I needed to write a weighted version of random.choice (each element in the list has a different probability for being selected). This is what I came up with:

```
def weightedChoice(choices):
"""Like random.choice, but each element can have a different chance of
being selected.
choices can be any iterable containing iterables with two items each.
Technically, they can have more than two items, the rest will just be
ignored. The first item is the thing being chosen, the second item is
its weight. The weights can be any numeric values, what matters is the
relative differences between them.
"""
space = {}
current = 0
for choice, weight in choices:
if weight > 0:
space[current] = choice
current += weight
rand = random.uniform(0, current)
for key in sorted(space.keys() + [current]):
if rand < key:
return choice
choice = space[key]
return None
```

This function seems overly complex to me, and ugly. I'm hoping everyone here can offer some suggestions on improving it or alternate ways of doing this. Efficiency isn't as important to me as code cleanliness and readability.

## Solution 1

Since version 1.7.0, NumPy has a `choice`

function that supports probability distributions.

```
from numpy.random import choice
draw = choice(list_of_candidates, number_of_items_to_pick,
p=probability_distribution)
```

Note that `probability_distribution`

is a sequence in the same order of `list_of_candidates`

. You can also use the keyword `replace=False`

to change the behavior so that drawn items are not replaced.

## Solution 2

Since Python 3.6 there is a method `choices`

from the `random`

module.

```
In [1]: import random
In [2]: random.choices(
...: population=[['a','b'], ['b','a'], ['c','b']],
...: weights=[0.2, 0.2, 0.6],
...: k=10
...: )
Out[2]:
[['c', 'b'],
['c', 'b'],
['b', 'a'],
['c', 'b'],
['c', 'b'],
['b', 'a'],
['c', 'b'],
['b', 'a'],
['c', 'b'],
['c', 'b']]
```

Note that `random.choices`

will sample *with replacement*, per the docs:

Return a

`k`

sized list of elements chosen from the population with replacement.

Note for completeness of answer:

When a sampling unit is drawn from a finite population and is returned to that population, after its characteristic(s) have been recorded, before the next unit is drawn, the sampling is said to be "with replacement". It basically means each element may be chosen more than once.

If you need to sample without replacement, then as @ronan-paixão's brilliant answer states, you can use `numpy.choice`

, whose `replace`

argument controls such behaviour.

## Solution 3

```
def weighted_choice(choices):
total = sum(w for c, w in choices)
r = random.uniform(0, total)
upto = 0
for c, w in choices:
if upto + w >= r:
return c
upto += w
assert False, "Shouldn't get here"
```

## Solution 4

- Arrange the weights into a cumulative distribution.
- Use
**random.random()**to pick a random float`0.0 <= x < total`

. - Search the
distribution using
**bisect.bisect**as shown in the example at http://docs.python.org/dev/library/bisect.html#other-examples.

```
from random import random
from bisect import bisect
def weighted_choice(choices):
values, weights = zip(*choices)
total = 0
cum_weights = []
for w in weights:
total += w
cum_weights.append(total)
x = random() * total
i = bisect(cum_weights, x)
return values[i]
>>> weighted_choice([("WHITE",90), ("RED",8), ("GREEN",2)])
'WHITE'
```

If you need to make more than one choice, split this into two functions, one to build the cumulative weights and another to bisect to a random point.

## Solution 5

If you don't mind using numpy, you can use numpy.random.choice.

For example:

```
import numpy
items = [["item1", 0.2], ["item2", 0.3], ["item3", 0.45], ["item4", 0.05]
elems = [i[0] for i in items]
probs = [i[1] for i in items]
trials = 1000
results = [0] * len(items)
for i in range(trials):
res = numpy.random.choice(items, p=probs) #This is where the item is selected!
results[items.index(res)] += 1
results = [r / float(trials) for r in results]
print "item\texpected\tactual"
for i in range(len(probs)):
print "%s\t%0.4f\t%0.4f" % (items[i], probs[i], results[i])
```

If you know how many selections you need to make in advance, you can do it without a loop like this:

```
numpy.random.choice(items, trials, p=probs)
```

## Solution 6

As of Python `v3.6`

, `random.choices`

could be used to return a `list`

of elements of specified size from the given population with optional weights.

