Table of Contents
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Tutorial can be read at http://www.labri.fr/perso/nrougier/teaching/numpy/numpy.html
NumPy is the fundamental package for scientific computing with Python. It contains among other things:
Besides its obvious scientific uses, NumPy can also be used as an efficient multidimensional container of generic data. Arbitrary datatypes can be defined and this allows NumPy to seamlessly and speedily integrate with a wide variety of projects. We are going to explore numpy through a simple example, implementing the Game of Life.
Numpy is slanted toward scientific computing and we'll consider in this section the game of life by John Conway which is one of the earliest example of cellular automata (see figure below). Those cellular automaton can be conveniently considered as array of cells that are connected together through the notion of neighbours. We'll show in the following sections implementation of this game using pure python and numpy in order to illustrate main differences with python and numpy.
Note
This is an excerpt from wikipedia entry on Cellular Automaton.
The Game of Life, also known simply as Life, is a cellular automaton devised by the British mathematician John Horton Conway in 1970. It is the bestknown example of a cellular automaton. The "game" is actually a zeroplayer game, meaning that its evolution is determined by its initial state, needing no input from human players. One interacts with the Game of Life by creating an initial configuration and observing how it evolves.
The universe of the Game of Life is an infinite twodimensional orthogonal grid of square cells, each of which is in one of two possible states, live or dead. Every cell interacts with its eight neighbours, which are the cells that are directly horizontally, vertically, or diagonally adjacent. At each step in time, the following transitions occur:
The initial pattern constitutes the 'seed' of the system. The first generation is created by applying the above rules simultaneously to every cell in the seed – births and deaths happen simultaneously, and the discrete moment at which this happens is sometimes called a tick. (In other words, each generation is a pure function of the one before.) The rules continue to be applied repeatedly to create further generations.
We'll first use a very simple setup and more precisely, we'll use the glider pattern that is known to move one step diagonally in 4 iterations as illustrated below:
t = 0 
t = 1 
t = 2 
t = 3 
t = 4 
This property will help us debug our scripts.
Note
We could have used the more efficient python array interface but people may be more familiar with the list object.
In pure python, we can code the Game of Life using a list of lists representing the board where cells are supposed to evolve:
>>> Z = [[0,0,0,0,0,0], [0,0,0,1,0,0], [0,1,0,1,0,0], [0,0,1,1,0,0], [0,0,0,0,0,0], [0,0,0,0,0,0]]
This board possesses a 0 border that allows to accelerate things a bit by avoiding to have specific tests for borders when counting the number of neighbours. First step is to count neighbours:
def compute_neigbours(Z): shape = len(Z), len(Z[0]) N = [[0,]*(shape[0]) for i in range(shape[1])] for x in range(1,shape[0]1): for y in range(1,shape[1]1): N[x][y] = Z[x1][y1]+Z[x][y1]+Z[x+1][y1] \ + Z[x1][y] +Z[x+1][y] \ + Z[x1][y+1]+Z[x][y+1]+Z[x+1][y+1] return N
To iterate one step in time, we then simply count the number of neighbours for each internal cell and we update the whole board according to the 4 rules:
def iterate(Z): N = compute_neighbours(Z) for x in range(1,shape[0]1): for y in range(1,shape[1]1): if Z[x][y] == 1 and (N[x][y] < 2 or N[x][y] > 3): Z[x][y] = 0 elif Z[x][y] == 0 and N[x][y] == 3: Z[x][y] = 1 return Z
Note
The show command is supplied witht he script.
Using a dedicated display function, we can check the program's correct:
>>> show(Z) [0, 0, 1, 0] [1, 0, 1, 0] [0, 1, 1, 0] [0, 0, 0, 0] >>> for i in range(4): iterate(Z) >>> show(Z) [0, 0, 0, 0] [0, 0, 0, 1] [0, 1, 0, 1] [0, 0, 1, 1]
You can download the full script here: gameoflifepython.py
Note
There exists many more different ways to create a numpy array.
