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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.
I have a set of randomized float values that are to be arranged into an array at the end of each loop that produces 67 of them, however, there are 64 total loops.
As an example, if I had 4 values per loop and 3 total loops of integers, I would like it to be like this:
values = [[0, 4, 5, 1],[6, 6, 5, 3],[0,0,0,7]]
such that I could identify them as separate arrays, however, I am unsure of the best way to append the values after they are created, but am aware of how to return them. Forgive me as I am unskilled with the logic.
import math
import random
funcs = []
coord = []
pi = math.pi
funcAmt = 0
coordAmt = 0
repeatAmt = 0
coordPass = 0
while funcAmt < 64:
while coordAmt < 67:
coordAmt += 1
uniform = round(random.uniform(-pi, pi), 2)
print("Coord [",coordAmt,"] {",uniform,"} Func:", funcAmt + 1)
if uniform in coord:
repeatAmt += 1
print("Repeat Found!")
coordAmt -= 1
print("Repeat [",repeatAmt,"] Resolved")
pass
else:
coordPass += 1
coord.append(uniform)
#<<<Append Here>>>
funcAmt += 1
coord.clear()
coordAmt = 0
In my given code above, it would be similar to:
func = [
[<67 items>],
...63 more times
]
Your "append here" logic should append the coordinate list and then clear that list for the next iteration of the outer loop:
funcs.append(coord[:]) # The slice notation makes a copy of the list
coord.clear() # or simply coord = []
You should learn to use a for loop. This will simplify your looping: you don't have to maintain the counts yourself. For instance:
for funcAmt in range(64):
for coordAmt in range(67):
...
You might also look up how to make a "list comprehension", which can reduce your process to a single line of code -- a long, involved line, but readable with proper white space.
Does that get you moving?
There are a couple of ways around this. Instead of using while lists and counters, you could just use for loops. Or at least do that for the outer loop, since it looks like you still want to check for repeats. Here's an example using your original dimensions of 3 and 4:
from math import pi
import random
coord_sets = 3
coords = 4
biglist = []
for i in range(coord_sets):
coords_set = []
non_repeating_coords = 0
while non_repeating_coords < coords:
new_coord = round(random.uniform(-1.0*pi, pi), 2)
if new_coord not in coords_set:
coords_set.append(new_coord)
non_repeating_coords += 1
biglist.append(coords_set)
print(biglist)
You can use sets because they don't allow duplicate values:
from math import pi
import random
funcs = []
funcAmt = 0
while funcAmt < 64: # This is the number of loops
myset = set()
while len(myset) < 67: # This is the length of each set
uniform = round(random.uniform(-pi, pi), 2)
myset.add(uniform)
funcs.append(list(myset)) # Append randomly generated set as a list
funcAmt += 1
print(funcs)
maybe you can benefit from arrays in numpy:
import numpy as np
funcs = np.random.uniform(-np.pi, np.pi, [63, 67])
This creates an array of shape (63, 67) from uniform random between -pi to pi.
I had an interview with a hedge fund company in New York a few months ago and unfortunately, I did not get the internship offer as a data/software engineer. (They also asked the solution to be in Python.)
I pretty much screwed up on the first interview problem...
Question: Given a string of a million numbers (Pi for example), write
a function/program that returns all repeating 3 digit numbers and number of
repetition greater than 1
For example: if the string was: 123412345123456 then the function/program would return:
123 - 3 times
234 - 3 times
345 - 2 times
They did not give me the solution after I failed the interview, but they did tell me that the time complexity for the solution was constant of 1000 since all the possible outcomes are between:
000 --> 999
Now that I'm thinking about it, I don't think it's possible to come up with a constant time algorithm. Is it?
You got off lightly, you probably don't want to be working for a hedge fund where the quants don't understand basic algorithms :-)
There is no way to process an arbitrarily-sized data structure in O(1) if, as in this case, you need to visit every element at least once. The best you can hope for is O(n) in this case, where n is the length of the string.