`random.choices(population, weights=None, *, cum_weights=None, k=1)`

*population*:`list`

containing unique observations. (If empty, raises`IndexError`

)*weights*: More precisely relative weights required to make selections.*cum_weights*: cumulative weights required to make selections.*k*: size(`len`

) of the`list`

to be outputted. (Default`len()=1`

)

*Few Caveats:*

1) It makes use of weighted sampling with replacement so the drawn items would be later replaced. The values in the weights sequence in itself do not matter, but their relative ratio does.

Unlike `np.random.choice`

which can only take on probabilities as weights and also which must ensure summation of individual probabilities upto 1 criteria, there are no such regulations here. As long as they belong to numeric types (`int/float/fraction`

except `Decimal`

type) , these would still perform.

```
>>> import random
# weights being integers
>>> random.choices(["white", "green", "red"], [12, 12, 4], k=10)
['green', 'red', 'green', 'white', 'white', 'white', 'green', 'white', 'red', 'white']
# weights being floats
>>> random.choices(["white", "green", "red"], [.12, .12, .04], k=10)
['white', 'white', 'green', 'green', 'red', 'red', 'white', 'green', 'white', 'green']
# weights being fractions
>>> random.choices(["white", "green", "red"], [12/100, 12/100, 4/100], k=10)
['green', 'green', 'white', 'red', 'green', 'red', 'white', 'green', 'green', 'green']
```

2) If neither *weights* nor *cum_weights* are specified, selections are made with equal probability. If a *weights* sequence is supplied, it must be the same length as the *population* sequence.

Specifying both *weights* and *cum_weights* raises a `TypeError`

.

```
>>> random.choices(["white", "green", "red"], k=10)
['white', 'white', 'green', 'red', 'red', 'red', 'white', 'white', 'white', 'green']
```

3) *cum_weights* are typically a result of `itertools.accumulate`

function which are really handy in such situations.

_{ From the documentation linked: }Internally, the relative weights are converted to cumulative weights before making selections, so supplying the cumulative weights saves work.

So, either supplying `weights=[12, 12, 4]`

or `cum_weights=[12, 24, 28]`

for our contrived case produces the same outcome and the latter seems to be more faster / efficient.

## Solution 7

Crude, but may be sufficient:

```
import random
weighted_choice = lambda s : random.choice(sum(([v]*wt for v,wt in s),[]))
```

Does it work?

```
# define choices and relative weights
choices = [("WHITE",90), ("RED",8), ("GREEN",2)]
# initialize tally dict
tally = dict.fromkeys(choices, 0)
# tally up 1000 weighted choices
for i in xrange(1000):
tally[weighted_choice(choices)] += 1
print tally.items()
```

Prints:

```
[('WHITE', 904), ('GREEN', 22), ('RED', 74)]
```

Assumes that all weights are integers. They don't have to add up to 100, I just did that to make the test results easier to interpret. (If weights are floating point numbers, multiply them all by 10 repeatedly until all weights >= 1.)

```
weights = [.6, .2, .001, .199]
while any(w < 1.0 for w in weights):
weights = [w*10 for w in weights]
weights = map(int, weights)
```

## Solution 8

If you have a weighted dictionary instead of a list you can write this

```
items = { "a": 10, "b": 5, "c": 1 }
random.choice([k for k in items for dummy in range(items[k])])
```

Note that `[k for k in items for dummy in range(items[k])]`

produces this list `['a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'c', 'b', 'b', 'b', 'b', 'b']`

