Suppose I have a simple list in python:
a = [0,1,2]
How do I generate unique arrays from this list so that I get:
arr_gen = [[0,1],
[1,2],
[0,2]]
I've tried using numpy.random.choice this way:
>>> np.random.choice(a, (3,2), replace = False)
But I get the following error below:
>>> ValueError: Cannot take a larger sample than population when 'replace=False'
You could try using the built-in itertools module.
>>> import itertools
>>> a = [0, 1, 2]
>>> list(itertools.combinations(a, 2)
[(0, 1), (0, 2), (1, 2)]
You can also solve this without using itertools, by using iterations and numpy
import numpy
def combinations(iterable, r):
char = tuple(iterable)
n = len(char)
if r > n:
return
index = numpy.arange(r)
# retruns the first sequence
yield tuple(char[i] for i in index)
while True:
for i in reversed(range(r)):
if index[i] != i + n - r:
break
else:
return
index[i] += 1
for j in range(i + 1, r):
index[j] = index[j-1] + 1
yield tuple(char[i] for i in index)
a = [0, 1, 2]
print([x for x in combinations(a, 2)])
Output :
[(0, 1), (0, 2), (1, 2)]
Method 1
Just import combinations from itertools module in python. So by using this code you can get what you wanted.
>>> from itertools import combinations
>>> print(list(map(list,combinations([0, 1, 2], 2))))
[[0, 1], [0, 2], [1, 2]]
Method 2 (To get unique list)
In this question you also said that you need to generate unique arrays also, So lets assume that your array is [0,1,1].You just can use this one,
>>> from itertools import combinations
>>> print(list(map(list,{*map(tuple,combinations([0, 1, 2], 2))})))
[[0, 1], [1, 1]]
In sets removes duplicate items. So after that you have to convert set to list also.
Method 3 (Recursive method)
def myComb(lst):
if len(lst) == 0:
return []
elif len(lst) == 1:
return [lst]
else:
l = []
for i in range(len(lst)):
x = lst[i]
xs = lst[:i] + lst[i+1:]
for p in myComb(xs):
l.append([x]+p)
return l
ans=[]
for p in myComb([0,1,2]):
p.pop()
p.sort()
if p not in ans:
ans.append(p)
print(ans)
ans=[[0, 1], [0, 2], [1, 2]]
