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how to make sure that two numbers next to each other in a list are different

I have a simple code that generates a list of random numbers.
x = [random.randrange(0,11) for i in range(10)]
The problem I'm having is that, since it's random, it sometimes produces duplicate numbers right next to each other. How do I change the code so that it never happens? I'm looking for something like this:
[1, 7, 2, 8, 7, 2, 8, 2, 6, 5]
So that every time I run the code, all the numbers that are next to each other are different.

x = []
while len(x) < 10:
r = random.randrange(0,11)
if not x or x[-1] != r:
x.append(r)
x[-1] contains the last inserted element, which we check not to be the same as the new random number. With not x we check that the array is not empty, as it would generate a IndexError during the first iteration of the loop

Here's an approach that doesn't rely on retrying:
>>> import random
>>> x = [random.choice(range(12))]
>>> for _ in range(9):
... x.append(random.choice([*range(x[-1]), *range(x[-1]+1, 12)]))
...
>>> x
[6, 2, 5, 8, 1, 8, 0, 4, 6, 0]
The idea is to choose each new number by picking from a list that excludes the previously picked number.
Note that having to re-generate a new list to pick from each time keeps this from actually being an efficiency improvement. If you were generating a very long list from a relatively short range, though, it might be worthwhile to generate different pools of numbers up front so that you could then select from the appropriate one in constant time:
>>> pool = [[*range(i), *range(i+1, 3)] for i in range(3)]
>>> x = [random.choice(random.choice(pool))]
>>> for _ in range(10000):
... x.append(random.choice(pool[x[-1]]))
...
>>> x
[0, 2, 0, 2, 0, 2, 1, 0, 1, 2, 0, 1, 2, 1, 0, ...]

O(n) solution by adding to the last element randomly from [1,stop) modulo stop
import random
x = [random.randrange(0,11)]
x.extend((x[-1]+random.randrange(1,11)) % 11 for i in range(9))
x
Output
[0, 10, 4, 5, 10, 1, 4, 8, 0, 9]

from random import randrange
from itertools import islice, groupby
# Make an infinite amount of randrange's results available
pool = iter(lambda: randrange(0, 11), None)
# Use groupby to squash consecutive values into one and islice to at most 10 in total
result = [v for v, _ in islice(groupby(pool), 10)]

Function solution that doesn't iterate to check for repeats, just checks each add against the last number in the list:
import random
def get_random_list_without_neighbors(lower_limit, upper_limit, length):
res = []
# add the first number
res.append(random.randrange(lower_limit, upper_limit))
while len(res) < length:
x = random.randrange(lower_limit, upper_limit)
# check that the new number x doesn't match the last number in the list
if x != res[-1]:
res.append(x)
return res
>>> print(get_random_list_without_neighbors(0, 11, 10)
[10, 1, 2, 3, 1, 8, 6, 5, 6, 2]

def random_sequence_without_same_neighbours(n, min, max):
x = [random.randrange(min, max + 1)]
uniq_value_count = max - min + 1
next_choises_count = uniq_value_count - 1
for i in range(n - 1):
circular_shift = random.randrange(0, next_choises_count)
x.append(min + (x[-1] + circular_shift + 1) % uniq_value_count)
return x
random_sequence_without_same_neighbours(n=10, min=0, max=10)

It's not to much pythonic but you can do something like this
import random
def random_numbers_generator(n):
"Generate a list of random numbers but without two duplicate numbers in a row "
result = []
for _ in range(n):
number = random.randint(1, n)
if result and number == result[-1]:
continue
result.append(number)
return result
print(random_numbers_generator(10))
Result:
3, 6, 2, 4, 2, 6, 2, 1, 4, 7]

Can you for loop completely through a range, but starting from the nth element?

I would like to know if there exists a base solution to do something like this:
for n in range(length=8, start_position= 3, direction= forward)
The problem I'm encountering is I would like the loop to continue past the final index, and pick up again at idx =0, then idx=1, etc. and stop at idx= 3, the start_position.
To give context, I seek all possible complete solutions to the n-queen problem.

Based on your latest edit, you need a "normal" range and the modulo operator:
for i in range(START, START + LEN):
do_something_with(i % LEN)

from itertools import chain
for n in chain(range(3,8), range(3)):
...
The chain() returns an iterator with 3, 4, ..., 7, 0, 1, 2

Another option for solving this is to use modular arithmetic. You could do something like this, for example:
for i in range(8)
idx = (i + 3) % 8
# use idx
This easily can be generalized to work with different lengths and offsets.

def loop_around_range(length, start_position, direction='forward'):
looped_range = [k % length for k in range(start_position, start_position+length)]
if direction == 'forward':
return looped_range
else:
return looped_range[::-1]

