如何获取python字典或大小为X的列表，并为每个元素分配字典本身的Y个随机索引值

Question

I'm not sure if it would be best to use a list or a dictionary for this algorithm. Assuming I use a dictionary, I want to create a dictionary of size X and randomly assign each element with Y index values of the dictionary itself.

Meaning I could take a dictionary of size 5, and assign each of those 5 elements with 2 index values ranging between 1-5.

The constraints would be that the index value can not be assigned to its own index, so the 2nd index can only be assigned values 1,3,4,5; And Y must always be less that X, in order to prevent assigning duplicate index values to the same index.

What I have so far is being done with a list rather than a dictionary, but I'm not sure if this is the best method. I'd like to keep the algorithm running at 0(n) speed as well, even if the size of the list/dictionary is huge. Either way, this is where I'm at.

So, I make X a list of size 5. I set Y equal to 3, meaning I want each of those 5 elements to contain 3 index values. In the for-loop I create another list excluding the index value I'm currently assigning values to.

X = range(5)[::1] # [0, 1, 2, 3, 4]
print(X)
Y = 3    
assigned = []

for k in range(0, len(X)):  
    XExcluded = [x for i,x in enumerate(X) if i!=k]   # if k==3 then [0, 1, 2, 4]
    print("EXcluded: {}" .format(XExcluded))

    assigned.append(list(random.sample(XExcluded, Y)))
    print("assigned: {}" .format(assigned))

Sample Output:

[0, 1, 2, 3, 4]
EXcluded: [1, 2, 3, 4]
assigned: [[1, 2, 3]]
EXcluded: [0, 2, 3, 4]
assigned: [[1, 2, 3], [3, 2, 4]]
EXcluded: [0, 1, 3, 4]
assigned: [[1, 2, 3], [3, 2, 4], [3, 4, 1]]
EXcluded: [0, 1, 2, 4]
assigned: [[1, 2, 3], [3, 2, 4], [3, 4, 1], [0, 1, 2]]
EXcluded: [0, 1, 2, 3]
assigned: [[1, 2, 3], [3, 2, 4], [3, 4, 1], [0, 1, 2], [2, 3, 1]]

One thing I would really like to implement is someway to average out which index values are being assigned over time, because right now the algorithm may assign certain index values more than others. This may be more apparent when starting with smaller lists, but I'd imagine this wont be as much of a problem when starting with a very large list since it would allow the randomly sampled index values to better average out over time.

Answer 1

To balance selections, in the case of your example, where X and Y are both small, a naive solution would be checking which value is missing in each iteration and adding that value to next round to add additional weighting for it during random sampling.

Below is just a simple workable example, without too much effort in efficiency optimization. You may consider using other data structure than set for optimization. (Diff in sets takes O(n). At the end it may get to O(n^2). Since this is aimed at short list case, set was chosen for its simplicity in code.)

import random

X = set(range(5)[::1]) # assume you have distinct values as shown in your example
Y = 3
assigned = []
tobeweighted=None

def assigned_and_missing(fullset, index):
  def sample(inlist, weighteditem, Y):
    if weighteditem is not None:
      inlist.append(weighteditem)
    print("Weighted item: {}" .format(weighteditem))
    print("Interim selection (weighted item appended): {}" .format(inlist))
    randselection = set(random.sample(inlist, Y))
    remain = fullset - set((index+1,)) - randselection   
    missingitem = remain.pop()
    if len(randselection) < Y:
      randselection.add(remain.pop())
    return randselection, missingitem
  return sample


for k in range(0, len(X)):
    weighted_random = assigned_and_missing(X, k)
    XExcluded = [x for i,x in enumerate(X) if i!=k]   # if k==3 then [0, 1, 2, 4]

    print()
    print("EXcluded: {}" .format(XExcluded))

    selection, tobeweighted = weighted_random(XExcluded, tobeweighted, Y)
    print("Final selection: {}" .format(selection))

    assigned.append(selection)
    print("assigned: {}" .format(assigned))
    print("Item needs to be weighted in next round: {}" .format(tobeweighted))