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Numpy 1 Degree of Freedom

In the following code I'm confused as to what the third line means. What does the ddof = 1 do. I tried looking it up, but I still don't quite understand the concept or the purpose. I would really appreciate it if somebody could point me in the right direction.
Thanks
data = stats.binom.rvs(n = 10, p = 0.3, size = 10000)
print "Mean: %g" % np.mean(data)
print "SD: %g" % np.std(data, **ddof=1**)

First read the documentation:
Means Delta Degrees of Freedom. The divisor used in
calculations is N - ddof, where N represents the number of elements.
By default ddof is zero.
Searching for Degrees of Freedom then explains the statistical concept (emphasis mine):
Estimates of statistical parameters can be based upon different
amounts of information or data. The number of independent pieces of
information that go into the estimate of a parameter are called the
degrees of freedom. In general, the degrees of freedom of an estimate
of a parameter are equal to the number of independent scores that go
into the estimate minus the number of parameters used as intermediate
steps in the estimation of the parameter itself (most of the time the
sample variance has N − 1 degrees of freedom, since it is computed
from N random scores minus the only 1 parameter estimated as
intermediate step, which is the sample mean).

Degrees of freedom is an important concept which you may want to look it up, but the computational difference is actually straight forward, consider these:
In [20]:
x = np.array([6,5,4,6,6,7,2])
In [21]:
np.std(x)
Out[21]:
1.5518257844571737
#default is ddof=0, what this actually does:
In [22]:
np.sqrt((((x-x.mean())**2)/len(x)).sum())
Out[22]:
1.5518257844571737
In [23]:
np.std(x, ddof=1)
Out[23]:
1.6761634196950517
#what ddof=1 does:
In [24]:
np.sqrt((((x-x.mean())**2)/(len(x)-1)).sum())
Out[24]:
1.6761634196950517
In most languages (R, SAS etc), the default is to return std of ddof=1. numpy's default is ddof=0, which something worth noting.

It refers to denominator degrees of freedom
In some cases (e.g. working with population level data) your denominator is N. In other cases (e.g. sample level data) your denominator is N-1 (or whatever your ddof value is set to).
So the difference here is do you want to divide by N, or divide by N-ddof? Where and when you divide by each is a more domain/context specific question
In numpy, the default is ddof = 0 (divide by N), so if you want a different denominator value you have to manually specify it

ddof = 1 refers to Degrees of Freedom.
It's a statistical concept.
The formula for Standard Deviation of population:
But most of the time, we are trying to use sample Standard Deviation to estimate the true Standard Deviation of population. The formula above is a downward-biased estimation, using N-1 instead of N gives us a correction.
Check wiki: https://en.wikipedia.org/wiki/Standard_deviation#Corrected_sample_standard_deviation for more information

Why does numpy std() give a different result to matlab std()?

I try to convert matlab code to numpy and figured out that numpy has a different result with the std function.
in matlab
std([1,3,4,6])
ans = 2.0817
in numpy
np.std([1,3,4,6])
1.8027756377319946
Is this normal? And how should I handle this?

The NumPy function np.std takes an optional parameter ddof: "Delta Degrees of Freedom". By default, this is 0. Set it to 1 to get the MATLAB result:
>>> np.std([1,3,4,6], ddof=1)
2.0816659994661326
To add a little more context, in the calculation of the variance (of which the standard deviation is the square root) we typically divide by the number of values we have.
But if we select a random sample of N elements from a larger distribution and calculate the variance, division by N can lead to an underestimate of the actual variance. To fix this, we can lower the number we divide by (the degrees of freedom) to a number less than N (usually N-1). The ddof parameter allows us change the divisor by the amount we specify.
Unless told otherwise, NumPy will calculate the biased estimator for the variance (ddof=0, dividing by N). This is what you want if you are working with the entire distribution (and not a subset of values which have been randomly picked from a larger distribution). If the ddof parameter is given, NumPy divides by N - ddof instead.
The default behaviour of MATLAB's std is to correct the bias for sample variance by dividing by N-1. This gets rid of some of (but probably not all of) of the bias in the standard deviation. This is likely to be what you want if you're using the function on a random sample of a larger distribution.
The nice answer by #hbaderts gives further mathematical details.

