Permutation without duplicates in Python - python

I have N positions, and each position can be either 0 or 1. I have fixed number of 1s, and I want to permutate these fixed number of 1s in these N positions.
from itertools import permutations
p = [0 for k in xrange(6)]
for k in xrange(0,3):
p[k] = 1
print(list(permutations(p)))
But above result contains four [0,0,0,1,1,1] in the list. I only want one of them. How can I get rid of these duplicates?


You could grab the positions of the 1s instead:
from itertools import combinations
def place_ones(size, count):
for positions in combinations(range(size), count):
p = [0] * size
for i in positions:
p[i] = 1
yield p
In action:
>>> list(place_ones(6, 3))
[
[1, 1, 1, 0, 0, 0],
[1, 1, 0, 1, 0, 0],
[1, 1, 0, 0, 1, 0],
[1, 1, 0, 0, 0, 1],
[1, 0, 1, 1, 0, 0],
[1, 0, 1, 0, 1, 0],
[1, 0, 1, 0, 0, 1],
[1, 0, 0, 1, 1, 0],
[1, 0, 0, 1, 0, 1],
[1, 0, 0, 0, 1, 1],
[0, 1, 1, 1, 0, 0],
[0, 1, 1, 0, 1, 0],
[0, 1, 1, 0, 0, 1],
[0, 1, 0, 1, 1, 0],
[0, 1, 0, 1, 0, 1],
[0, 1, 0, 0, 1, 1],
[0, 0, 1, 1, 1, 0],
[0, 0, 1, 1, 0, 1],
[0, 0, 1, 0, 1, 1],
[0, 0, 0, 1, 1, 1],
]


Set is perfect for this, as set does not not contain any duplicated element:
set(permutations(p))

Related

Generating binary entries array in python

I would like to generate an array as follows:
[[0,0,0],
[0,0,1],
[0,1,0],
[0,1,1],
[1,0,0],
[1,0,1],
[1,1,0]
[1,1,1]]
I tried to achieve this by setting 3 for loops, but I wish to go further to 4, 5, and higher bit-numbers, so the last method would not scale easly to these numbers.
Is there any simple way for doing this?

I can't figure out why you want this, but here goes:
For 3:
>>> [[int(x) for x in "{0:03b}".format(y)] for y in range(8)]
[[0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 1, 1], [1, 0, 0], [1, 0, 1], [1, 1, 0], [1, 1, 1]]
>>>
For 5:
>>> [[int(x) for x in "{0:05b}".format(y)] for y in range(32)]
[[0, 0, 0, 0, 0], [0, 0, 0, 0, 1], [0, 0, 0, 1, 0], [0, 0, 0, 1, 1], [0, 0, 1, 0, 0], [0, 0, 1, 0, 1], [0, 0, 1, 1, 0], [0, 0, 1, 1, 1], [0, 1, 0, 0, 0], [0, 1, 0, 0, 1], [0, 1, 0, 1, 0], [0, 1, 0, 1, 1], [0, 1, 1, 0, 0], [0, 1, 1, 0, 1], [0, 1, 1, 1, 0], [0, 1, 1, 1, 1], [1, 0, 0, 0, 0], [1, 0, 0, 0, 1], [1, 0, 0, 1, 0], [1, 0, 0, 1, 1], [1, 0, 1, 0, 0], [1, 0, 1, 0, 1], [1, 0, 1, 1, 0], [1, 0, 1, 1, 1], [1, 1, 0, 0, 0], [1, 1, 0, 0, 1], [1, 1, 0, 1, 0], [1, 1, 0, 1, 1], [1, 1, 1, 0, 0], [1, 1, 1, 0, 1], [1, 1, 1, 1, 0], [1, 1, 1, 1, 1]]
>>>
Matching your formatting is harder.

You can use itertools.product to do this.
>>> import itertools
>>> list(itertools.product([0,1], repeat=3))
[(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)]
https://docs.python.org/3/library/itertools.html#itertools.product

You can use a recursive function like the following:
def generate_binary_entries(n, t=[[]]): # n: length of bit number
if n == 0:
return t
new_t = []
for entry in t:
new_t.append(entry + [0])
new_t.append(entry + [1])
return generate_binary_entries(n - 1, new_t)
Then
generate_binary_entries(4)
generates
[[0, 0, 0, 0],
[0, 0, 0, 1],
[0, 0, 1, 0],
[0, 0, 1, 1],
[0, 1, 0, 0],
[0, 1, 0, 1],
[0, 1, 1, 0],
[0, 1, 1, 1],
[1, 0, 0, 0],
[1, 0, 0, 1],
[1, 0, 1, 0],
[1, 0, 1, 1],
[1, 1, 0, 0],
[1, 1, 0, 1],
[1, 1, 1, 0],
[1, 1, 1, 1]]

