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how to generate 100000 random numbers in python [duplicate]

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?

This will return a list of 10 numbers selected from the range 0 to 99, without duplicates.
import random
random.sample(range(100), 10)

You can use the shuffle function from the random module like this:
import random
nums = list(range(1, 100)) # list of integers from 1 to 99
# adjust this boundaries to fit your needs
random.shuffle(nums)
print(nums) # <- 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.

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.

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, ...]

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

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

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().

Here is a very small function I made, hope this helps!
import random
numbers = list(range(0, 100))
random.shuffle(numbers)

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]
"""

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]...

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]

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.

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]

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.

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.

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)

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.

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.

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

Edit: ignore my answer here. use python's random.shuffle or random.sample, as mentioned in other answers.
to sample integers without replacement between `minval` and `maxval`:
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]

From the CLI in win xp:
python -c "import random; print(sorted(set([random.randint(6,49) for i in range(7)]))[:6])"
In Canada we have the 6/49 Lotto. I just wrap the above code in lotto.bat and run C:\home\lotto.bat or just C:\home\lotto.
Because random.randint often repeats a number, I use set with range(7) and then shorten it to a length of 6.
Occasionally if a number repeats more than 2 times the resulting list length will be less than 6.
EDIT: However, random.sample(range(6,49),6) is the correct way to go.

How can I make all possible list combinations using reversible seeds?

I want to have a list made out of 100 items, each item being a value from 0-31 inclusive. I then want to be able to take one of these lists, and know what seed/input is necessary to randomly generate that exact list.

Using some suitable Linear Congruential Generator:
You could use this research paper by Pierre L'Ecuyer:
Tables_of_linear_congruential_generators_of_different_sizes_and_good_lattice_structure
The lowest power of 2 modulus for which the paper gives examples of (decently pseudo-random) LCGs is 230, which is close to 1 billion. See table 4 of the paper. Just pick one of those LCGs, say:
un+1 = ((116646453 * un) + 5437) mod 230
Each of your items is exactly 5 bits wide. If you decide to group your items 6 by 6, each group is exactly 30 bits wide, so can be considered as one state of this modulo 230 LCG.
From a initial group of 6 items, one step of the LCG will generate the next group, that is the next 6 items. And the paper tells you that the serie will look reasonably random overall.
Hence, you can regard the first 6 items as your “seed”, as you can reconstruct the whole list from its leftmost 6 items.
Even assuming that for the sake of obfuscation you started the visible part of the list after the seed, you would still have only about one billion possible seeds to worry about. Any decent computer would be able to find the left-hidden seed by brute force within a handful of seconds, by simulating the LCG for every possible seed and comparing with the actual list.
Sample Python code:
One can start by creating a class that, given a seed, supplies an unlimited serie of items between 0 and 31:
class LEcuyer30:
def __init__ (self, seed):
self.seed = seed & ((1 << 30) - 1)
self.currGroup = self.seed
self.itemCount = 6
def __toNextGroup(self):
nextGroup = ((116646453 * self.currGroup) + 5437) % (1 << 30)
self.currGroup = nextGroup
self.itemCount = 6
def getItem(self):
if (self.itemCount <= 0):
self.__toNextGroup()
# extract 5 bits:
word = self.currGroup >> (5 * (self.itemCount - 1))
self.itemCount -= 1
return (word & 31)
Test code:
We can create a sequence of 20 items and print it:
# Main program:
randomSeed = 514703103
rng = LEcuyer30(randomSeed)
itemList = []
for k in range(20):
item = rng.getItem()
itemList.append(item)
print("itemList = %s" % itemList)
Program output:
itemList = [15, 10, 27, 15, 23, 31, 1, 10, 5, 15, 16, 8, 4, 16, 24, 31, 7, 5, 8, 19]

Python randint generates numbers with the same number of digits

for i in range(number_of_numbers):
randomNumber = randint(1, pow(10, numberPower))
numbers.append(randomNumber)
numbersIncreasing.append(randomNumber)
So I'm making a little math game that generates n amount of numbers from user input. This is the code that generates the numbers and appends them to a list. The only problem is, is it gives numbers with the same amount of digits, which is whatever numberPower is set to (12). An example would be two numbers like 9,564,713,666,123 or 1,256,286,233,753, and all the numbers have 12 digits. I was expecting more very mixed numbers, that could literally be anywhere between 1 and 10^12.