The standard deviation is the square root of the variance. The variance of a random variable X is defined as
An estimator for the variance would therefore be
where denotes the sample mean. For randomly selected , it can be shown that this estimator does not converge to the real variance, but to
If you randomly select samples and estimate the sample mean and variance, you will have to use a corrected (unbiased) estimator
which will converge to . The correction term is also called Bessel's correction.
Now by default, MATLABs std calculates the unbiased estimator with the correction term n-1. NumPy however (as #ajcr explained) calculates the biased estimator with no correction term by default. The parameter ddof allows to set any correction term n-ddof. By setting it to 1 you get the same result as in MATLAB.
Similarly, MATLAB allows to add a second parameter w, which specifies the "weighing scheme". The default, w=0, results in the correction term n-1 (unbiased estimator), while for w=1, only n is used as correction term (biased estimator).

For people who aren't great with statistics, a simplistic guide is:
Include ddof=1 if you're calculating np.std() for a sample taken from your full dataset.
Ensure ddof=0 if you're calculating np.std() for the full population
The DDOF is included for samples in order to counterbalance bias that can occur in the numbers.

Random Number Generator Explanation

from random import *
def main():
t = 0
for i in range(1000): # thousand
t += random()
print(t/1000)
main()
I was looking at the source code for a sample program my professor gave me and I came across this RNG. can anyone explain how this RNG works?

If you plotted the points, you would see that this actually produces a Gaussian ("normal") distribution about the mean of the random function.
Generate random numbers following a normal distribution in C/C++ talks about random number generation; it's a pretty common technique to do this if all you have is a uniform number generator like in standard C.
What I've given you here is a histogram of 100,000 values drawn from your function (of course, returned not printed, if you aren't familiar with python). The y axis is the frequency that the value appears, the x axis is the bin of the value. As you can see, the average value is 1/2, and by 3 standard deviations (99.7 percent of the data) we have almost no values in the range. That should be intuitive; we "usually" get 1/2, and very rarely get .99999

Have a look at the documentation. Its quite well written:
https://docs.python.org/2/library/random.html
The idea is that that program generates a random number 1000 times which is sufficiently enough to get mean as 0.5

The program is using the Central Limit Theorem - sums of independent and identically distributed random variables X with finite variance asymptotically converge to a normal (a.k.a. Gaussian) distribution whose mean is the sum of the means, and variance is the sum of the variances. Scaling this by N, the number of X's summed, gives the sample mean (a.k.a. average). If the expected value of X is μ and the variance of X is σ2, the expected value of the sample mean is also μ and it has variance σ2 / N.
Since a Uniform(0,1) has mean 0.5 and variance 1/12, your algorithm will generate results that are pretty close to normally distributed with a mean of 0.5 and a variance of 1/12000. Consequently 99.7% of the outcomes should fall within +/-3 standard deviations of the mean, i.e., in the range 0.5+/-0.0274.
This is a ridiculously inefficient way to generate normals. Better alternatives include the Box-Muller method, Polar method, or ziggurat method.

The thing making this random is the random() function being called. random() will generate 1 (for most practical purposes) random float between 0 and 1.
>>>random()
0.1759916412898097
>>>random()
0.5489228122596088
etc.
The rest of it is just adding each random to a total and then dividing by the number of randoms, essentially finding the average of all 1000 randoms, which as Cyber pointed out is actually not a random number at all.

Python: Numpy standard deviation error

This is a simple test
import numpy as np
data = np.array([-1,0,1])
print data.std()
>> 0.816496580928
I don't understand how this result been generated? Obviously:
( (1^0.5 + 1^0.5 + 0^0.5)/(3-1) )^0.5 = 1
and in matlab it gives me std([-1,0,1]) = 1. Could you help me get understand how numpy.std() works?