How to define an array with all possible combinations of numbers

I want to define an array with a given number of columns (let's say n=5) and in each cell of the array, the value can be either 0 or 1. And I would like to create all possibilities of ones and zeros, which means, that each row would represent one possible vector with n elements.
In other words, I want the table to look like this:
I know that create the vector of ones and zeros is quite easy but how can I ensure that the vectors would not repeat in the table and that there will be all possible combinations included (If my math is correct the table should have 2**5 = 32 rows)
How can I do it in Python? Thank you very much

Easy with itertools:
itertools.product(*[[0, 1]] * 3)
results in
[(0, 0, 0),
(0, 0, 1),
(0, 1, 0),
(0, 1, 1),
(1, 0, 0),
(1, 0, 1),
(1, 1, 0),
(1, 1, 1)]

You could generate all the numbers up to 32, and convert each to binary representation using bit shifts.
combs = [[(n >> p) & 1 for p in range(4, -1, -1)] for n in range(32)]
which gives combs as:
[
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 1],
[0, 0, 0, 1, 0],
[0, 0, 0, 1, 1],
[0, 0, 1, 0, 0],
[0, 0, 1, 0, 1],
[0, 0, 1, 1, 0],
[0, 0, 1, 1, 1],
[0, 1, 0, 0, 0],
[0, 1, 0, 0, 1],
[0, 1, 0, 1, 0],
[0, 1, 0, 1, 1],
[0, 1, 1, 0, 0],
[0, 1, 1, 0, 1],
[0, 1, 1, 1, 0],
[0, 1, 1, 1, 1],
[1, 0, 0, 0, 0],
[1, 0, 0, 0, 1],
[1, 0, 0, 1, 0],
[1, 0, 0, 1, 1],
[1, 0, 1, 0, 0],
[1, 0, 1, 0, 1],
[1, 0, 1, 1, 0],
[1, 0, 1, 1, 1],
[1, 1, 0, 0, 0],
[1, 1, 0, 0, 1],
[1, 1, 0, 1, 0],
[1, 1, 0, 1, 1],
[1, 1, 1, 0, 0],
[1, 1, 1, 0, 1],
[1, 1, 1, 1, 0],
[1, 1, 1, 1, 1]
]
Alternatively, you could use a recursive generation function:
def gimme_combs(n):
if n == 1: return [[0], [1]]
lower_combs = gimme_combs(n - 1)
return [[0] + c for c in lower_combs] + \
[[1] + c for c in lower_combs]
which would give the same result when called with:
combs = gimme_combs(5)

Turning a list into list of lists

I am writing a function which takes columns=c and rows=r (both can be unequal!) and that should a list of lists, where each row is a list containing c elements, all rows within a list. How do I create such sublists given the list below?
list = [0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1]
should return:
[[0, 0, 0, 0, 0], [1, 1, 0, 1, 1], [0, 0, 1, 1, 1], [1, 1, 1, 1, 0], [0, 1, 0, 1, 1]]
I tried to use split() however it seems like it works for strings only.

Numpy:
import numpy
c, r = 4, 5
list_ = [0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0]
numpy.array(list_).reshape(c, r).tolist()
#out (shortened example list to avoid 5x5):
[[0, 0, 0, 0, 0], [1, 1, 0, 1, 1], [0, 0, 1, 1, 1], [1, 1, 1, 1, 0]]
However, if your goal is to create "an cxr array with zeroes and ones", you should better use:
numpy.random.randint(0, high=2, size=(c, r))
# out
array([[1, 1, 1, 0, 0],
[1, 1, 0, 0, 0],
[0, 1, 1, 1, 0],
[1, 0, 0, 1, 0]])

Use itertools.islice: (Also don't use list as a variable name. It replaces the builtin function)
from itertools import islice
def chunker(data, rows, cols):
d = iter(data)
return [list(islice(d, cols)) for row in range(rows)]
data = [0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1]
result = chunker(data, 4, 5)
Result:
[[0, 0, 0, 0, 0],
[1, 1, 0, 1, 1],
[0, 0, 1, 1, 1],
[1, 1, 1, 1, 0]]