You could implement this for an arbitrary iterable by using itertools.cycle.
from itertools import cycle
def circular_iterator(iterable, skip=0, length=None, reverse=False):
"""Produces a full cycle of #iterable#, skipping the first #skip# elements
then tacking them on to the end.
if #iterable# does not implement #__len__#, you must provide #length#
"""
if reverse:
iterable = reversed(iterable)
cyc_iter = cycle(iterable)
for _ in range(skip):
next(cyc_iter, None)
if length:
total_length = length
else:
total_length = len(iterable)
for _ in range(total_length):
yield next(cyc_iter, None)
>>> lst = [x for x in range(1, 9)]
# [1, 2, 3, 4, 5, 6, 7, 8]
>>> list(circular_iterator(lst, skip=3))
[4, 5, 6, 7, 8, 1, 2, 3]

Adding values from loop to a tuple results in nested tuples instead of a flat tuple or list

I'm trying to perform the following:
tup1 = ()
for i in range(1, 10, 2):
tup1 = (tup1, i)
print tup1
I expect the output to be the sequence 1 to 10.
However, I end up with the following:
((((((), 0), 2), 4), 6), 8)
What would be a correct way to meet the requirement?

If you just want an iterable with the even numbers 1 to 10 then the simplest way to do it:
seq = range(2, 11, 2)
If you are doing this as a means of learning Python and you want to build up your own data structure, use a list:
l = []
for i in range(2, 11, 2):
l.append(i)
The above for loop can be rewritten as a list comprehension:
l = [i for i in range(2, 11, 2)]
or using an if clause in the loop comprehension:
l = [ i for i in range(1, 11) if i % 2 == 0]

You can append an item to a tuple using the += operator.
tup1=()
for i in range(1,10,2):
tup1+= (i,)
print tup1
This prints (1, 3, 5, 7, 9)

Tuples are immutable objects in Python. Thus means you can't modify them. What you're doing right now is creating a new tuple with the previous one inside
You could do:
lst = []
for i in range(1,10,2):
lst.append(i)
tup = tuple(lst) #If you really want a tuple
print tup
But lst = range(1,10,2) or tup = tuple(range(1,10,2)) is much better (Unless you want to use append for some reason)

Read about List Comprehension
tuple(i for i in range(1, 10, 2))
Or
tup1 = ()
for i in range(1, 10, 2):
tup1 += (i,)
print tup1

it's something like this:
print range(1, 11)

You are skipping by two by using for i in range(1,10,2): if you use for i in range(1,11): if will increment by 1. As for tup1=(tup1,i) you are constantly adding a tuple to each other which is creating the weird output. You could use a list if you want to store them. Otherwise using will do it just fine:
print(range(10))

List item
For appending into list or tuple you can use append() function or you can use += operator which does the same.
s=()
for sequence of numbers from 1 to 10
for i in range(1,11):
s+=(i,)
print(s) #(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
for sequence of numbers from 1 to 10 with step size 2
x=()
for i in range(1,11,2):
x+=(i,)
print(x) #odd nos from 1-9 (1, 3, 5, 7, 9)
x=()
for i in range(2,11,2):
x+=(i,)
print(x) #even nos from 2-10 (2, 4, 6, 8, 10)
List item

Storing values from loop in a list or tuple in Python by following ways -
-> By appending the value in the list (here new_data1) as join will not work here.
new_data1 = []
for line in all_words:
new_data=' '.join(lemmatize_sentence(line))
new_data1.append(new_data)
#print (new_data)
print (new_data1)
P.S. - This is just a snapshot of a code just for hint .
Hope this helps!!

Create list [(1,1),(1,2), .. ,(1,n)], for any given n. Use this list then to create all fractions 1.0/k k from 1 to n

completely new to python tried doing this but did not work.
>>> list (range (1, 9))
[1, 2, 3, 4, 5, 6, 7, 8]
>>> args = [1,9]
>>> list (range (args))
Any help will be appreciated

create first list:
[(1, x) for x in range(1, n + 1)]
create second list:
[1.0 / x for x in range(1, n + 1)]

First, the list of (1,i) pairs:
pairs = [(1, x) for x in range(1, n + 1)]
Next, the list of ratios:
divs = [num/denom for num, denom in pairs]
(you may also need from __future__ import division if you're using python 2.x)

Python shuffle such that position will never repeat

I'd like to do a random shuffle of a list but with one condition: an element can never be in the same original position after the shuffle.
Is there a one line way to do such in python for a list?
Example:
list_ex = [1,2,3]
each of the following shuffled lists should have the same probability of being sampled after the shuffle:
list_ex_shuffled = [2,3,1]
list_ex_shuffled = [3,1,2]
but the permutations [1,2,3], [1,3,2], [2,1,3] and [3,2,1] are not allowed since all of them repeat one of the elements positions.
NOTE: Each element in the list_ex is a unique id. No repetition of the same element is allowed.