The first thing to do is to create the proper numpy array to hold the cells. This can be done very easily with:
>>> import numpy as np >>> Z = np.array([[0,0,0,0,0,0], [0,0,0,1,0,0], [0,1,0,1,0,0], [0,0,1,1,0,0], [0,0,0,0,0,0], [0,0,0,0,0,0]])
Note
For a complete review on numpy data types, check the documentation.
Note that we did not specify the data type of the array and thus, numpy will choose one for us. Since all elements are integers, numpy will then choose an integer data type. This can be easily checked using:
>>> print Z.dtype int64
We can also check the shape of the array to make sure it is 6x6:
>>> print Z.shape (6, 6)
Each element of Z can be accessed using a row and a column index (in that order):
>>> print Z[0,5] 0
Note
This element access is actually called indexing and this is very powerful tool for vectorized computation.
But even better, we can also access a subpart of the array using the slice notation:
>>> print Z[1:5,1:5] [[0 0 1 0] [1 0 1 0] [0 1 1 0] [0 0 0 0]]
In the example above, we actually extract a subpart of Z ranging from rows 1 to 5 and columns 1 to 5. It is important to understand at this point that this is really a subpart of Z in the sense that any change to this subpart will have immediate impact on Z:
>>> A = Z[1:5,1:5] >>> A[0,0] = 9 >>> print A [[9 0 1 0] [1 0 1 0] [0 1 1 0] [0 0 0 0]]>>> print Z [[0 0 0 0 0 0] [0 9 0 1 0 0] [0 1 0 1 0 0] [0 0 1 1 0 0] [0 0 0 0 0 0] [0 0 0 0 0 0]]
We set the value of A[0,0] to 9 and we see immediate change in Z[1,1] because A[0,0] actually corresponds to Z[1,1]. This may seem trivial with such simple arrays, but things can become much more complex (we'll see that later). If in doubt, you can check easily if an array is part of another one:
>>> print Z.base is None True >>> print A.base is Z True
Note
It is not always possible to vectorize computations and it requires generally some experience. You'll acquire this experience by using numpy (of course) but also by asking questions on the mailing list
We now need a function to count the neighbours. We could do it the same way as for the python version, but this would make things very slow because of the nested loops. We would prefer to act on the whole array whenever possible, this is called vectorization.
Ok, let's start then...
First, you need to know that you can manipulate Z as if (and only as if) it was a regular scalar:
>>> print 1+(2*Z+3) [[4 4 4 4 4 4] [4 4 4 6 4 4] [4 6 4 6 4 4] [4 4 6 6 4 4] [4 4 4 4 4 4] [4 4 4 4 4 4]]
If you look carefully at the output, you may realize that the ouptut corresponds to the formula above applied individually to each element. Said differently, we have (1+(2*Z+3))[i,j] == (1+(2*Z[i,j]+3)) for any i,j.
Ok, so far, so good. Now what happens if we add Z with one of its subpart, let's say Z[1:1,1:1] ?
>>> Z + Z[1:1,1:1] Traceback (most recent call last): File "<stdin>", line 1, in <module> ValueError: operands could not be broadcast together with shapes (6,6) (4,4)
This raises a Value Error, but more interestingly, numpy complains about the impossibility of broadcasting the two arrays together. Broadcasting is a very powerful feature of numpy and most of the time, it saves you a lot of hassle. Let's consider for example the following code:
>>> print Z+1 [[1 1 1 1 1 1] [1 1 1 2 1 1] [1 2 1 2 1 1] [1 1 2 2 1 1] [1 1 1 1 1 1] [1 1 1 1 1 1]]
Note
See also the broadcasting section in the numpy documentation.
How can a matrix and a scalar be added together ? Well, they can't. But numpy is smart enough to guess that you actually want to add 1 to each of the element of Z. This concept of broadcasting is quite powerful and it will take you some time before masterizing it fully (if even possible).
