'How do I create a list of random numbers without duplicates?
I tried using random.randint(0, 100), but some numbers were the same. Is there a method/module to create a list unique random numbers?
Note: The following code is based on an answer and has been added after the answer was posted. It is not a part of the question; it is the solution.
def getScores():
# open files to read and write
f1 = open("page.txt", "r");
p1 = open("pgRes.txt", "a");
gScores = [];
bScores = [];
yScores = [];
# run 50 tests of 40 random queries to implement "bootstrapping" method
for i in range(50):
# get 40 random queries from the 50
lines = random.sample(f1.readlines(), 40);
Solution 1:[1]
You can use the shuffle function from the random module like this:
import random
my_list = list(xrange(1,100)) # list of integers from 1 to 99
# adjust this boundaries to fit your needs
random.shuffle(my_list)
print my_list # <- List of unique random numbers
Note here that the shuffle method doesn't return any list as one may expect, it only shuffle the list passed by reference.
Solution 2:[2]
You can first create a list of numbers from a to b, where a and b are respectively the smallest and greatest numbers in your list, then shuffle it with Fisher-Yates algorithm or using the Python's random.shuffle method.
Solution 3:[3]
The solution presented in this answer works, but it could become problematic with memory if the sample size is small, but the population is huge (e.g. random.sample(insanelyLargeNumber, 10)).
To fix that, I would go with this:
answer = set()
sampleSize = 10
answerSize = 0
while answerSize < sampleSize:
r = random.randint(0,100)
if r not in answer:
answerSize += 1
answer.add(r)
# answer now contains 10 unique, random integers from 0.. 100
Solution 4:[4]
Linear Congruential Pseudo-random Number Generator
O(1) Memory
O(k) Operations
This problem can be solved with a simple Linear Congruential Generator. This requires constant memory overhead (8 integers) and at most 2*(sequence length) computations.
All other solutions use more memory and more compute! If you only need a few random sequences, this method will be significantly cheaper. For ranges of size N, if you want to generate on the order of N unique k-sequences or more, I recommend the accepted solution using the builtin methods random.sample(range(N),k) as this has been optimized in python for speed.
Code
# Return a randomized "range" using a Linear Congruential Generator
# to produce the number sequence. Parameters are the same as for
# python builtin "range".
# Memory -- storage for 8 integers, regardless of parameters.
# Compute -- at most 2*"maximum" steps required to generate sequence.
#
def random_range(start, stop=None, step=None):
import random, math
# Set a default values the same way "range" does.
if (stop == None): start, stop = 0, start
if (step == None): step = 1
# Use a mapping to convert a standard range into the desired range.
mapping = lambda i: (i*step) + start
# Compute the number of numbers in this range.
maximum = (stop - start) // step
# Seed range with a random integer.
value = random.randint(0,maximum)
#
# Construct an offset, multiplier, and modulus for a linear
# congruential generator. These generators are cyclic and
# non-repeating when they maintain the properties:
#
# 1) "modulus" and "offset" are relatively prime.
# 2) ["multiplier" - 1] is divisible by all prime factors of "modulus".
# 3) ["multiplier" - 1] is divisible by 4 if "modulus" is divisible by 4.
#
offset = random.randint(0,maximum) * 2 + 1 # Pick a random odd-valued offset.
multiplier = 4*(maximum//4) + 1 # Pick a multiplier 1 greater than a multiple of 4.
modulus = int(2**math.ceil(math.log2(maximum))) # Pick a modulus just big enough to generate all numbers (power of 2).
# Track how many random numbers have been returned.
found = 0
while found < maximum:
# If this is a valid value, yield it in generator fashion.
if value < maximum:
found += 1
yield mapping(value)
# Calculate the next value in the sequence.
value = (value*multiplier + offset) % modulus
Usage
The usage of this function "random_range" is the same as for any generator (like "range"). An example:
# Show off random range.
print()
for v in range(3,6):
v = 2**v
l = list(random_range(v))
print("Need",v,"found",len(set(l)),"(min,max)",(min(l),max(l)))
print("",l)
print()
Sample Results
Required 8 cycles to generate a sequence of 8 values.