The crux of this problem is that you need to divide by N (3), not N-1 (2). As Iarsmans pointed out, numpy will use the population variance, not the sample variance.
So the real answer is sqrt(2/3) which is exactly that: 0.8164965...
If you happen to be trying to deliberately use a different value (than the default of 0) for the degrees of freedom, use the keyword argument ddofwith a positive value other than 0:
np.std(data, ddof=1)
... but doing so here would reintroduce your original problem as numpy will divide by N - ddof.

It is worth reading the help page for the function/method before suggesting it is incorrect. The method does exactly what the doc-string says it should be doing, divides by 3, because By default ddofis zero.:
In [3]: numpy.std?
String form: <function std at 0x104222398>
File: /System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/numpy/core/fromnumeric.py
Definition: numpy.std(a, axis=None, dtype=None, out=None, ddof=0, keepdims=False)
Docstring:
Compute the standard deviation along the specified axis.
...
ddof : int, optional
Means Delta Degrees of Freedom. The divisor used in calculations
is ``N - ddof``, where ``N`` represents the number of elements.
By default `ddof` is zero.

When getting into NumPy from Matlab, you'll probably want to keep the docs for both handy. They're similar but often differ in small but important details. Basically, they calculate the standard deviation differently. I would strongly recommend checking the documentation for anything you use that calculates standard deviation, whether a pocket calculator or a programming language, since the default is not (sorry!) standardized.
Numpy STD: http://docs.scipy.org/doc/numpy/reference/generated/numpy.std.html
Matlab STD: http://www.mathworks.com/help/matlab/ref/std.html
The Numpy docs for std are a bit opaque, IMHO, especially considering that NumPy docs are generally fairly clear. If you read far enough: The average squared deviation is normally calculated as x.sum() / N, where N = len(x). If, however, ddof is specified, the divisor N - ddof is used instead. In standard statistical practice, ddof=1 provides an unbiased estimator of the variance of the infinite population. (In english, default is pop std dev, set ddof=1 for sample std dev).
OTOH, the Matlab docs make clear the difference that's tripping you up:
There are two common textbook definitions for the standard deviation s of a data vector X. [equations omitted] n is the number of elements in the sample. The two forms of the equation differ only in n – 1 versus n in the divisor.
So, by default, Matlab calculates the sample standard deviation (N-1 in the divisor, so bigger to compensate for the fact this is a sample) and Numpy calculates the population standard deviation (N in the divisor). You use the ddof parameter to switch to the sample standard, or any other denominator you want (which goes beyond my statistics knowledge).
Lastly, it doesn't help on this problem, but you'll probably find this helpful at some point. Link

Scipy standard deviation

I'm trying to calculate standard deviation for some distribution and keep getting two different results from two paths. It doesn't make much sense to me - could someone explain why is this happening?
scipy.stats.binom(189, 100/189).std()
6.8622115305451707
scipy.stats.tstd([1]*100 + [0]*89)
0.50047821327986164
Why aren't those two numbers equal?

The basic reason is that you're taking the standard deviation of two quite different things there. I think you're misunderstanding what scipy.stats.binom does. From the documentation:
The probability mass function for binom is:
binom.pmf(k) = choose(n,k) * p**k * (1-p)**(n-k)
for k in {0,1,...,n}.
binom takes n and p as shape parameters.
When you do binom(189, 100/189), you are creating a distribution that could take on any value from 0 to 189. This distribution unsurprisingly has a much larger variance than the other sample data you're using, which is restricted to values of either zero or one.
It looks like what you want would be scipy.stats.binom(1, 100/189).std(). However, you still can't expect the exact same value as what you're getting with your sample data, because the binom.std is computing the standard deviation of the overall distribution, whereas the other version (scipy.stats.tstd([1]*100 + [0]*89)) is computing the standard deviation only of a sample. If you increase the size of your sample (e.g., do scipy.stats.tstd([1]*1000 + [0]*890)), the sample standard deviation will approach the value you're getting from binom.std.
You can also get the population (not sample) std by using scipy.std or numpy.std instead of scipy.stats.tstd. scipy.stats.tstd doesn't have a ddof option to let you choose the degrees of freedom, and always computes a sample std.