Generate random array of floats between a range


I haven’t been able to find a function to generate an array of random floats of a given length between a certain range.

I’ve looked at Random sampling but no function seems to do what I need.

random.uniform comes close but it only returns a single element, not a specific number.

This is what I’m after:

ran_floats = some_function(low=0.5, high=13.3, size=50)

which would return an array of 50 random non-unique floats (ie: repetitions are allowed) uniformly distributed in the range [0.5, 13.3].

Is there such a function?

Asked By: Gabriel



Why not use a list comprehension?

In Python 2

ran_floats = [random.uniform(low,high) for _ in xrange(size)]

In Python 3, range works like xrange(ref)

ran_floats = [random.uniform(low,high) for _ in range(size)]
Answered By: isedev

Why not to combine random.uniform with a list comprehension?

>>> def random_floats(low, high, size):
...    return [random.uniform(low, high) for _ in xrange(size)]
>>> random_floats(0.5, 2.8, 5)
[2.366910411506704, 1.878800401620107, 1.0145196974227986, 2.332600336488709, 1.945869474662082]
Answered By: pkacprzak

There may already be a function to do what you’re looking for, but I don’t know about it (yet?).
In the meantime, I would suggess using:

ran_floats = numpy.random.rand(50) * (13.3-0.5) + 0.5

This will produce an array of shape (50,) with a uniform distribution between 0.5 and 13.3.

You could also define a function:

def random_uniform_range(shape=[1,],low=0,high=1):
    Random uniform range

    Produces a random uniform distribution of specified shape, with arbitrary max and
    min values. Default shape is [1], and default range is [0,1].
    return numpy.random.rand(shape) * (high - min) + min

EDIT: Hmm, yeah, so I missed it, there is numpy.random.uniform() with the same exact call you want!
Try import numpy; help(numpy.random.uniform) for more information.

Answered By: PhilMacKay

np.random.uniform fits your use case:

sampl = np.random.uniform(low=0.5, high=13.3, size=(50,))

Update Oct 2019:

While the syntax is still supported, it looks like the API changed with NumPy 1.17 to support greater control over the random number generator. Going forward the API has changed and you should look at

The enhancement proposal is here:

Answered By: JoshAdel

This is the simplest way

Answered By: George Gee

The for loop in list comprehension takes time and makes it slow.
It is better to use numpy parameters (low, high, size, ..etc)

import numpy as np
import time
rang = 10000
tic = time.time()
for i in range(rang):
    sampl = np.random.uniform(low=0, high=2, size=(182))
print("it took: ", time.time() - tic)

tic = time.time()
for i in range(rang):
    ran_floats = [np.random.uniform(0,2) for _ in range(182)]
print("it took: ", time.time() - tic)

sample output:

(‘it took: ‘, 0.06406784057617188)

(‘it took: ‘, 1.7253198623657227)

Answered By: Mohamed Ibrahim

Alternatively you could use SciPy

from scipy import stats
stats.uniform(0.5, 13.3).rvs(50)

and for the record to sample integers it’s

stats.randint(10, 20).rvs(50)
Answered By: Stuart Hallows

np.random.random_sample(size) will generate random floats in the half-open interval [0.0, 1.0).

Answered By: shivaraj karki

Alternatively, if you are OK with a list of real numbers instead, you can use the standard random.randrange:

def some_function(low, high, size):
    low_int = int(low * 1000)
    high_int = int(high *1000)
    return [random.randrange(low_int, high_int, size)/1000 for _ in range(size)]
Answered By: Paul

This should work for your example

sample = (np.random.random([50, ]) * 13.3) - 0.5
Answered By: Marnim Galib
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