Need 8 found 8 (min,max) (0, 7)
[1, 0, 7, 6, 5, 4, 3, 2]
Required 16 cycles to generate a sequence of 9 values.
Need 9 found 9 (min,max) (0, 8)
[3, 5, 8, 7, 2, 6, 0, 1, 4]
Required 16 cycles to generate a sequence of 16 values.
Need 16 found 16 (min,max) (0, 15)
[5, 14, 11, 8, 3, 2, 13, 1, 0, 6, 9, 4, 7, 12, 10, 15]
Required 32 cycles to generate a sequence of 17 values.
Need 17 found 17 (min,max) (0, 16)
[12, 6, 16, 15, 10, 3, 14, 5, 11, 13, 0, 1, 4, 8, 7, 2, ...]
Required 32 cycles to generate a sequence of 32 values.
Need 32 found 32 (min,max) (0, 31)
[19, 15, 1, 6, 10, 7, 0, 28, 23, 24, 31, 17, 22, 20, 9, ...]
Required 64 cycles to generate a sequence of 33 values.
Need 33 found 33 (min,max) (0, 32)
[11, 13, 0, 8, 2, 9, 27, 6, 29, 16, 15, 10, 3, 14, 5, 24, ...]
Solution 5:[5]
If you need to sample extremely large numbers, you cannot use range
random.sample(range(10000000000000000000000000000000), 10)
because it throws:
OverflowError: Python int too large to convert to C ssize_t
Also, if random.sample cannot produce the number of items you want due to the range being too small
random.sample(range(2), 1000)
it throws:
ValueError: Sample larger than population
This function resolves both problems:
import random
def random_sample(count, start, stop, step=1):
def gen_random():
while True:
yield random.randrange(start, stop, step)
def gen_n_unique(source, n):
seen = set()
seenadd = seen.add
for i in (i for i in source() if i not in seen and not seenadd(i)):
yield i
if len(seen) == n:
break
return [i for i in gen_n_unique(gen_random,
min(count, int(abs(stop - start) / abs(step))))]
Usage with extremely large numbers:
print('\n'.join(map(str, random_sample(10, 2, 10000000000000000000000000000000))))
Sample result:
7822019936001013053229712669368
6289033704329783896566642145909
2473484300603494430244265004275
5842266362922067540967510912174
6775107889200427514968714189847
9674137095837778645652621150351
9969632214348349234653730196586
1397846105816635294077965449171
3911263633583030536971422042360
9864578596169364050929858013943
Usage where the range is smaller than the number of requested items:
print(', '.join(map(str, random_sample(100000, 0, 3))))
Sample result:
2, 0, 1
It also works with with negative ranges and steps:
print(', '.join(map(str, random_sample(10, 10, -10, -2))))
print(', '.join(map(str, random_sample(10, 5, -5, -2))))
Sample results:
2, -8, 6, -2, -4, 0, 4, 10, -6, 8
-3, 1, 5, -1, 3
Solution 6:[6]
If the list of N numbers from 1 to N is randomly generated, then yes, there is a possibility that some numbers may be repeated.
If you want a list of numbers from 1 to N in a random order, fill an array with integers from 1 to N, and then use a Fisher-Yates shuffle or Python's random.shuffle().
Solution 7:[7]
Here is a very small function I made, hope this helps!
import random
numbers = list(range(0, 100))
random.shuffle(numbers)
Solution 8:[8]
A very simple function that also solves your problem
from random import randint
data = []
def unique_rand(inicial, limit, total):
data = []
i = 0
while i < total:
number = randint(inicial, limit)
if number not in data:
data.append(number)
i += 1
return data
data = unique_rand(1, 60, 6)
print(data)
"""
prints something like
[34, 45, 2, 36, 25, 32]
"""
Solution 9:[9]
The answer provided here works very well with respect to time as well as memory but a bit more complicated as it uses advanced python constructs such as yield. The simpler answer works well in practice but, the issue with that answer is that it may generate many spurious integers before actually constructing the required set. Try it out with populationSize = 1000, sampleSize = 999. In theory, there is a chance that it doesn't terminate.