You can use a list comprehension:
c, r = 4, 5
list = [0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1]
list_of_lists = [list[i - c: i] for i in range(c, len(list), c)]

l= [0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1]
print([L[i:i+4] for i in range(0,len(L),4)])
output:
[[0, 0, 0, 0], [0, 1, 1, 0], [1, 1, 0, 0], [1, 1, 1, 1], [1, 1, 1, 0], [0, 1, 0, 1], [1]]

using slicing and list comprehension.
new_list=[list[i:i+5] for i in range(len(list)//5)]
just do this like it,it will be done.
a sample usage screenshot

Try this:
ls = [0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1]
[ls[i*5:i*5+5] for i in range(len(ls)//5)]
Out[1]:
[[0, 0, 0, 0, 0],
[0, 0, 0, 0, 1],
[0, 0, 0, 1, 1],
[0, 0, 1, 1, 0],
[0, 1, 1, 0, 1]]
Or as a function:
def split_list(list, length):
return [list[i*length:i*length+length] for i in range((len(list)//length))]
split_list(ls, 5)

Error shuffling list into new list

I am creating a list by shifting an old list out_g, item by item, and appending the result to the new one, new_sets. As I am iterating, I check the resulting shift, and it is correct. After this is complete, I print out the new list, and it is all a single object repeated. What am I missing?
The error occurs during the for loop at the end, where I append the results to new_sets.
#!/usr/bin/python
import math
def LFSR(register, feedback, output):
"""
https://natronics.github.io/blag/2014/gps-prn/
:param list feedback: which positions to use as feedback (1 indexed)
:param list output: which positions are output (1 indexed)
:returns output of shift register:
"""
# calculate output
out = [register[i-1] for i in output]
if len(out) > 1:
out = sum(out) % 2
else:
out = out[0]
# modulo 2 add feedback
fb = sum([register[i-1] for i in feedback]) % 2
# shift to the right
for i in reversed(range(len(register[1:]))):
register[i+1] = register[i]
# put feedback in position 1
register[0] = fb
return out
def shiftInPlace(l, n):
# https://stackoverflow.com/questions/2150108/efficient-way-to-shift-a-list-in-python
n = n % len(l)
head = l[:n]
l[:n] = []
l.extend(head)
return l
##########
## Main ##
##########
n = 3
# init register states
if n == 5 :
LFSR_A = [1,1,1,1,0]
LFSR_B = [1,1,1,0,1]
LFSR_A_TAPS =[5,4,3,2]
LFSR_B_TAPS =[5,3]
elif n == 7:
LFSR_A = [1,0,0,1,0,1,0]
LFSR_B = [1,0,0,1,1,1,0]
LFSR_A_TAPS = [7,3,2,1]
LFSR_B_TAPS = [7,3]
elif n == 3:
LFSR_A = [1,0,1]
LFSR_B = [0,1,1]
LFSR_A_TAPS = [3,2]
LFSR_B_TAPS = [3,1]
output_reg = [n]
N = 2**n-1
out_g = []
for i in range(0,N): #replace N w/ spread_fact
a = (LFSR(LFSR_A, LFSR_A_TAPS, output_reg))
b = (LFSR(LFSR_B, LFSR_B_TAPS, output_reg))
out_g.append(a ^ b)
# FOR BALANCED GOLD CODES NUMBER OF ONES MUST BE ONE MORE THAN NUMBER
# OF ZEROS
nzeros = sum(x == 0 for x in out_g)
nones = sum(x == 1 for x in out_g)
print "Gold Code Output Period[%d] of length %d -- {%d} 0's, {%d} 1's" % (N,N,nzeros,nones)
# produce all time shifted versions of the code
new_sets = []
for i in range(0,N-1):
new_sets.append(shiftInPlace(out_g,1))
# a=shiftInPlace(out_g,1)
# new_sets.append(a)
print new_sets[i]
print new_sets
My output :
Gold Code Output Period[7] of length 7 -- {3} 0's, {4} 1's
[1, 1, 0, 1, 0, 1, 0]
[1, 0, 1, 0, 1, 0, 1]
[0, 1, 0, 1, 0, 1, 1]
[1, 0, 1, 0, 1, 1, 0]
[0, 1, 0, 1, 1, 0, 1]
[1, 0, 1, 1, 0, 1, 0]
[[1, 0, 1, 1, 0, 1, 0], [1, 0, 1, 1, 0, 1, 0], [1, 0, 1, 1, 0, 1, 0], [1, 0, 1, 1, 0, 1, 0], [1, 0, 1, 1, 0, 1, 0], [1, 0, 1, 1, 0, 1, 0]]
Correct values are printing on the iteration, but the final list has all the same values.

