问题
I have a list of items. Each of these items has its own probability.
Can anyone suggest an algorithm to pick an item based on its probability?
回答1:
So with each item store a number that marks its relative probability, for example if you have 3 items one should be twice as likely to be selected as either of the other two then your list will have:
[{A,1},{B,1},{C,2}]
Then sum the numbers of the list (i.e. 4 in our case). Now generate a random number and choose that index. int index = rand.nextInt(4); return the number such that the index is in the correct range.
Java code:
class Item {
int relativeProb;
String name;
//Getters Setters and Constructor
}
...
class RandomSelector {
List<Item> items = new List();
Random rand = new Random();
int totalSum = 0;
RandomSelector() {
for(Item item : items) {
totalSum = totalSum + item.relativeProb;
}
}
public Item getRandom() {
int index = rand.nextInt(totalSum);
int sum = 0;
int i=0;
while(sum < index ) {
sum = sum + items.get(i++).relativeProb;
}
return items.get(Math.max(0,i-1));
}
}
回答2:
- Generate a uniformly distributed random number.
- Iterate through your list until the cumulative probability of the visited elements is greater than the random number
Sample code:
double p = Math.random();
double cumulativeProbability = 0.0;
for (Item item : items) {
cumulativeProbability += item.probability();
if (p <= cumulativeProbability) {
return item;
}
}
回答3:
pretend that we have the following list
Item A 25%
Item B 15%
Item C 35%
Item D 5%
Item E 20%
Lets pretend that all the probabilities are integers, and assign each item a "range" that calculated as follows.
Start - Sum of probability of all items before
End - Start + own probability
The new numbers are as follows
Item A 0 to 25
Item B 26 to 40
Item C 41 to 75
Item D 76 to 80
Item E 81 to 100
Now pick a random number from 0 to 100. Lets say that you pick 32. 32 falls in Item B's range.
mj
回答4:
You can try the Roulette Wheel Selection.
First, add all the probabilities, then scale all the probabilities in the scale of 1, by dividing each one by the sum. Suppose the probabilities are A(0.4), B(0.3), C(0.25) and D(0.05). Then you can generate a random floating-point number in the range [0, 1]. Now you can decide like this:
random number between 0.00 and 0.40 -> pick A
between 0.40 and 0.70 -> pick B
between 0.70 and 0.95 -> pick C
between 0.95 and 1.00 -> pick D
You can also do it with random integers - say you generate a random integer between 0 to 99 (inclusive), then you can make decision like the above.
回答5:
Algorithm described in Ushman's, Brent's and @kaushaya's answers are implemented in Apache commons-math library.
Take a look at EnumeratedDistribution class (groovy code follows):
def probabilities = [
new Pair<String, Double>("one", 25),
new Pair<String, Double>("two", 30),
new Pair<String, Double>("three", 45)]
def distribution = new EnumeratedDistribution<String>(probabilities)
println distribution.sample() // here you get one of your values
Note that sum of probabilities doesn't need to be equal to 1 or 100 - it will be normalized automatically.
回答6:
My method is pretty simple. Generate a random number. Now since the probabilities of your items are known,simply iterate through the sorted list of probability and pick the item whose probability is lesser than the randomly generated number.
For more details,read my answer here.
回答7:
A slow but simple way to do it is to have every member to pick a random number based on its probability and pick the one with highest value.
Analogy:
Imagine 1 of 3 people needs to be chosen but they have different probabilities. You give them die with different amount of faces. First person's dice has 4 face, 2nd person's 6, and the third person's 8. They roll their die and the one with the biggest number wins.
Lets say we have the following list:
[{A,50},{B,100},{C,200}]
Pseudocode:
A.value = random(0 to 50);
B.value = random(0 to 100);
C.value = random (0 to 200);
We pick the one with the highest value.
This method above does not exactly map the probabilities. For example 100 will not have twice the chance of 50. But we can do it in a by tweaking the method a bit.
Method 2
Instead of picking a number from 0 to the weight we can limit them from the upper limit of previous variable to addition of the current variable.
[{A,50},{B,100},{C,200}]
Pseudocode:
A.lowLimit= 0; A.topLimit=50;
B.lowLimit= A.topLimit+1; B.topLimit= B.lowLimit+100
C.lowLimit= B.topLimit+1; C.topLimit= C.lowLimit+200
resulting limits
A.limits = 0,50
B.limits = 51,151
C.limits = 152,352
Then we pick a random number from 0 to 352 and compare it to each variable's limits to see whether the random number is in its limits.
I believe this tweak has better performance since there is only 1 random generation.
There is a similar method in other answers but this method does not require the total to be 100 or 1.00.
回答8:
Brent's answer is good, but it doesn't account for the possibility of erroneously choosing an item with a probability of 0 in cases where p = 0. That's easy enough to handle by checking the probability (or perhaps not adding the item in the first place):
double p = Math.random();
double cumulativeProbability = 0.0;
for (Item item : items) {
cumulativeProbability += item.probability();
if (p <= cumulativeProbability && item.probability() != 0) {
return item;
}
}
回答9:
If you don't mind adding a third party dependency in your code you can use the MockNeat.probabilities() method.
For example:
String s = mockNeat.probabilites(String.class)
.add(0.1, "A") // 10% chance to pick A
.add(0.2, "B") // 20% chance to pick B
.add(0.5, "C") // 50% chance to pick C
.add(0.2, "D") // 20% chance to pick D
.val();
Disclaimer: I am the author of the library, so I might be biased when I am recommending it.
回答10:
You could use the Julia code:
function selrnd(a::Vector{Int})
c = a[:]
sumc = c[1]
for i=2:length(c)
sumc += c[i]
c[i] += c[i-1]
end
r = rand()*sumc
for i=1:length(c)
if r <= c[i]
return i
end
end
end
This function returns the index of an item efficiently.
来源：https://stackoverflow.com/questions/9330394/how-to-pick-an-item-by-its-probability