import numpy as np
(1)np.random.random_sample
help(np.random.random_sample)
Help on built-in function random_sample:
random_sample(...) method of mtrand.RandomState instance
random_sample(size=None)
Return random floats in the half-open interval [0.0, 1.0).
Results are from the "continuous uniform" distribution over the
stated interval. To sample :math:`Unif[a, b), b > a` multiply
the output of `random_sample` by `(b-a)` and add `a`::
(b - a) * random_sample() + a
Parameters
----------
size : int or tuple of ints, optional
Output shape. If the given shape is, e.g., ``(m, n, k)``, then
``m * n * k`` samples are drawn. Default is None, in which case a
single value is returned.
Returns
-------
out : float or ndarray of floats
Array of random floats of shape `size` (unless ``size=None``, in which
case a single float is returned).
Examples
--------
>>> np.random.random_sample()
0.47108547995356098
>>> type(np.random.random_sample())
<type 'float'>
>>> np.random.random_sample((5,))
array([ 0.30220482, 0.86820401, 0.1654503 , 0.11659149, 0.54323428])
Three-by-two array of random numbers from [-5, 0):
>>> 5 * np.random.random_sample((3, 2)) - 5
array([[-3.99149989, -0.52338984],
[-2.99091858, -0.79479508],
[-1.23204345, -1.75224494]])
np.random.random_sample((m,n))给出m*n的均匀分布的随机数组。
np.random.random_sample((3,3))
array([[0.29173378, 0.29419769, 0.47702224],
[0.67796941, 0.12930519, 0.06604015],
[0.41514769, 0.61113887, 0.2035459 ]])
如果要得到[a,b)的均匀分布的随机数组，只需要(b-a)*np.random.random_sample((m,n))+a:
(5-2)*np.random.random_sample((3,3))+2 #2 ~5均匀分布随机数
array([[4.19892161, 4.66196903, 2.90223592],
[4.70942939, 2.73620934, 2.22161209],
[3.46348355, 4.58905218, 2.33377181]])
（2）np.random.rand
help(np.random.rand)
Help on built-in function rand:
rand(...) method of mtrand.RandomState instance
rand(d0, d1, ..., dn)
Random values in a given shape.
Create an array of the given shape and populate it with
random samples from a uniform distribution
over ``[0, 1)``.
Parameters
----------
d0, d1, ..., dn : int, optional
The dimensions of the returned array, should all be positive.
If no argument is given a single Python float is returned.
Returns
-------
out : ndarray, shape ``(d0, d1, ..., dn)``
Random values.
See Also
--------
random
Notes
-----
This is a convenience function. If you want an interface that
takes a shape-tuple as the first argument, refer to
np.random.random_sample .
Examples
--------
>>> np.random.rand(3,2)
array([[ 0.14022471, 0.96360618], #random
[ 0.37601032, 0.25528411], #random
[ 0.49313049, 0.94909878]]) #random
np.random.rand（d0,d1..)是np.random.random_sample的简洁用法，也是形成d0*d1*...维度的[0,1)的随机数组
np.random.rand(3,3)
array([[0.30821699, 0.64856611, 0.17575119],
[0.99049116, 0.60985425, 0.01740196],
[0.49243809, 0.98859105, 0.71051433]])
(3)np.random.randn
help(np.random.randn)
Help on built-in function randn:
randn(...) method of mtrand.RandomState instance
randn(d0, d1, ..., dn)
Return a sample (or samples) from the "standard normal" distribution.
If positive, int_like or int-convertible arguments are provided,
`randn` generates an array of shape ``(d0, d1, ..., dn)``, filled
with random floats sampled from a univariate "normal" (Gaussian)
distribution of mean 0 and variance 1 (if any of the :math:`d_i` are
floats, they are first converted to integers by truncation). A single
float randomly sampled from the distribution is returned if no
argument is provided.
This is a convenience function. If you want an interface that takes a
tuple as the first argument, use `numpy.random.standard_normal` instead.
Parameters
----------
d0, d1, ..., dn : int, optional
The dimensions of the returned array, should be all positive.
If no argument is given a single Python float is returned.
Returns
-------
Z : ndarray or float
A ``(d0, d1, ..., dn)``-shaped array of floating-point samples from
the standard normal distribution, or a single such float if
no parameters were supplied.
See Also
--------
standard_normal : Similar, but takes a tuple as its argument.
Notes
-----
For random samples from :math:`N(mu, sigma^2)`, use:
``sigma * np.random.randn(...) + mu``
Examples
--------
>>> np.random.randn()
2.1923875335537315 #random
Two-by-four array of samples from N(3, 6.25):
>>> 2.5 * np.random.randn(2, 4) + 3
array([[-4.49401501, 4.00950034, -1.81814867, 7.29718677], #random
[ 0.39924804, 4.68456316, 4.99394529, 4.84057254]]) #random
np.random.randn(d0,d1...dn)是形成(d0*d1...dn)维度的均值为0，均方差为1的标准正太分布随机数组.
(4) np.random.randint
help(np.random.randint)
Help on built-in function randint:
randint(...) method of mtrand.RandomState instance
randint(low, high=None, size=None, dtype='l')
Return random integers from `low` (inclusive) to `high` (exclusive).