Although, as an aside, a nominal O(n) algorithm will be O(1) for a fixed input size so, technically, they may have been correct here. However, that's not usually how people use complexity analysis.
It appears to me you could have impressed them in a number of ways.
First, by informing them that it's not possible to do it in O(1), unless you use the "suspect" reasoning given above.
Second, by showing your elite skills by providing Pythonic code such as:
inpStr = '123412345123456'
# O(1) array creation.
freq = [0] * 1000
# O(n) string processing.
for val in [int(inpStr[pos:pos+3]) for pos in range(len(inpStr) - 2)]:
freq[val] += 1
# O(1) output of relevant array values.
print ([(num, freq[num]) for num in range(1000) if freq[num] > 1])
This outputs:
[(123, 3), (234, 3), (345, 2)]
though you could, of course, modify the output format to anything you desire.
And, finally, by telling them there's almost certainly no problem with an O(n) solution, since the code above delivers results for a one-million-digit string in well under half a second. It seems to scale quite linearly as well, since a 10,000,000-character string takes 3.5 seconds and a 100,000,000-character one takes 36 seconds.
And, if they need better than that, there are ways to parallelise this sort of stuff that can greatly speed it up.
Not within a single Python interpreter of course, due to the GIL, but you could split the string into something like (overlap indicated by vv is required to allow proper processing of the boundary areas):
vv
123412 vv
123451
5123456
You can farm these out to separate workers and combine the results afterwards.
The splitting of input and combining of output are likely to swamp any saving with small strings (and possibly even million-digit strings) but, for much larger data sets, it may well make a difference. My usual mantra of "measure, don't guess" applies here, of course.
This mantra also applies to other possibilities, such as bypassing Python altogether and using a different language which may be faster.
For example, the following C code, running on the same hardware as the earlier Python code, handles a hundred million digits in 0.6 seconds, roughly the same amount of time as the Python code processed one million. In other words, much faster:
#include <stdio.h>
#include <string.h>
int main(void) {
static char inpStr[100000000+1];
static int freq[1000];
// Set up test data.
memset(inpStr, '1', sizeof(inpStr));
inpStr[sizeof(inpStr)-1] = '\0';
// Need at least three digits to do anything useful.
if (strlen(inpStr) <= 2) return 0;
// Get initial feed from first two digits, process others.
int val = (inpStr[0] - '0') * 10 + inpStr[1] - '0';
char *inpPtr = &(inpStr[2]);
while (*inpPtr != '\0') {
// Remove hundreds, add next digit as units, adjust table.
val = (val % 100) * 10 + *inpPtr++ - '0';
freq[val]++;
}
// Output (relevant part of) table.
for (int i = 0; i < 1000; ++i)
if (freq[i] > 1)
printf("%3d -> %d\n", i, freq[i]);
return 0;
}
Constant time isn't possible. All 1 million digits need to be looked at at least once, so that is a time complexity of O(n), where n = 1 million in this case.
For a simple O(n) solution, create an array of size 1000 that represents the number of occurrences of each possible 3 digit number. Advance 1 digit at a time, first index == 0, last index == 999997, and increment array[3 digit number] to create a histogram (count of occurrences for each possible 3 digit number). Then output the content of the array with counts > 1.
A million is small for the answer I give below. Expecting only that you have to be able to run the solution in the interview, without a pause, then The following works in less than two seconds and gives the required result:
from collections import Counter
def triple_counter(s):
c = Counter(s[n-3: n] for n in range(3, len(s)))
for tri, n in c.most_common():
if n > 1:
print('%s - %i times.' % (tri, n))
else:
break
if __name__ == '__main__':
import random
s = ''.join(random.choice('0123456789') for _ in range(1_000_000))
triple_counter(s)
Hopefully the interviewer would be looking for use of the standard libraries collections.Counter class.
Parallel execution version
I wrote a blog post on this with more explanation.
The simple O(n) solution would be to count each 3-digit number:
for nr in range(1000):
cnt = text.count('%03d' % nr)
if cnt > 1:
print '%03d is found %d times' % (nr, cnt)
This would search through all 1 million digits 1000 times.