The answer below addresses both issues, as it is deterministic and somewhat efficient though currently not as efficient as the other two.
def randomSample(populationSize, sampleSize):
populationStr = str(populationSize)
dTree, samples = {}, []
for i in range(sampleSize):
val, dTree = getElem(populationStr, dTree, '')
samples.append(int(val))
return samples, dTree
where the functions getElem, percolateUp are as defined below
import random
def getElem(populationStr, dTree, key):
msd = int(populationStr[0])
if not key in dTree.keys():
dTree[key] = range(msd + 1)
idx = random.randint(0, len(dTree[key]) - 1)
key = key + str(dTree[key][idx])
if len(populationStr) == 1:
dTree[key[:-1]].pop(idx)
return key, (percolateUp(dTree, key[:-1]))
newPopulation = populationStr[1:]
if int(key[-1]) != msd:
newPopulation = str(10**(len(newPopulation)) - 1)
return getElem(newPopulation, dTree, key)
def percolateUp(dTree, key):
while (dTree[key] == []):
dTree[key[:-1]].remove( int(key[-1]) )
key = key[:-1]
return dTree
Finally, the timing on average was about 15ms for a large value of n as shown below,
In [3]: n = 10000000000000000000000000000000
In [4]: %time l,t = randomSample(n, 5)
Wall time: 15 ms
In [5]: l
Out[5]:
[10000000000000000000000000000000L,
5731058186417515132221063394952L,
85813091721736310254927217189L,
6349042316505875821781301073204L,
2356846126709988590164624736328L]
Solution 10:[10]
In order to obtain a program that generates a list of random values without duplicates that is deterministic, efficient and built with basic programming constructs consider the function extractSamples defined below,
def extractSamples(populationSize, sampleSize, intervalLst) :
import random
if (sampleSize > populationSize) :
raise ValueError("sampleSize = "+str(sampleSize) +" > populationSize (= " + str(populationSize) + ")")
samples = []
while (len(samples) < sampleSize) :
i = random.randint(0, (len(intervalLst)-1))
(a,b) = intervalLst[i]
sample = random.randint(a,b)
if (a==b) :
intervalLst.pop(i)
elif (a == sample) : # shorten beginning of interval
intervalLst[i] = (sample+1, b)
elif ( sample == b) : # shorten interval end
intervalLst[i] = (a, sample - 1)
else :
intervalLst[i] = (a, sample - 1)
intervalLst.append((sample+1, b))
samples.append(sample)
return samples
The basic idea is to keep track of intervals intervalLst for possible values from which to select our required elements from. This is deterministic in the sense that we are guaranteed to generate a sample within a fixed number of steps (solely dependent on populationSize and sampleSize).
To use the above function to generate our required list,
In [3]: populationSize, sampleSize = 10**17, 10**5
In [4]: %time lst1 = extractSamples(populationSize, sampleSize, [(0, populationSize-1)])
CPU times: user 289 ms, sys: 9.96 ms, total: 299 ms
Wall time: 293 ms
We may also compare with an earlier solution (for a lower value of populationSize)
In [5]: populationSize, sampleSize = 10**8, 10**5
In [6]: %time lst = random.sample(range(populationSize), sampleSize)
CPU times: user 1.89 s, sys: 299 ms, total: 2.19 s
Wall time: 2.18 s
In [7]: %time lst1 = extractSamples(populationSize, sampleSize, [(0, populationSize-1)])
CPU times: user 449 ms, sys: 8.92 ms, total: 458 ms
Wall time: 442 ms
Note that I reduced populationSize value as it produces Memory Error for higher values when using the random.sample solution (also mentioned in previous answers here and here). For above values, we can also observe that extractSamples outperforms the random.sample approach.
P.S. : Though the core approach is similar to my earlier answer, there are substantial modifications in implementation as well as approach alongwith improvement in clarity.
Solution 11:[11]
The problem with the set based approaches ("if random value in return values, try again") is that their runtime is undetermined due to collisions (which require another "try again" iteration), especially when a large amount of random values are returned from the range.
An alternative that isn't prone to this non-deterministic runtime is the following:
import bisect
import random
def fast_sample(low, high, num):
""" Samples :param num: integer numbers in range of
[:param low:, :param high:) without replacement
by maintaining a list of ranges of values that
are permitted.
This list of ranges is used to map a random number
of a contiguous a range (`r_n`) to a permissible
number `r` (from `ranges`).