## Solution 9

Here's is the version that is being included in the standard library for Python 3.6:

```
import itertools as _itertools
import bisect as _bisect
class Random36(random.Random):
"Show the code included in the Python 3.6 version of the Random class"
def choices(self, population, weights=None, *, cum_weights=None, k=1):
"""Return a k sized list of population elements chosen with replacement.
If the relative weights or cumulative weights are not specified,
the selections are made with equal probability.
"""
random = self.random
if cum_weights is None:
if weights is None:
_int = int
total = len(population)
return [population[_int(random() * total)] for i in range(k)]
cum_weights = list(_itertools.accumulate(weights))
elif weights is not None:
raise TypeError('Cannot specify both weights and cumulative weights')
if len(cum_weights) != len(population):
raise ValueError('The number of weights does not match the population')
bisect = _bisect.bisect
total = cum_weights[-1]
return [population[bisect(cum_weights, random() * total)] for i in range(k)]
```

Source: https://hg.python.org/cpython/file/tip/Lib/random.py#l340

## Solution 10

A very basic and easy approach for a weighted choice is the following:

```
np.random.choice(['A', 'B', 'C'], p=[0.3, 0.4, 0.3])
```

## Solution 11

```
import numpy as np
w=np.array([ 0.4, 0.8, 1.6, 0.8, 0.4])
np.random.choice(w, p=w/sum(w))
```

## Solution 12

I'm probably too late to contribute anything useful, but here's a simple, short, and very efficient snippet:

```
def choose_index(probabilies):
cmf = probabilies[0]
choice = random.random()
for k in xrange(len(probabilies)):
if choice <= cmf:
return k
else:
cmf += probabilies[k+1]
```

No need to sort your probabilities or create a vector with your cmf, and it terminates once it finds its choice. Memory: O(1), time: O(N), with average running time ~ N/2.

If you have weights, simply add one line:

```
def choose_index(weights):
probabilities = weights / sum(weights)
cmf = probabilies[0]
choice = random.random()
for k in xrange(len(probabilies)):
if choice <= cmf:
return k
else:
cmf += probabilies[k+1]
```

## Solution 13

If your list of weighted choices is relatively static, and you want frequent sampling, you can do one O(N) preprocessing step, and then do the selection in O(1), using the functions in this related answer.

```
# run only when `choices` changes.
preprocessed_data = prep(weight for _,weight in choices)
# O(1) selection
value = choices[sample(preprocessed_data)][0]
```

## Solution 14

If you happen to have Python 3, and are afraid of installing `numpy`

or writing your own loops, you could do:

```
import itertools, bisect, random
def weighted_choice(choices):
weights = list(zip(*choices))[1]
return choices[bisect.bisect(list(itertools.accumulate(weights)),
random.uniform(0, sum(weights)))][0]
```

Because you can build *anything* out of a bag of plumbing adaptors! Although... I must admit that Ned's answer, while slightly longer, is easier to understand.

## Solution 15

I looked the pointed other thread and came up with this variation in my coding style, this returns the index of choice for purpose of tallying, but it is simple to return the string ( commented return alternative):

```
import random
import bisect
try:
range = xrange
except:
pass
def weighted_choice(choices):
total, cumulative = 0, []
for c,w in choices:
total += w
cumulative.append((total, c))
r = random.uniform(0, total)
# return index
return bisect.bisect(cumulative, (r,))
# return item string
#return choices[bisect.bisect(cumulative, (r,))][0]
# define choices and relative weights
choices = [("WHITE",90), ("RED",8), ("GREEN",2)]
tally = [0 for item in choices]
n = 100000
# tally up n weighted choices
for i in range(n):
tally[weighted_choice(choices)] += 1
print([t/sum(tally)*100 for t in tally])
```

## Solution 16

A general solution:

```
import random
def weighted_choice(choices, weights):
total = sum(weights)
treshold = random.uniform(0, total)
for k, weight in enumerate(weights):
total -= weight
if total < treshold:
return choices[k]
```

## Solution 17

Here is another version of weighted_choice that uses numpy. Pass in the weights vector and it will return an array of 0's containing a 1 indicating which bin was chosen. The code defaults to just making a single draw but you can pass in the number of draws to be made and the counts per bin drawn will be returned.