Traversing the digits only once:
counts = [0] * 1000
for idx in range(len(text)-2):
counts[int(text[idx:idx+3])] += 1
for nr, cnt in enumerate(counts):
if cnt > 1:
print '%03d is found %d times' % (nr, cnt)
Timing shows that iterating only once over the index is twice as fast as using count.
Here is a NumPy implementation of the "consensus" O(n) algorithm: walk through all triplets and bin as you go. The binning is done by upon encountering say "385", adding one to bin[3, 8, 5] which is an O(1) operation. Bins are arranged in a 10x10x10 cube. As the binning is fully vectorized there is no loop in the code.
def setup_data(n):
import random
digits = "0123456789"
return dict(text = ''.join(random.choice(digits) for i in range(n)))
def f_np(text):
# Get the data into NumPy
import numpy as np
a = np.frombuffer(bytes(text, 'utf8'), dtype=np.uint8) - ord('0')
# Rolling triplets
a3 = np.lib.stride_tricks.as_strided(a, (3, a.size-2), 2*a.strides)
bins = np.zeros((10, 10, 10), dtype=int)
# Next line performs O(n) binning
np.add.at(bins, tuple(a3), 1)
# Filtering is left as an exercise
return bins.ravel()
def f_py(text):
counts = [0] * 1000
for idx in range(len(text)-2):
counts[int(text[idx:idx+3])] += 1
return counts
import numpy as np
import types
from timeit import timeit
for n in (10, 1000, 1000000):
data = setup_data(n)
ref = f_np(**data)
print(f'n = {n}')
for name, func in list(globals().items()):
if not name.startswith('f_') or not isinstance(func, types.FunctionType):
continue
try:
assert np.all(ref == func(**data))
print("{:16s}{:16.8f} ms".format(name[2:], timeit(
'f(**data)', globals={'f':func, 'data':data}, number=10)*100))
except:
print("{:16s} apparently crashed".format(name[2:]))
Unsurprisingly, NumPy is a bit faster than #Daniel's pure Python solution on large data sets. Sample output:
# n = 10
# np 0.03481400 ms
# py 0.00669330 ms
# n = 1000
# np 0.11215360 ms
# py 0.34836530 ms
# n = 1000000
# np 82.46765980 ms
# py 360.51235450 ms
I would solve the problem as follows:
def find_numbers(str_num):
final_dict = {}
buffer = {}
for idx in range(len(str_num) - 3):
num = int(str_num[idx:idx + 3])
if num not in buffer:
buffer[num] = 0
buffer[num] += 1
if buffer[num] > 1:
final_dict[num] = buffer[num]
return final_dict
Applied to your example string, this yields:
>>> find_numbers("123412345123456")
{345: 2, 234: 3, 123: 3}
This solution runs in O(n) for n being the length of the provided string, and is, I guess, the best you can get.
As per my understanding, you cannot have the solution in a constant time. It will take at least one pass over the million digit number (assuming its a string). You can have a 3-digit rolling iteration over the digits of the million length number and increase the value of hash key by 1 if it already exists or create a new hash key (initialized by value 1) if it doesn't exists already in the dictionary.
The code will look something like this:
def calc_repeating_digits(number):
hash = {}
for i in range(len(str(number))-2):
current_three_digits = number[i:i+3]
if current_three_digits in hash.keys():
hash[current_three_digits] += 1
else:
hash[current_three_digits] = 1
return hash
You can filter down to the keys which have item value greater than 1.
As mentioned in another answer, you cannot do this algorithm in constant time, because you must look at at least n digits. Linear time is the fastest you can get.
However, the algorithm can be done in O(1) space. You only need to store the counts of each 3 digit number, so you need an array of 1000 entries. You can then stream the number in.
My guess is that either the interviewer misspoke when they gave you the solution, or you misheard "constant time" when they said "constant space."
Here's my answer:
from timeit import timeit
from collections import Counter
import types
import random
def setup_data(n):
digits = "0123456789"
return dict(text = ''.join(random.choice(digits) for i in range(n)))
def f_counter(text):
c = Counter()
for i in range(len(text)-2):
ss = text[i:i+3]
c.update([ss])
return (i for i in c.items() if i[1] > 1)
def f_dict(text):
d = {}
for i in range(len(text)-2):
ss = text[i:i+3]
if ss not in d:
d[ss] = 0
d[ss] += 1
return ((i, d[i]) for i in d if d[i] > 1)
def f_array(text):
a = [[[0 for _ in range(10)] for _ in range(10)] for _ in range(10)]
for n in range(len(text)-2):
i, j, k = (int(ss) for ss in text[n:n+3])
a[i][j][k] += 1
for i, b in enumerate(a):
for j, c in enumerate(b):
for k, d in enumerate(c):
if d > 1: yield (f'{i}{j}{k}', d)
for n in (1E1, 1E3, 1E6):
n = int(n)
data = setup_data(n)
print(f'n = {n}')
results = {}
for name, func in list(globals().items()):
if not name.startswith('f_') or not isinstance(func, types.FunctionType):
continue
print("{:16s}{:16.8f} ms".format(name[2:], timeit(
'results[name] = f(**data)', globals={'f':func, 'data':data, 'results':results, 'name':name}, number=10)*100))
for r in results:
print('{:10}: {}'.format(r, sorted(list(results[r]))[:5]))
The array lookup method is very fast (even faster than #paul-panzer's numpy method!). Of course, it cheats since it isn't technicailly finished after it completes, because it's returning a generator. It also doesn't have to check every iteration if the value already exists, which is likely to help a lot.
n = 10
counter 0.10595780 ms
dict 0.01070654 ms
array 0.00135370 ms
f_counter : []
f_dict : []
f_array : []
n = 1000
counter 2.89462101 ms
dict 0.40434612 ms
array 0.00073838 ms
f_counter : [('008', 2), ('009', 3), ('010', 2), ('016', 2), ('017', 2)]
f_dict : [('008', 2), ('009', 3), ('010', 2), ('016', 2), ('017', 2)]
f_array : [('008', 2), ('009', 3), ('010', 2), ('016', 2), ('017', 2)]
n = 1000000
counter 2849.00500992 ms
dict 438.44007806 ms
array 0.00135370 ms
f_counter : [('000', 1058), ('001', 943), ('002', 1030), ('003', 982), ('004', 1042)]
f_dict : [('000', 1058), ('001', 943), ('002', 1030), ('003', 982), ('004', 1042)]
f_array : [('000', 1058), ('001', 943), ('002', 1030), ('003', 982), ('004', 1042)]
Image as answer:
Looks like a sliding window.
Here is my solution:
from collections import defaultdict
string = "103264685134845354863"
d = defaultdict(int)
for elt in range(len(string)-2):
d[string[elt:elt+3]] += 1
d = {key: d[key] for key in d.keys() if d[key] > 1}
With a bit of creativity in for loop(and additional lookup list with True/False/None for example) you should be able to get rid of last line, as you only want to create keys in dict that we visited once up to that point.
Hope it helps :)
-Telling from the perspective of C.
-You can have an int 3-d array results[10][10][10];
-Go from 0th location to n-4th location, where n being the size of the string array.
-On each location, check the current, next and next's next.
-Increment the cntr as resutls[current][next][next's next]++;
-Print the values of
results[1][2][3]
results[2][3][4]
results[3][4][5]
results[4][5][6]
results[5][6][7]
results[6][7][8]
results[7][8][9]
-It is O(n) time, there is no comparisons involved.
-You can run some parallel stuff here by partitioning the array and calculating the matches around the partitions.
inputStr = '123456123138276237284287434628736482376487234682734682736487263482736487236482634'
count = {}
for i in range(len(inputStr) - 2):
subNum = int(inputStr[i:i+3])
if subNum not in count:
count[subNum] = 1
else:
count[subNum] += 1
print count
I am running my own little experiment and need a little help with the code.
I am creating a list that stores 100 sets in index locations 0-99, with each stored set storing random numbers ranging from 1 to 100 that came from a randomly generated list containing 100 numbers.
For each set of numbers, I use the set() command to filter out any duplicates before appending this set to a list...so basically I have a list of 100 sets which contain numbers between 1-100.
I wrote a little bit of code to check the length of each set - I noticed that my sets were often 60-69 elements in length! Basically, 1/3 of all numbers is a duplicate.
The code:
from random import randint
sets = []
#Generate list containing 100 sets of sets.
#sets contain numbers between 1 and 100 and no duplicates.
for i in range(0, 100):
nums = []
for x in range(1, 101):
nums.append(randint(1, 100))
sets.append(set(nums))
#print sizes of each set
for i in range(0, len(sets)):
print(len(sets[i]))
#I now want to create a final set
#using the data stored within all sets to
#see if there is any unique value.
So here is the bit I can't get my head around...I want to see if there is a unique number in all of those sets! What I can't work out is how I go about doing that.
I know I can directly compare a set with another set if they are stored in their own variables...but I can't work out an efficient way of looping through a list of sets and compare them all to create a new set which, I hope, might contain just one unique value!
I have seen this code in the documentation...
s.symmetric_difference_update(t)
But I can't work out how I might apply that to my code.
Any help would be greatly appreciated!!
You could use a Counter dict to count the occurrences keeping values that only have a value of 1 across all sets:
from collections import Counter
sets = [{randint(1, 100) for _ in range(100)} for i in range(100)]
from itertools import chain
cn = Counter(chain.from_iterable(sets))
unique = [k for k, v in cn.items() if v == 1] # use {} to get a set
print(unique)
For an element to only be unique to any set the count of the element must be 1 across all sets in your list.
If we use a simple example where we add a value definitely outside our range:
In [27]: from random import randint
In [28]: from collections import Counter
In [29]: from itertools import chain
In [30]: sets = [{randint(1, 100) for _ in range(100)} for i in range(0, 100)]+ [{1, 2, 102},{3,4,103}]
In [31]: cn = Counter(chain.from_iterable(sets))
In [32]: unique = [k for k, v in cn.items() if v == 1]
In [33]: print(unique)
[103, 102]
If you want to find the sets that contain any of those elements:
In [34]: for st in sets:
....: if not st.isdisjoint(unique):
....: print(st)
....:
set([1, 2, 102])
set([3, 4, 103])
For your edited part of the question you can still use a Counter dict using Counter.most_common to get the min and max occurrence:
from collections import Counter
cn = Counter()
identified_sets = 0
sets = ({randint(1, MAX) for _ in range(MAX)} for i in range(MAX))
for i, st in enumerate(sets):
cn.update(st)
if len(st) < 60 or len(st) > 70:
print("Set {} Set Length: {}, Duplicates discarded: {:.0f}% *****".
format(i, len(st), (float((MAX - len(st)))/MAX)*100))
identified_sets += 1
else:
print("Set {} Set Length: {}, Duplicates discarded: {:.0f}%".
format(i, len(st), (float((MAX - len(st)))/MAX)*100))
#print lowest fequency
comm = cn.most_common()
print("key {} : count {}".format(comm[-1][0],comm[-1][1]))
#print highest frequency
print("key {} : count {}".format(comm[0][0], comm[0][1]))
print("Count of identified sets: {}, {:.0f}%".
format(identified_sets, (float(identified_sets)/MAX)*100))
If you call random.seed(0) before you create the sets in this and your own code you will see they both return identical numbers.
well you can do:
result = set()
for s in sets:
result.symmetric_difference_update(s)
After looking through the comments I decided to do things a little differently to accomplish my goal. Essentially, I realised I just wanted to check the frequency of numbers generated by a random number generator after all duplicates have been removed. I thought I could do this by using sets to remove duplicates and then using a set to remove duplicates found in sets...but this actually doesn't work!!
I also noticed that with 100 sets containing a maximum 100 possible numbers, on average the number of duplicated numbers was around 30-40%. As you increase the maximum number of sets and, thus the maximum number of numbers generated, the % of duplicated numbers discarded decreases by a clear pattern.
After further investigation you can work out the % of discarded numbers - its all down to probability of hitting the same number once a number has been generated...
Anyway...thanks for the help!
The code updated:
from random import randint
sets = []
identified_sets = 0
MAX = 100
for i in range(0, MAX):
nums = []
for x in range(1, MAX + 1):
nums.append(randint(1, MAX))
nums.sort()
print("Set %i" % i)
print(nums)
print()
sets.append(set(nums))
for i in range(0, len(sets)):
#Only relevant when using MAX == 100
if len(sets[i]) < 60 or len(sets[i]) > 70:
print("Set %i Set Length: %i, Duplicates discarded: %.0f%s *****" %
(i, len(sets[i]), (float((MAX - len(sets[i])))/MAX)*100, "%"))
identified_sets += 1
else:
print("Set %i Set Length: %i, Duplicates discarded: %.0f%s" %
(i, len(sets[i]), (float((MAX - len(sets[i])))/MAX)*100, "%"))
#dictionary of numbers
count = {}
for i in range(1, MAX + 1):
count[i] = 0
#count occurances of numbers
for s in sets:
for e in s:
count[int(e)] += 1
#print lowest fequency
print("key %i : count %i" %
(min(count, key=count.get), count[min(count, key=count.get)]))
#print highest frequency
print("key %i : count %i" %
(max(count, key=count.get), count[max(count, key=count.get)]))
#print identified sets <60 and >70 in length as these appear less often
print("Count of identified sets: %i, %.0f%s" %
(identified_sets, (float(identified_sets)/MAX)*100, "%"))
You can keep the reversed matrix as well, which is a mapping from numbers to the set of set indexes where this number has places in. This mapping should be a dict (from numbers to sets) in gerenal, but a simple list of sets can do the trick here.
(We could use Counter too, instead of keeping the whole reversed matrix)
from random import randint
sets = [set() for _ in range(100)]
byNum = [set() for _ in range(100)]
#Generate list containing 100 sets of sets.
#sets contain numbers between 1 and 100 and no duplicates.
for setIndex in range(0, 100):
for numIndex in range(100):
num = randint(1, 100)
byNum[num].add(setIndex)
sets[setIndex].add(num)
#print sizes of each set
for setIndex, _set in enumerate(sets):
print(setIndex, len(_set))
#I now want to create a final set
#using the data stored within all sets to
#see if there is any unique value.
for num, setIndexes in enumerate(byNum)[1:]:
if len(setIndexes) == 100:
print 'number %s has appeared in all the random sets'%num
Instead of a complete shuffle, I am looking for a partial shuffle function in python.
Example : "string" must give rise to "stnrig", but not "nrsgit"
It would be better if I can define a specific "percentage" of characters that have to be rearranged.
Purpose is to test string comparison algorithms. I want to determine the "percentage of shuffle" beyond which an(my) algorithm will mark two (shuffled) strings as completely different.
Update :
Here is my code. Improvements are welcome !
import random
percent_to_shuffle = int(raw_input("Give the percent value to shuffle : "))
to_shuffle = list(raw_input("Give the string to be shuffled : "))
num_of_chars_to_shuffle = int((len(to_shuffle)*percent_to_shuffle)/100)
for i in range(0,num_of_chars_to_shuffle):
x=random.randint(0,(len(to_shuffle)-1))
y=random.randint(0,(len(to_shuffle)-1))
z=to_shuffle[x]
to_shuffle[x]=to_shuffle[y]
to_shuffle[y]=z
print ''.join(to_shuffle)
This is a problem simpler than it looks. And the language has the right tools not to stay between you and the idea,as usual:
import random
def pashuffle(string, perc=10):
data = list(string)
for index, letter in enumerate(data):
if random.randrange(0, 100) < perc/2:
new_index = random.randrange(0, len(data))
data[index], data[new_index] = data[new_index], data[index]
return "".join(data)
Your problem is tricky, because there are some edge cases to think about:
Strings with repeated characters (i.e. how would you shuffle "aaaab"?)