"""
ranges = [high]
high_ = high - 1
while len(ranges) - 1 < num:
# generate a random number from an ever decreasing
# contiguous range (which we'll map to the true
# random number).
# consider an example with low=0, high=10,
# part way through this loop with:
#
# ranges = [0, 2, 3, 7, 9, 10]
#
# r_n :-> r
# 0 :-> 1
# 1 :-> 4
# 2 :-> 5
# 3 :-> 6
# 4 :-> 8
r_n = random.randint(low, high_)
range_index = bisect.bisect_left(ranges, r_n)
r = r_n + range_index
for i in xrange(range_index, len(ranges)):
if ranges[i] <= r:
# as many "gaps" we iterate over, as much
# is the true random value (`r`) shifted.
r = r_n + i + 1
elif ranges[i] > r_n:
break
# mark `r` as another "gap" of the original
# [low, high) range.
ranges.insert(i, r)
# Fewer values possible.
high_ -= 1
# `ranges` happens to contain the result.
return ranges[:-1]
Solution 12:[12]
I found a quite faster way than having to use the range function (very slow), and without using random function from python (I don´t like the random built-in library because when you seed it, it repeats the pattern of the random numbers generator)
import numpy as np
nums = set(np.random.randint(low=0, high=100, size=150)) #generate some more for the duplicates
nums = list(nums)[:100]
This is quite fast.
Solution 13:[13]
You can use Numpy library for quick answer as shown below -
Given code snippet lists down 6 unique numbers between the range of 0 to 5. You can adjust the parameters for your comfort.
import numpy as np
import random
a = np.linspace( 0, 5, 6 )
random.shuffle(a)
print(a)
Output
[ 2. 1. 5. 3. 4. 0.]
It doesn't put any constraints as we see in random.sample as referred here.
Solution 14:[14]
One straightforward alternative is to use np.random.choice() as shown below
np.random.choice(range(10), size=3, replace=False)
This results in three integer numbers that are different from each other. e.g., [1, 3, 5], [2, 5, 1]...
Solution 15:[15]
import random
sourcelist=[]
resultlist=[]
for x in range(100):
sourcelist.append(x)
for y in sourcelist:
resultlist.insert(random.randint(0,len(resultlist)),y)
print (resultlist)
Solution 16:[16]
Try using...
import random
LENGTH = 100
random_with_possible_duplicates = [random.randrange(-3, 3) for _ in range(LENGTH)]
random_without_duplicates = list(set(random_with_possible_duplicates)) # This removes duplicates
Advatages
Fast, efficient and readable.
Possible Issues
This method can change the length of the list if there are duplicates.
Solution 17:[17]
If you wish to ensure that the numbers being added are unique, you could use a Set object
if using 2.7 or greater, or import the sets module if not.
As others have mentioned, this means the numbers are not truly random.
Solution 18:[18]
If the amount of numbers you want is random, you can do something like this. In this case, length is the highest number you want to choose from.
If it notices the new random number was already chosen, itll subtract 1 from count (since a count was added before it knew whether it was a duplicate or not). If its not in the list, then do what you want with it and add it to the list so it cant get picked again.
import random
def randomizer():
chosen_number=[]
count=0
user_input = int(input("Enter number for how many rows to randomly select: "))
numlist=[]
#length = whatever the highest number you want to choose from
while 1<=user_input<=length:
count=count+1
if count>user_input:
break
else:
chosen_number = random.randint(0, length)
if line_number in numlist:
count=count-1
continue
if chosen_number not in numlist:
numlist.append(chosen_number)
#do what you want here
Solution 19:[19]
Edit: ignore my answer here. use python's random.shuffle or random.sample, as mentioned in other answers.
import numpy as np
minval, maxval, n_samples = -50, 50, 10
generator = np.random.default_rng(seed=0)
samples = generator.permutation(np.arange(minval, maxval))[:n_samples]
# or, if minval is 0,
samples = generator.permutation(maxval)[:n_samples]
with jax:
import jax
minval, maxval, n_samples = -50, 50, 10
key = jax.random.PRNGKey(seed=0)
samples = jax.random.shuffle(key, jax.numpy.arange(minval, maxval))[:n_samples]
Sources
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