If the weights vector does not sum to 1, it will be normalized so that it does.

```
import numpy as np
def weighted_choice(weights, n=1):
if np.sum(weights)!=1:
weights = weights/np.sum(weights)
draws = np.random.random_sample(size=n)
weights = np.cumsum(weights)
weights = np.insert(weights,0,0.0)
counts = np.histogram(draws, bins=weights)
return(counts[0])
```

## Solution 18

It depends on how many times you want to sample the distribution.

Suppose you want to sample the distribution K times. Then, the time complexity using `np.random.choice()`

each time is `O(K(n + log(n)))`

when `n`

is the number of items in the distribution.

In my case, I needed to sample the same distribution multiple times of the order of 10^3 where n is of the order of 10^6. I used the below code, which precomputes the cumulative distribution and samples it in `O(log(n))`

. Overall time complexity is `O(n+K*log(n))`

.

```
import numpy as np
n,k = 10**6,10**3
# Create dummy distribution
a = np.array([i+1 for i in range(n)])
p = np.array([1.0/n]*n)
cfd = p.cumsum()
for _ in range(k):
x = np.random.uniform()
idx = cfd.searchsorted(x, side='right')
sampled_element = a[idx]
```

## Solution 19

There is lecture on this by Sebastien Thurn in the free Udacity course AI for Robotics. Basically he makes a circular array of the indexed weights using the mod operator `%`

, sets a variable beta to 0, randomly chooses an index,
for loops through N where N is the number of indices and in the for loop firstly increments beta by the formula:

beta = beta + uniform sample from {0...2* Weight_max}

and then nested in the for loop, a while loop per below:

```
while w[index] < beta:
beta = beta - w[index]
index = index + 1
select p[index]
```

Then on to the next index to resample based on the probabilities (or normalized probability in the case presented in the course).

On Udacity find Lesson 8, video number 21 of Artificial Intelligence for Robotics where he is lecturing on particle filters.

## Solution 20

Another way of doing this, assuming we have weights at the same index as the elements in the element array.

```
import numpy as np
weights = [0.1, 0.3, 0.5] #weights for the item at index 0,1,2
# sum of weights should be <=1, you can also divide each weight by sum of all weights to standardise it to <=1 constraint.
trials = 1 #number of trials
num_item = 1 #number of items that can be picked in each trial
selected_item_arr = np.random.multinomial(num_item, weights, trials)
# gives number of times an item was selected at a particular index
# this assumes selection with replacement
# one possible output
# selected_item_arr
# array([[0, 0, 1]])
# say if trials = 5, the the possible output could be
# selected_item_arr
# array([[1, 0, 0],
# [0, 0, 1],
# [0, 0, 1],
# [0, 1, 0],
# [0, 0, 1]])
```

Now let's assume, we have to sample out 3 items in 1 trial. You can assume that there are three balls R,G,B present in large quantity in ratio of their weights given by weight array, the following could be possible outcome:

```
num_item = 3
trials = 1
selected_item_arr = np.random.multinomial(num_item, weights, trials)
# selected_item_arr can give output like :
# array([[1, 0, 2]])
```

you can also think number of items to be selected as number of binomial/ multinomial trials within a set. So, the above example can be still work as

```
num_binomial_trial = 5
weights = [0.1,0.9] #say an unfair coin weights for H/T
num_experiment_set = 1
selected_item_arr = np.random.multinomial(num_binomial_trial, weights, num_experiment_set)
# possible output
# selected_item_arr
# array([[1, 4]])
# i.e H came 1 time and T came 4 times in 5 binomial trials. And one set contains 5 binomial trails.
```

## Solution 21

One way is to randomize on the total of all the weights and then use the values as the limit points for each var. Here is a crude implementation as a generator.

```
def rand_weighted(weights):
"""
Generator which uses the weights to generate a
weighted random values
"""
sum_weights = sum(weights.values())
cum_weights = {}
current_weight = 0
for key, value in sorted(weights.iteritems()):
current_weight += value
cum_weights[key] = current_weight
while True:
sel = int(random.uniform(0, 1) * sum_weights)
for key, value in sorted(cum_weights.iteritems()):
if sel < value:
break
yield key
```

## Solution 22

Using numpy

```
def choice(items, weights):
return items[np.argmin((np.cumsum(weights) / sum(weights)) < np.random.rand())]
```

## Solution 23

I needed to do something like this really fast really simple, from searching for ideas i finally built this template. The idea is receive the weighted values in a form of a json from the api, which here is simulated by the dict.