Return random integers from the "discrete uniform" distribution of
the specified dtype in the "half-open" interval [`low`, `high`). If
`high` is None (the default), then results are from [0, `low`).
Parameters
----------
low : int
Lowest (signed) integer to be drawn from the distribution (unless
``high=None``, in which case this parameter is one above the
*highest* such integer).
high : int, optional
If provided, one above the largest (signed) integer to be drawn
from the distribution (see above for behavior if ``high=None``).
size : int or tuple of ints, optional
Output shape. If the given shape is, e.g., ``(m, n, k)``, then
``m * n * k`` samples are drawn. Default is None, in which case a
single value is returned.
dtype : dtype, optional
Desired dtype of the result. All dtypes are determined by their
name, i.e., 'int64', 'int', etc, so byteorder is not available
and a specific precision may have different C types depending
on the platform. The default value is 'np.int'.
.. versionadded:: 1.11.0
Returns
-------
out : int or ndarray of ints
`size`-shaped array of random integers from the appropriate
distribution, or a single such random int if `size` not provided.
See Also
--------
random.random_integers : similar to `randint`, only for the closed
interval [`low`, `high`], and 1 is the lowest value if `high` is
omitted. In particular, this other one is the one to use to generate
uniformly distributed discrete non-integers.
Examples
--------
>>> np.random.randint(2, size=10)
array([1, 0, 0, 0, 1, 1, 0, 0, 1, 0])
>>> np.random.randint(1, size=10)
array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0])
Generate a 2 x 4 array of ints between 0 and 4, inclusive:
>>> np.random.randint(5, size=(2, 4))
array([[4, 0, 2, 1],
[3, 2, 2, 0]])
randint(low, high=None, size=None, dtype='l')是得到size规模的[low,high)的随机数组，如果high=None，那么，形成size规模的[0,low)的随机数组；
random_integers(low,high=None,size=None,dtype='1')与randint类似，唯一的区别是[low,high]。
（5）np.random.binomial
help(np.random.binomial)
Help on built-in function binomial:
binomial(...) method of mtrand.RandomState instance
binomial(n, p, size=None)
Draw samples from a binomial distribution.
Samples are drawn from a binomial distribution with specified
parameters, n trials and p probability of success where
n an integer >= 0 and p is in the interval [0,1]. (n may be
input as a float, but it is truncated to an integer in use)
Parameters
----------
n : int or array_like of ints
Parameter of the distribution, >= 0. Floats are also accepted,
but they will be truncated to integers.
p : float or array_like of floats
Parameter of the distribution, >= 0 and <=1.
size : int or tuple of ints, optional
Output shape. If the given shape is, e.g., ``(m, n, k)``, then
``m * n * k`` samples are drawn. If size is ``None`` (default),
a single value is returned if ``n`` and ``p`` are both scalars.
Otherwise, ``np.broadcast(n, p).size`` samples are drawn.
Returns
-------
out : ndarray or scalar
Drawn samples from the parameterized binomial distribution, where
each sample is equal to the number of successes over the n trials.
See Also
--------
scipy.stats.binom : probability density function, distribution or
cumulative density function, etc.
Notes
-----
The probability density for the binomial distribution is
.. math:: P(N) = inom{n}{N}p^N(1-p)^{n-N},
where :math:`n` is the number of trials, :math:`p` is the probability
of success, and :math:`N` is the number of successes.
When estimating the standard error of a proportion in a population by
using a random sample, the normal distribution works well unless the
product p*n <=5, where p = population proportion estimate, and n =
number of samples, in which case the binomial distribution is used
instead. For example, a sample of 15 people shows 4 who are left
handed, and 11 who are right handed. Then p = 4/15 = 27%. 0.27*15 = 4,
so the binomial distribution should be used in this case.
References
----------
.. [1] Dalgaard, Peter, "Introductory Statistics with R",
Springer-Verlag, 2002.
.. [2] Glantz, Stanton A. "Primer of Biostatistics.", McGraw-Hill,
Fifth Edition, 2002.
.. [3] Lentner, Marvin, "Elementary Applied Statistics", Bogden
and Quigley, 1972.
.. [4] Weisstein, Eric W. "Binomial Distribution." From MathWorld--A
Wolfram Web Resource.
http://mathworld.wolfram.com/BinomialDistribution.html
.. [5] Wikipedia, "Binomial distribution",
http://en.wikipedia.org/wiki/Binomial_distribution
Examples
--------
Draw samples from the distribution:
>>> n, p = 10, .5 # number of trials, probability of each trial
>>> s = np.random.binomial(n, p, 1000)
# result of flipping a coin 10 times, tested 1000 times.
A real world example. A company drills 9 wild-cat oil exploration
wells, each with an estimated probability of success of 0.1. All nine
wells fail. What is the probability of that happening?
Let's do 20,000 trials of the model, and count the number that
generate zero positive results.
>>> sum(np.random.binomial(9, 0.1, 20000) == 0)/20000.
# answer = 0.38885, or 38%.
binomial(n, p, size=None) 形成符合二项分布概率的size的[0,n]的随机数组，其中二项分布的单次成功概率为p
np.random.binomial(5,0.1,4)
array([0, 2, 0, 0])
5 in np.random.binomial(5,0.1,10000)
False
5 in np.random.binomial(5,0.1,1000000)
True