However, in the present case (counting neighbours if you remember), we won't use broadcasting (uh ?). But we'll use vectorize computation using the following code:
>>> N = np.zeros(Z.shape, dtype=int) >>> N[1:1,1:1] += (Z[ :2, :2] + Z[ :2,1:1] + Z[ :2,2:] + Z[1:1, :2] + Z[1:1,2:] + Z[2: , :2] + Z[2: ,1:1] + Z[2: ,2:])
To understand this code, have a look at the figure below:
What we actually did with the above code is to add all the darker blue squares together. Since they have been chosen carefully, the result will be exactly what we expected. If you want to convince yourself, consider a cell in the lighter blue area of the central subfigure and check what will the result for a given cell.
Note
Note the use of the ravel function that flatten an array. This is necessary since the argwhere function returns flattened indices.
In a first approach, we can write the iterate function using the argwhere method that will give us the indices where a given condition is True.
def iterate(Z): # Iterate the game of life : naive version # Count neighbours N = np.zeros(Z.shape, int) N[1:1,1:1] += (Z[0:2,0:2] + Z[0:2,1:1] + Z[0:2,2:] + Z[1:1,0:2] + Z[1:1,2:] + Z[2: ,0:2] + Z[2: ,1:1] + Z[2: ,2:]) N_ = N.ravel() Z_ = Z.ravel() # Apply rules R1 = np.argwhere( (Z_==1) & (N_ < 2) ) R2 = np.argwhere( (Z_==1) & (N_ > 3) ) R3 = np.argwhere( (Z_==1) & ((N_==2)  (N_==3)) ) R4 = np.argwhere( (Z_==0) & (N_==3) ) # Set new values Z_[R1] = 0 Z_[R2] = 0 Z_[R3] = Z_[R3] Z_[R4] = 1 # Make sure borders stay null Z[0,:] = Z[1,:] = Z[:,0] = Z[:,1] = 0
Even if this first version does not use nested loops, it is far from optimal because of the use of the 4 argwhere calls that may be quite slow. We can instead take advantages of numpy features the following way.
def iterate_2(Z): # Count neighbours N = (Z[0:2,0:2] + Z[0:2,1:1] + Z[0:2,2:] + Z[1:1,0:2] + Z[1:1,2:] + Z[2: ,0:2] + Z[2: ,1:1] + Z[2: ,2:]) # Apply rules birth = (N==3) & (Z[1:1,1:1]==0) survive = ((N==2)  (N==3)) & (Z[1:1,1:1]==1) Z[...] = 0 Z[1:1,1:1][birth  survive] = 1 return Z
If you look at the birth and survive lines, you'll see that these two variables are indeed arrays. The righthand side of these two expressions are in fact logical expressions that will result in boolean arrays (just print them to check). We then set all Z values to 0 (all cells become dead) and we use the birth and survive arrays to conditionally set Z values to 1. And we're done ! Let's test this:
>>> print Z [[0 0 0 0 0 0] [0 0 0 1 0 0] [0 1 0 1 0 0] [0 0 1 1 0 0] [0 0 0 0 0 0] [0 0 0 0 0 0]] >>> for i in range(4): iterate_2(Z) >>> print Z [[0 0 0 0 0 0] [0 0 0 0 0 0] [0 0 0 0 1 0] [0 0 1 0 1 0] [0 0 0 1 1 0] [0 0 0 0 0 0]]
You can download the full script here: gameoflifenumpy.py
While numpy works perfectly with very small arrays, you'll really benefit from numpy power with big to very big arrays. So let us reconsider the game of life with a bigger array. First, we won't initalize the array by hand but initalize it randomly:
>>> Z = np.random.randint(0,2,(256,512))
and we simply iterate as previously:
>>> for i in range(100): iterate(Z)
and display result:
>>> size = np.array(Z.shape) >>> dpi = 72.0 >>> figsize= size[1]/float(dpi),size[0]/float(dpi) >>> fig = plt.figure(figsize=figsize, dpi=dpi, facecolor="white") >>> fig.add_axes([0.0, 0.0, 1.0, 1.0], frameon=False) >>> plt.imshow(Z,interpolation='nearest', cmap=plt.cm.gray_r) >>> plt.xticks([]), plt.yticks([]) >>> plt.show()
Easy enough, no ?