This is because most numbers between 1 and 10**12 have 12 digits. Specifically, 90% of numbers between 1 and 10^12 have 12 digits. Then 9% have 11 digits; only 1% have 10 digits or less.
9 numbers have 1 digit (1-9);
90 numbers have 2 digits (10-99);
900 numbers have 3 digits (100-999);
9000 numbers have 4 digits (1000-9999);
...
9 * 10**10 numbers have 11 digits (10000000000-99999999999);
9 * 10**11 numbers have 12 digits (100000000000-999999999999);
1 number has 13 digits (1000000000000).
If you want, you can try to draw a histogram showing how many numbers have how many digits, when randomly generating a large number of numbers between 1 and 10**12:
import random
import matplotlib.pyplot as plt
X = [random.randint(1,10**12) for _ in range(10000)]
nb_digits = [len(str(x)) for x in X]
plt.hist(nb_digits, bins=range(1,14), density=True)
plt.show()
Another way to think about it: leading zeroes
Generating a number uniformly at random between 0 and 10**12 - 1 is the same as generating 12 digits uniformly at random between 0 and 9, and concatenating them to write a number in base 10.
The "number of digits" of the resulting number will be 12 minus the number of leading zeroes.
For instance, the string 000000049608 counts as a 5-digit number, because the first 7 digits are zeroes.
Thus, if the leftmost digit generated is not a 0, then the resulting number will count as a 12-digit number. This happens whenever the first digit is 1,2,3,4,5,6,7,8 or 9, which is 90% of the time.
What is the probability to get a number with 5 digits or less, like above? It's the probability that the leftmost 7 digits are all zeroes. The probability that this happens is (1/10)**7 = 0.00001%. As you can see, most numbers don't have a lot of leading zeroes.
If you want to skew the distribution to get more diversity in number of digits
Maybe you don't want your random numbers to follow a uniform distribution? Try this:
import random
number_of_numbers = 100
numberPower = 12
numbers = []
for _ in range(100):
randomNumberOfDigits = random.randint(1, numberPower)
randomNumber = random.randint(10**(randomNumberOfDigits - 1), 10**randomNumberOfDigits - 1)
numbers.append(randomNumber)
print(numbers)
# [4, 42, 6004, 2, 46111, 2023179648, 41991462, 525, 2412, 221700769, 47, 5304427, 102, 23, 54323975, 906, 894363079057, 556, 675354, 9365725, 838, 490947621, 82648961, 9232330, 2885693, 28889965, 31601, 505250081, 534874, 85, 4630, 5, 8088582, 67921960, 722774315903, 16, 33, 7752943, 16628713055, 74, 2561429212, 67626556043, 155962, 596, 8655352595, 4735, 7, 8296258, 17722, 78, 93044214350, 17247469, 58295330, 622, 332, 450232321717, 70292459, 11, 447064711120, 895793, 8, 540110, 43032518210, 7, 20361, 606, 9, 9943927, 983, 54327, 40473362991, 2358, 7102932592, 15960846, 99385316379, 88654316391, 219692155, 3732404, 5474, 6343805291, 74, 844532493412, 2, 45228, 186785825, 10704, 4833, 744164008354, 976046952, 7836, 49344611762, 34985277, 362, 49577, 868571104, 237506, 2, 920019513002, 83847, 77908803]
Note: I chose to write ** for power instead of ^, because we're illustrating with python code and in python, ** means power while ^ means something entirely different.