Then translate it into a list in which each value repeats proportionally to it's weight, and just use random.choice to select a value from the list.

I tried it running with 10, 100 and 1000 iterations. The distribution seems pretty solid.

```
def weighted_choice(weighted_dict):
"""Input example: dict(apples=60, oranges=30, pineapples=10)"""
weight_list = []
for key in weighted_dict.keys():
weight_list += [key] * weighted_dict[key]
return random.choice(weight_list)
```

## Solution 24

I didn't love the syntax of any of those. I really wanted to just specify what the items were and what the weighting of each was. I realize I could have used `random.choices`

but instead I quickly wrote the class below.

```
import random, string
from numpy import cumsum
class randomChoiceWithProportions:
'''
Accepts a dictionary of choices as keys and weights as values. Example if you want a unfair dice:
choiceWeightDic = {"1":0.16666666666666666, "2": 0.16666666666666666, "3": 0.16666666666666666
, "4": 0.16666666666666666, "5": .06666666666666666, "6": 0.26666666666666666}
dice = randomChoiceWithProportions(choiceWeightDic)
samples = []
for i in range(100000):
samples.append(dice.sample())
# Should be close to .26666
samples.count("6")/len(samples)
# Should be close to .16666
samples.count("1")/len(samples)
'''
def __init__(self, choiceWeightDic):
self.choiceWeightDic = choiceWeightDic
weightSum = sum(self.choiceWeightDic.values())
assert weightSum == 1, 'Weights sum to ' + str(weightSum) + ', not 1.'
self.valWeightDict = self._compute_valWeights()
def _compute_valWeights(self):
valWeights = list(cumsum(list(self.choiceWeightDic.values())))
valWeightDict = dict(zip(list(self.choiceWeightDic.keys()), valWeights))
return valWeightDict
def sample(self):
num = random.uniform(0,1)
for key, val in self.valWeightDict.items():
if val >= num:
return key
```

## Solution 25

Provide random.choice() with a pre-weighted list:

**Solution & Test:**

```
import random
options = ['a', 'b', 'c', 'd']
weights = [1, 2, 5, 2]
weighted_options = [[opt]*wgt for opt, wgt in zip(options, weights)]
weighted_options = [opt for sublist in weighted_options for opt in sublist]
print(weighted_options)
# test
counts = {c: 0 for c in options}
for x in range(10000):
counts[random.choice(weighted_options)] += 1
for opt, wgt in zip(options, weights):
wgt_r = counts[opt] / 10000 * sum(weights)
print(opt, counts[opt], wgt, wgt_r)
```

**Output:**

```
['a', 'b', 'b', 'c', 'c', 'c', 'c', 'c', 'd', 'd']
a 1025 1 1.025
b 1948 2 1.948
c 5019 5 5.019
d 2008 2 2.008
```

## Solution 26

In case you don't define in advance how many items you want to pick (so, you don't do something like `k=10`

) and you just have probabilities, you can do the below. Note that your probabilities do not need to add up to 1, they can be independent of each other:

```
soup_items = ['pepper', 'onion', 'tomato', 'celery']
items_probability = [0.2, 0.3, 0.9, 0.1]
selected_items = [item for item,p in zip(soup_items,items_probability) if random.random()<p]
print(selected_items)
>>>['pepper','tomato']
```

## Solution 27

**Step-1:** Generate CDF `F`

in which you're interesting

**Step-2:** Generate u.r.v. `u`

**Step-3:** Evaluate `z=F^{-1}(u)`

This modeling is described in course of probability theory or stochastic processes. This is applicable just because you have easy CDF.