The problem should be obvious from your output - you are seeing the same list because you are appending the same list. Consider - you even name your function "shift in place", so that returns a mutated version of the same list you passed in, and then you append that same list. So one quick fix is to make a copy which you end up appending:
new_sets = []
for i in range(0,N-1):
new_sets.append(shiftInPlace(out_g,1)[:]) # append copy
# a=shiftInPlace(out_g,1)
# new_sets.append(a)
print new_sets[i]
This gives the output:
Gold Code Output Period[7] of length 7 -- {3} 0's, {4} 1's
[1, 1, 0, 1, 0, 1, 0]
[1, 0, 1, 0, 1, 0, 1]
[0, 1, 0, 1, 0, 1, 1]
[1, 0, 1, 0, 1, 1, 0]
[0, 1, 0, 1, 1, 0, 1]
[1, 0, 1, 1, 0, 1, 0]
[[1, 1, 0, 1, 0, 1, 0], [1, 0, 1, 0, 1, 0, 1], [0, 1, 0, 1, 0, 1, 1], [1, 0, 1, 0, 1, 1, 0], [0, 1, 0, 1, 1, 0, 1], [1, 0, 1, 1, 0, 1, 0]]
As an aside, for efficient in-place rotations, consider changing your data-structure to a collections.deque, which implements a doubly-linked list:
In [10]: from collections import deque
...: d = deque([1, 1, 0, 1, 0, 1, 0])
...: print(d)
...: for i in range(0, N-1):
...: d.rotate(-1)
...: print(d)
...:
deque([1, 1, 0, 1, 0, 1, 0])
deque([1, 0, 1, 0, 1, 0, 1])
deque([0, 1, 0, 1, 0, 1, 1])
deque([1, 0, 1, 0, 1, 1, 0])
deque([0, 1, 0, 1, 1, 0, 1])
deque([1, 0, 1, 1, 0, 1, 0])
deque([0, 1, 1, 0, 1, 0, 1])

You might try creating your list of rotations like this:
>>> li=[1,0,1,1,0,0]
>>> [li[r:]+li[:r] for r in range(len(li))]
[[1, 0, 1, 1, 0, 0], [0, 1, 1, 0, 0, 1], [1, 1, 0, 0, 1, 0], [1, 0, 0, 1, 0, 1], [0, 0, 1, 0, 1, 1], [0, 1, 0, 1, 1, 0]]

