Generating random lists in Python query - list

Question

I want my program to take numbers from 1 to X and randomly distribute those numbers between X/2 numbers of lists Y times. I don't want a number to be repeated during one shuffle, also I don't want the lists to repeat at all. So if there is list [1,2] there shouldn't be another list containing 1 or 2 in same shuffle and there shouldn't be another [1,2] or [2,1] in the whole result.
This is what I came up with, however, it keeps on repeating numbers. Any adice?
import random
def Shuffler():
amount = int(raw_input("Numbers up to: "))
times = int(raw_input("Number of shuffles: "))
numberslist = range(1,amount+1)
twos = []
thisshuffle = []
final = []
while len(final) < (amount/2)*times:
twos = []
thisshuffle = []
while len(twos) < 2:
if len(numberslist)!=0:
randomnumber = random.choice(numberslist)
if (randomnumber in twos) or (randomnumber in thisshuffle):
numberslist.remove(randomnumber)
else:
twos.append(randomnumber)
thisshuffle.append(randomnumber)
numberslist.remove(randomnumber)
else:
numberslist = range(1,amount+1)
if (twos or list(reversed(twos))) not in final:
final.append(twos)
k=0
for i in range(times): #this shit prints shit
print "%s:" % (i+1), final[k:k+amount/2]
print
k = k + amount/2
Shuffler()
Shuffler()

Answer 1

As ccf pointed out, your requirements aren't trivial. A few more steps and you'd have a sudoku generator :)
I tried a few solutions but they either didn't produce random output or were rather inefficient. Ccf's solution is clearly well written but it seems to have the same problem; it produces ordered output (e.g. [1, 2], [1, 3], [1, 4], [1, 5], [1, 6]).
#cff - wouldn't it be better to use itertools.combinations rather than itertools.permutations to avoid generating repetitions?
Here's a "solution" that's quite similar to ccf's (doesn't produce random output either):
import itertools
def Shuffler():
amount = int(raw_input("Numbers up to: "))
times = int(raw_input("Number of shuffles: "))
rng = range(1, amount+1)
perms = list(itertools.combinations(rng, 2))
lst_single = []
lst_all = []
for p in perms:
if len(lst_all) >= times:
for i, lst in enumerate(lst_all):
print str(i+1) + ": ", lst
break
if len(lst_single) == amount/2:
lst_all.append(lst_single)
lst_single = []
elif p[0] < p[1]:
p = list(p)
lst_single.append(p)
Shuffler()
Output
Numbers up to: 6
Number of shuffles: 3
1: [[1, 2], [1, 3], [1, 4]]
2: [[1, 6], [2, 3], [2, 4]]
3: [[2, 6], [3, 4], [3, 5]]
And here's a slightly hackish solution that seems to produce the output you want but in an inefficient way. It relies on a set for filtering out unwanted combinations but still wastes resources producing them in the first place.
import random
def Shuffler():
amount = int(raw_input("Numbers up to: "))
times = int(raw_input("Number of shuffles: "))
rng = range(1, amount+1)
final = []
lst_len = amount/2
combos_unique = set()
while len(combos_unique) < lst_len*times:
combo_rand = random.sample(rng, 2)
if combo_rand[0] < combo_rand[1]:
combos_unique.add(tuple(combo_rand))
tmp = []
for combo in combos_unique:
tmp.append(list(combo))
if len(tmp) >= lst_len:
final.append(tmp)
tmp = []
for i, lst in enumerate(final):
print str(i+1) + ": ", lst
Shuffler()
Output
Numbers up to: 6
Number of shuffles: 3
1: [[2, 6], [4, 6], [5, 6]]
2: [[4, 5], [1, 3], [1, 6]]
3: [[3, 4], [2, 4], [3, 5]]

Answer 2

You don't want any repeated numbers in one shuffle, and any repeated list ... and so on. This is not an easy task. Plus another fact is that, the unique sets of numbers are fixed, which cannot be set too high. For example, if you set "Numbers up to:" 5, and "Number of shuffles: " 20, for sure you will get repeated numbers.
The issue with your code, I see, is in this if statement:
if (twos or list(reversed(twos))) not in final:
final.append(twos)
(twos or list(reversed(twos))) is logical OR, the result is twos, because twos is not empty. I suggest you change the if statement to:
if (twos not in final) and (list(reversed(twos)) not in final):
final.append(twos)
The following code (python 2.7x) uses permutations and shuffle to generate numbers list. Next, make the list unique (e.g., no [1,2] and [2,1] in same list). then, divide them into groups based on number of shuffles specified by the user. Press any letter, script will exit. Hope it helps:
from itertools import permutations
from random import shuffle
def Shuffler():
try:
amount = input("Numbers up to: ")
p = list(permutations(range(1, amount + 1), 2))
p_uniq = [list(x) for x in p if x[::-1] in p and x[0]<=x[1]]
shuf_max = len(p_uniq) /(amount / 2)
times = shuf_max + 1 # set a higher value to trigger prompt
while times > shuf_max:
shuffle(p_uniq) # shuffle the unique list in place
times = input("Number of shuffles (MAX %s): " % (shuf_max))
else:
for i, group in enumerate(list(zip(*[iter(p_uniq[: (amount /2) * times + 1])]* (amount/2)))):
print "%i: " % (i + 1), list(group)
Shuffler()
except:
print 'quitting...'
Shuffler()
Output：
Numbers up to: 10
Number of shuffles (MAX 9): 8
1: [[6, 7], [1, 9], [2, 5], [5, 9], [9, 10]]
2: [[1, 10], [3, 8], [4, 10], [8, 10], [1, 5]]
3: [[1, 4], [6, 8], [3, 6], [2, 4], [4, 7]]
4: [[2, 10], [5, 8], [3, 9], [1, 7], [4, 9]]
5: [[1, 2], [7, 9], [1, 3], [6, 9], [1, 6]]
6: [[2, 9], [4, 8], [3, 5], [8, 9], [7, 10]]
7: [[2, 7], [2, 3], [7, 8], [3, 7], [3, 10]]
8: [[3, 4], [2, 6], [5, 6], [5, 7], [4, 6]]
Numbers up to: