Get random sample from list while maintaining ordering of items

Question

I have a sorted list  let say   its not really just numbers  its a list of objects that are sorted with a complicated time consuming algorithm   mylist     1   2   3   4   5   6   7   8  9    10     Is there some python function that will give me N of the items  but will keep the order   Example   randomList   getRandom mylist 4    randomList     3   6  7   9   randomList   getRandom mylist 4    randomList     1   2   4   8     etc

User · Answer

Apparently random sample was introduced in python 2 3  so for version under that  we can use shuffle  example for 4 items    myRange    range 0 len mylist    shuffle myRange  coupons     bestCoupons i  for i in sorted myRange  4

User · Answer

random sample implement it    gt  gt  gt  random sample  1  2  3  4  5    3      Three samples without replacement  4  1  5

User · Answer

Maybe you can just generate the sample of indices and then collect the items from your list   randIndex   random sample range len mylist    sample size  randIndex sort   rand    mylist i  for i in randIndex

User · Answer

Simple-to-code O N   K log K   way  Take a random sample without replacement of the indices  sort the indices  and take them from the original   indices   random sample range len myList    K   myList i  for i in sorted indices     Or more concisely    x 1  for x in sorted random sample enumerate myList  K        Optimized O N -time  O 1 -auxiliary-space way  You can alternatively use a math trick and iteratively go through myList from left to right  picking numbers with dynamically-changing probability  N-numbersPicked   total-numbersVisited   The advantage of this approach is that it s an O N  algorithm since it doesn t involve sorting   from   future   import division  def orderedSampleWithoutReplacement seq  k       if not 0 lt  k lt  len seq           raise ValueError  Required that 0  lt   sample size  lt   population size        numbersPicked   0     for i number in enumerate seq           prob    k-numbersPicked   len seq -i          if random random    lt  prob              yield number             numbersPicked    1   Proof of concept and test that probabilities are correct   Simulated with 1 trillion pseudorandom samples over the course of 5 hours    gt  gt  gt  Counter          tuple orderedSampleWithoutReplacement  0 1 2 3   2           for   in range 10  9        Counter        0  3   166680161        1  2   166672608        0  2   166669915        2  3   166667390        1  3   166660630        0  1   166649296      Probabilities diverge from true probabilities by less a factor of 1 0001  Running this test again resulted in a different order meaning it isn t biased towards one ordering  Running the test with fewer samples for  0 1 2 3 4   k 3 and  0 1 2 3 4 5   k 4 had similar results   edit  Not sure why people are voting up wrong comments or afraid to upvote    NO  there is nothing wrong with this method       Also a useful note from user tegan in the comments  If this is python2  you will want to use xrange  as usual  if you really care about extra space    edit  Proof  Considering the uniform distribution  without replacement  of picking a subset of k out of a population seq of size len seq   we can consider a partition at an arbitrary point i into  left   0 1     i-1  and  right   i i 1     len seq    Given that we picked numbersPicked from the left known subset  the remaining must come from the same uniform distribution on the right unknown subset  though the parameters are now different  In particular  the probability that seq i  contains a chosen element is  remainingToChoose  remainingToChooseFrom  or  k-numbersPicked   len seq -i   so we simulate that and recurse on the result   This must terminate since if  remainingToChoose     remainingToChooseFrom  then all remaining probabilities are 1   This is similar to a probability tree that happens to be dynamically generated  Basically you can simulate a uniform probability distribution by conditioning on prior choices  as you grow the probability tree  you pick the probability of the current branch such that it is aposteriori the same as prior leaves  i e  conditioned on prior choices  this will work because this probability is uniformly exactly N k    edit  Timothy Shields mentions Reservoir Sampling  which is the generalization of this method when len seq  is unknown  such as with a generator expression   Specifically the one noted as  algorithm R  is O N  and O 1  space if done in-place  it involves taking the first N element and slowly replacing them  a hint at an inductive proof is also given   There are also useful distributed variants and miscellaneous variants of reservoir sampling to be found on the wikipedia page   edit  Here s another way to code it below in a more semantically obvious manner   from   future   import division import random  def orderedSampleWithoutReplacement seq  sampleSize       totalElems   len seq      if not 0 lt  sampleSize lt  totalElems          raise ValueError  Required that 0  lt   sample size  lt   population size        picksRemaining   sampleSize     for elemsSeen element in enumerate seq           elemsRemaining   totalElems - elemsSeen         prob   picksRemaining elemsRemaining         if random random    lt  prob              yield element             picksRemaining -  1  from collections import Counter          Counter      tuple orderedSampleWithoutReplacement  0 1 2 3   2       for   in range 10  5

User · Answer

Following code will generate a random sample of size 4   import random  sample size   4 sorted sample         mylist i  for i in sorted random sample range len mylist    sample size        note  with Python 2  better use xrange instead of range   Explanation  random sample range len mylist    sample size    generates a random sample of the indices of the original list   These indices then get sorted to preserve the ordering of elements in the original list   Finally  the list comprehension pulls out the actual elements from the original list  given the sampled indices

[python] Get random sample from list while maintaining ordering of items?

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