How do you measure chained character swaps or re arranging blocks?
In any case, the metric defined to shuffle strings up to a certain percentage is likely to be the same you are using in your algorithm to see how close they are.
My code to shuffle n characters:
import random
def shuffle_n(s, n):
idx = range(len(s))
random.shuffle(idx)
idx = idx[:n]
mapping = dict((idx[i], idx[i-1]) for i in range(n))
return ''.join(s[mapping.get(x,x)] for x in range(len(s)))
Basically chooses n positions to swap at random, and then exchanges each of them with the next in the list... This way it ensures that no inverse swaps are generated and exactly n characters are swapped (if there are characters repeated, bad luck).
Explained run with 'string', 3 as input:
idx is [0, 1, 2, 3, 4, 5]
we shuffle it, now it is [5, 3, 1, 4, 0, 2]
we take just the first 3 elements, now it is [5, 3, 1]
those are the characters that we are going to swap
s t r i n g
^ ^ ^
t (1) will be i (3)
i (3) will be g (5)
g (5) will be t (1)
the rest will remain unchanged
so we get 'sirgnt'
The bad thing about this method is that it does not generate all the possible variations, for example, it could not make 'gnrits' from 'string'. This could be fixed by making partitions of the indices to be shuffled, like this:
import random
def randparts(l):
n = len(l)
s = random.randint(0, n-1) + 1
if s >= 2 and n - s >= 2: # the split makes two valid parts
yield l[:s]
for p in randparts(l[s:]):
yield p
else: # the split would make a single cycle
yield l
def shuffle_n(s, n):
idx = range(len(s))
random.shuffle(idx)
mapping = dict((x[i], x[i-1])
for i in range(len(x))
for x in randparts(idx[:n]))
return ''.join(s[mapping.get(x,x)] for x in range(len(s)))
import random
def partial_shuffle(a, part=0.5):
# which characters are to be shuffled:
idx_todo = random.sample(xrange(len(a)), int(len(a) * part))
# what are the new positions of these to-be-shuffled characters:
idx_target = idx_todo[:]
random.shuffle(idx_target)
# map all "normal" character positions {0:0, 1:1, 2:2, ...}
mapper = dict((i, i) for i in xrange(len(a)))
# update with all shuffles in the string: {old_pos:new_pos, old_pos:new_pos, ...}
mapper.update(zip(idx_todo, idx_target))
# use mapper to modify the string:
return ''.join(a[mapper[i]] for i in xrange(len(a)))
for i in xrange(5):
print partial_shuffle('abcdefghijklmnopqrstuvwxyz', 0.2)
prints
abcdefghljkvmnopqrstuxwiyz
ajcdefghitklmnopqrsbuvwxyz
abcdefhwijklmnopqrsguvtxyz
aecdubghijklmnopqrstwvfxyz
abjdefgcitklmnopqrshuvwxyz
Evil and using a deprecated API:
import random
# adjust constant to taste
# 0 -> no effect, 0.5 -> completely shuffled, 1.0 -> reversed
# Of course this assumes your input is already sorted ;)
''.join(sorted(
'abcdefghijklmnopqrstuvwxyz',
cmp = lambda a, b: cmp(a, b) * (-1 if random.random() < 0.2 else 1)
))
maybe like so:
>>> s = 'string'
>>> shufflethis = list(s[2:])
>>> random.shuffle(shufflethis)
>>> s[:2]+''.join(shufflethis)
'stingr'
Taking from fortran's idea, i'm adding this to collection. It's pretty fast:
def partial_shuffle(st, p=20):
p = int(round(p/100.0*len(st)))
idx = range(len(s))
sample = random.sample(idx, p)
res=str()
samptrav = 1
for i in range(len(st)):
if i in sample:
res += st[sample[-samptrav]]
samptrav += 1
continue
res += st[i]
return res