Randomize in a loop and keep rejecting the results until your condition is satisfied:
import random
def shuffle_list(some_list):
randomized_list = some_list[:]
while True:
random.shuffle(randomized_list)
for a, b in zip(some_list, randomized_list):
if a == b:
break
else:
return randomized_list

I'd describe such shuffles as 'permutations with no fixed points'. They're also known as derangements.
The probability that a random permutation is a derangement is approximately 1/e (fun to prove). This is true however long the list. Thus an obvious algorithm to give a random derangement is to shuffle the cards normally, and keep shuffling until you have a derangement. The expected number of necessary shuffles is about 3, and it's rare you'll have to shuffle more than ten times.
(1-1/e)**11 < 1%
Suppose there are n people at a party, each of whom brought an umbrella. At the end of the party, each person takes an umbrella at random from the basket. What is the probability that no-one holds their own umbrella?

You could generate all possible valid shufflings:
>>> list_ex = [1,2,3]
>>> import itertools
>>> list(itertools.ifilter(lambda p: not any(i1==i2 for i1,i2 in zip(list_ex, p)),
... itertools.permutations(list_ex, len(list_ex))))
[(2, 3, 1), (3, 1, 2)]
For some other sequence:
>>> list_ex = [7,8,9,0]
>>> list(itertools.ifilter(lambda p: not any(i1==i2 for i1,i2 in zip(list_ex, p)),
... itertools.permutations(list_ex, len(list_ex))))
[(8, 7, 0, 9), (8, 9, 0, 7), (8, 0, 7, 9), (9, 7, 0, 8), (9, 0, 7, 8), (9, 0, 8, 7), (0, 7, 8, 9), (0, 9, 7, 8), (0, 9, 8, 7)]
You could also make this a bit more efficient by short-circuiting the iterator if you just want one result:
>>> list_ex = [1,2,3]
>>> i = itertools.ifilter(lambda p: not any(i1==i2 for i1,i2 in zip(list_ex, p)),
... itertools.permutations(list_ex, len(list_ex)))
>>> next(i)
(2, 3, 1)
But, it would not be a random choice. You'd have to generate all of them and choose one for it to be an actual random result:
>>> list_ex = [1,2,3]
>>> i = itertools.ifilter(lambda p: not any(i1==i2 for i1,i2 in zip(list_ex, p)),
... itertools.permutations(list_ex, len(list_ex)))
>>> import random
>>> random.choice(list(i))
(2, 3, 1)

Here is another take on this. You can pick one solution or another depending on your needs. This is not a one liner but shuffles the indices of elements instead of the elements themselves. Thus, the original list may have duplicate values or values of types that cannot be compared or may be expensive to compare.
#! /usr/bin/env python
import random
def shuffled_values(data):
list_length = len(data)
candidate = range(list_length)
while True:
random.shuffle(candidate)
if not any(i==j for i,j in zip(candidate, range(list_length))):
yield [data[i] for i in candidate]
list_ex = [1, 2, 3]
list_gen = shuffled_values(list_ex)
for i in range(0, 10):
print list_gen.next()
This gives:
[2, 3, 1]
[3, 1, 2]
[3, 1, 2]
[2, 3, 1]
[3, 1, 2]
[3, 1, 2]
[2, 3, 1]
[2, 3, 1]
[3, 1, 2]
[2, 3, 1]
If list_ex is [2, 2, 2], this method will keep yielding [2, 2, 2] over and over. The other solutions will give you empty lists. I am not sure what you want in this case.

Use Knuth-Durstenfeld to shuffle the list. As long as it is found to be in the original position during the shuffling process, a new shuffling process is started from the beginning until it returns to a qualified arrangement. The time complexity of this algorithm is the smallest constant term:
def _random_derangement(x: list, randint: Callable[[int, int], int]) -> None:
'''
Random derangement list x in place, and return None.
An element can never be in the same original position after the shuffle. provides uniform distribution over permutations.
The formal parameter randint requires a callable object such as rand_int(b, a) that generates a random integer within the specified closed interval.
'''
from collections import namedtuple
sequence_type = namedtuple('sequence_type', ('sequence_number', 'elem'))
x_length = len(x)
if x_length > 1:
for i in range(x_length):
x[i] = sequence_type(sequence_number = i, elem = x[i])
end_label = x_length - 1
while True:
for i in range(end_label, 0, -1):
random_location = randint(i, 0)
if x[random_location].sequence_number != i:
x[i], x[random_location] = x[random_location], x[i]
else:
break
else:
if x[0].sequence_number != 0: break
for i in range(x_length):
x[i] = x[i].elem
complete_shuffle

Here's another algorithm. Take cards at random. If your ith card is card i, put it back and try again. Only problem, what if when you get to the last card it's the one you don't want. Swap it with one of the others.
I think this is fair (uniformally random).
import random
def permutation_without_fixed_points(n):
if n == 1:
raise ArgumentError, "n must be greater than 1"
result = []
remaining = range(n)
i = 0
while remaining:
if remaining == [n-1]:
break
x = i
while x == i:
j = random.randrange(len(remaining))
x = remaining[j]
remaining.pop(j)
result.append(x)
i += 1
if remaining == [n-1]:
j = random.randrange(n-1)
result.append(result[j])
result[j] = n
return result