find if a number divisible by the input numbers

Given two numbers a and b, we have to find the nth number which is divisible by a or b.
The format looks like below:
Input :
First line consists of an integer T, denoting the number of test cases.
Second line contains three integers a, b and N
Output :
For each test case, print the Nth
number in a new line.
Constraints :
1≤t≤105
1≤a,b≤104
1≤N≤10
Sample Input
1
2 3 10
Sample Output
15
Explanation
The numbers which are divisible by 2
or 3 are: 2,3,4,6,8,9,10,12,14,15 and the 10th number is 15
My code
test_case=input()
if int(test_case)<=100000 and int(test_case)>=1:
for p in range(int(test_case)):
count=1
j=1
inp=list(map(int,input().strip('').split()))
if inp[0]<=10000 and inp[0]>=1 and inp[1]<=10000 and inp[1]>=1 and inp[1]<=1000000000 and inp[1]>=1:
while(True ):
if count<=inp[2] :
k=j
if j%inp[0]==0 or j%inp[1] ==0:
count=count+1
j=j+1
else :
j=j+1
else:
break
print(k)
else:
break
Problem Statement:
For single test case input 2000 3000 100000 it is taking more than one second to complete.I want if i can get the results in less than 1 second. Is there a time efficient approach to this problem,may be if we can use some data structure and algorithms here??

For every two numbers there will be number k such that k=a*b. There will only be so many multiples of a and b under k. This set can be created like so:
s = set(a*1, b*1, ... a*(b-1), b*(a-1), a*b)
Say we take the values a=2, b=3 then s = (2,3,4,6). These are the possible values of c:
[1 - 4] => (2,3,4,6)
[5 - 8] => 6 + (2,3,4,6)
[9 - 12] => 6*2 + (2,3,4,6)
...
Notice that the values repeat with a predictable pattern. To get the row you can take the value of c and divide by length of the set s (call it n). The set index is the mod of c by n. Subtract 1 for 1 indexing used in the problem.
row = floor((c-1)/n)
column = `(c-1) % n`
result = (a*b)*row + s(column)
Python impl:
a = 2000
b = 3000
c = 100000
s = list(set([a*i for i in range(1, b+1)] + [b*i for i in range(1, a+1)]))
print((((c-1)//len(s)) * (a*b)) + s[(c - 1)%len(s)])

I'm not certain to grasp exactly what you're trying to accomplish. But if I get it right, isn't the answer simply b*(N/2)? since you are listing the multiples of both numbers the Nth will always be the second you list times N/2.
In your initial example that would be 3*10/2=15.
In the code example, it would be 3000*100000/2=150'000'000
Update:
Code to compute the desired values using set's and lists to speed up the calculation process. I'm still wondering what the recurrence for the odd indexes could be if anyone happens to stumble upon it...
a = 2000
b = 3000
c = 100000
a_list = [a*x for x in range(1, c)]
b_list = [b*x for x in range(1, c)]
nums = set(a_list)
nums.update(b_list)
nums = sorted(nums)
print(nums[c-1])
This code runs in 0.14s on my laptop. Which is significantly below the requested threshold. Nonetheless, this values will depend on the machine the code is run on.

Memory efficient way to read an array of integers from single line of input in python2.7

I want to read a single line of input containing integers separated by spaces.
Currently I use the following.
A = map(int, raw_input().split())
But now the N is around 10^5 and I don't need the whole array of integers, I just need to read them 1 at a time, in the same sequence as the input.
Can you suggest an efficient way to do this in Python2.7

Use generators:
numbers = '1 2 5 18 10 12 16 17 22 50'
gen = (int(x) for x in numbers.split())
for g in gen:
print g
1
5
6
8
10
12
68
13
the generator object would use one item at a time, and won't construct a whole list.

You could parse the data a character at a time, this would reduce memory usage:
data = "1 50 30 1000 20 4 1 2"
number = []
numbers = []
for c in data:
if c == ' ':
if number:
numbers.append(int(''.join(number)))
number = []
else:
number.append(c)
if number:
numbers.append(int(''.join(number)))
print numbers
Giving you:
[1, 50, 30, 1000, 20, 4, 1, 2]
Probably quite a bit slower though.
Alternatively, you could use itertools.groupby() to read groups of digits as follows:
from itertools import groupby
data = "1 50 30 1000 20 4 1 2"
numbers = []
for k, g in groupby(data, lambda c: c.isdigit()):
if k:
numbers.append(int(''.join(g)))
print numbers

If you're able to destroy the original string, split accepts a parameter for the maximum number of breaks.
See docs for more details and examples.