, 13, 12, 11, 10, 9, 8, 5, 4, 2)

print(p.index([8, 1, 10, 5, 12, 0, 13, 7, 14, 9])) # 5897286400


for perm in p[100:105]: print(perm)
(0, 1, 2, 3, 4, 5, 6, 9, 10, 13)
(0, 1, 2, 3, 4, 5, 6, 9, 10, 14)
(0, 1, 2, 3, 4, 5, 6, 9, 11, 7)
(0, 1, 2, 3, 4, 5, 6, 9, 11, 8)
(0, 1, 2, 3, 4, 5, 6, 9, 11, 10)

Note that this Permutations class used on a list of non-unique values will produce duplicate permutations