... following up on my comment to juanpa's answer ...
When you append in this fashion, you append a reference to the in-place list. Your two-line code with variable a works the same way. You've appended 6 copies of the same variable reference; every time you shift the list, you shift the underlying object. All of the appended references point to that object.
Here's detailed output tracing your program. Note how all of the elements of new_sets change on every iteration. In my repair, I used the two-line assignment, but added a copy like this: new_sets.append(a[:])
Gold Code Output Period[7] of length 7 -- {3} 0's, {4} 1's
TRACE out_g = [0, 1, 1, 0, 1, 0, 1]
ENTER shiftInPlace, l= [0, 1, 1, 0, 1, 0, 1]
LEAVE shiftInPlace, head= [0] l= [1, 1, 0, 1, 0, 1, 0]
TRACE a= [1, 1, 0, 1, 0, 1, 0] new_sets= [[1, 1, 0, 1, 0, 1, 0]]
TRACE out_g = [1, 1, 0, 1, 0, 1, 0]
ENTER shiftInPlace, l= [1, 1, 0, 1, 0, 1, 0]
LEAVE shiftInPlace, head= [1] l= [1, 0, 1, 0, 1, 0, 1]
TRACE a= [1, 0, 1, 0, 1, 0, 1] new_sets= [[1, 0, 1, 0, 1, 0, 1], [1, 0, 1, 0, 1, 0, 1]]
TRACE out_g = [1, 0, 1, 0, 1, 0, 1]
ENTER shiftInPlace, l= [1, 0, 1, 0, 1, 0, 1]
LEAVE shiftInPlace, head= [1] l= [0, 1, 0, 1, 0, 1, 1]
TRACE a= [0, 1, 0, 1, 0, 1, 1] new_sets= [[0, 1, 0, 1, 0, 1, 1], [0, 1, 0, 1, 0, 1, 1], [0, 1, 0, 1, 0, 1, 1]]
TRACE out_g = [0, 1, 0, 1, 0, 1, 1]
ENTER shiftInPlace, l= [0, 1, 0, 1, 0, 1, 1]
LEAVE shiftInPlace, head= [0] l= [1, 0, 1, 0, 1, 1, 0]
TRACE a= [1, 0, 1, 0, 1, 1, 0] new_sets= [[1, 0, 1, 0, 1, 1, 0], [1, 0, 1, 0, 1, 1, 0], [1, 0, 1, 0, 1, 1, 0], [1, 0, 1, 0, 1, 1, 0]]
TRACE out_g = [1, 0, 1, 0, 1, 1, 0]
ENTER shiftInPlace, l= [1, 0, 1, 0, 1, 1, 0]
LEAVE shiftInPlace, head= [1] l= [0, 1, 0, 1, 1, 0, 1]
TRACE a= [0, 1, 0, 1, 1, 0, 1] new_sets= [[0, 1, 0, 1, 1, 0, 1], [0, 1, 0, 1, 1, 0, 1], [0, 1, 0, 1, 1, 0, 1], [0, 1, 0, 1, 1, 0, 1], [0, 1, 0, 1, 1, 0, 1]]
TRACE out_g = [0, 1, 0, 1, 1, 0, 1]
ENTER shiftInPlace, l= [0, 1, 0, 1, 1, 0, 1]
LEAVE shiftInPlace, head= [0] l= [1, 0, 1, 1, 0, 1, 0]
TRACE a= [1, 0, 1, 1, 0, 1, 0] new_sets= [[1, 0, 1, 1, 0, 1, 0], [1, 0, 1, 1, 0, 1, 0], [1, 0, 1, 1, 0, 1, 0], [1, 0, 1, 1, 0, 1, 0], [1, 0, 1, 1, 0, 1, 0], [1, 0, 1, 1, 0, 1, 0]]
[[1, 0, 1, 1, 0, 1, 0], [1, 0, 1, 1, 0, 1, 0], [1, 0, 1, 1, 0, 1, 0], [1, 0, 1, 1, 0, 1, 0], [1, 0, 1, 1, 0, 1, 0], [1, 0, 1, 1, 0, 1, 0]]

How can I use a Matrix as a dataset on PyBran?

I´m using pybrain in order to train a simple neural network in which the input is going to be a 7x5 Matrix.
The following are the inputs:
A = [[0, 0, 1, 0, 0],
[0, 1, 1, 0, 0],
[0, 1, 0, 1, 0],
[0, 1, 0, 1, 0],
[1, 1, 1, 1, 1],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1]]
E = [[1, 1, 1, 1, 1],
[1, 0, 0, 0, 0],
[1, 0, 0, 0, 0],
[1, 1, 1, 1, 0],
[1, 0, 0, 0, 0],
[1, 0, 0, 0, 0],
[1, 1, 1, 1, 1]]
I = [[0, 0, 1, 0, 0],
[0, 0, 1, 0, 0],
[0, 0, 1, 0, 0],
[0, 0, 1, 0, 0],
[0, 0, 1, 0, 0],
[0, 0, 1, 0, 0],
[0, 0, 1, 0, 0]]
O = [[1, 1, 1, 1, 0],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[1, 1, 1, 1, 0]]
U = [[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[1, 0, 0, 0, 1],
[0, 1, 0, 0, 1],
[0, 0, 1, 1, 0]]
I thought writing something like:
ds = SupervisedDataSet(1, 1)
ds.addSample((A), ("A",))
might work, but I´m getting:
ValueError: cannot copy sequence with size 7 to array axis with dimension 1
Is there any way I can give this datasets to pyBrain?

First you have to know that SupervisedDataSet works with list, so you will need to convert the 2D arrays into a list. You can do it with something like this:
def convertToList (matrix):
list = [ y for x in matrix for y in x]
return list
Then you will need to give the new list to the method SupervisedDataSet.
Also if you would like to use that info to make the network you should use some number to identify the letter like A = 1, E = 2, I = 3, O = 4, U = 5. So to do this, the second parameter for SupervisedDataSet should be just a number 1. In this way you are saying something like "For a list with 35 elements use these numbers to identify a single number".
Finally your code should look like this:
ds = SupervisedDataSet(35, 1)
A2 = convertToList(A)
ds.addSample(A2, (1,))
E2 = convertToList(E)
ds.addSample(E2, (2,))
I2 = convertToList(I)
ds.addSample(I2, (3,))
O2 = convertToList(O)
ds.addSample(O2, (4,))
U2 = convertToList(U)
ds.addSample(U2, (5,))
Hope this could help.

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