We have reviewed the very basics of numpy so let's move on to more complex (and more fun) things.
Note
Description taken from the GrayScott homepage
Reaction and diffusion of chemical species can produce a variety of patterns, reminiscent of those often seen in nature. The Gray Scott equations model such a reaction. For more information on this chemical system see the article Complex Patterns in a Simple System, John E. Pearson, Science, Volume 261, 9 July 1993.
Let's consider two chemical species U and V with respective concentrations u and v and diffusion rates r_{u} and r_{v}. V is converted into P with a rate of conversion k. f represents the rate of the process that feeds U and drains U, V and P. We can now write:
Chemical reaction  Equations 



Obviously, you may think we need two arrays, one for U and for V. But since U and V are tighly linked, it may be indeed better to use a single array. Numpy allows to do that with the notion of structured array:
>>> n = 200 >>> Z = np.zeros((n+2,n+2), [('U', np.double), ('V', np.double)]) >>> print Z.dtype [('U', '<f8'), ('V', '<f8')]
The size of the array is (n+2,n+2) since we need the borders when computing the neighbours. However, we'll compute differential equation only in the center part, so we can already creates some useful views of this array:
>>> U,V = Z['U'], Z['V'] >>> u,v = U[1:1,1:1], V[1:1,1:1]
Next, we need to compute the Laplacian and we'll use a discrete approximation obtained via the finite difference method using the same vectorization as for the Game of Life:
def laplacian(Z): return ( Z[0:2,1:1] + Z[1:1,0:2]  4*Z[1:1,1:1] + Z[1:1,2:] + Z[2: ,1:1] )
Finally, we can iterate the computation after havong choosed some interesting parameters:
for i in xrange(25000): Lu = laplacian(U) Lv = laplacian(V) uvv = u*v*v u += (Du*Lu  uvv + F *(1u)) v += (Dv*Lv + uvv  (F+k)*v )
And we're done !
You can download the full script here: grayscott.py
Here are some exercises, try to do them without looking at the solution (just highligh the blank part to see it).
Import the numpy package under the name np
import numpy as np
Print the numpy version and the configuration.
print np.__version__ np.__config__.show()
Hint
See np.zeros
Create a null vector of size 10
Z = np.zeros(10)
Create a null vector of size 10 but the fifth value which is 1
Z = np.zeros(10) Z[4] = 1
Hint
See np.arange
Create a vector with values ranging from 10 to 99
Z = 10 + np.arange(90)
Create a 3x3 matrix with values ranging from 0 to 8
Z = np.arange(9).reshape(3,3)
Hint
See np.nonzero
Find indices of nonzero elements from [1,2,0,0,4,0]
print np.nonzero([1,2,0,0,4,0])
Declare a 3x3 identity matrix
Z = np.eye(3)
Declare a 5x5 matrix with values 1,2,3,4 just below the diagonal
Z = np.diag(1+np.arange(4),k=1)
Hint
See Random sampling
Declare a 10x10x10 array with random values
Z = np.random.random((10,10,10))
Declare a 8x8 matrix and fill it with a checkerboard pattern
Z = np.zeros((8,8)) Z[1::2,::2] = 1 Z[::2,1::2] = 1
Declare a 10x10 array with random values and find the minimum and maximum values
Z = np.random.random((10,10,10)) Zmin, Zmax = Z.min(), Z.max()
Create a checkerboard 8x8 matrix using the tile function
Z = np.tile( np.array([[0,1],[1,0]]), (4,4))
Normalize a 5x5 random matrix (between 0 and 1)
Z = np.random.random((5,5)) Zmax,Zmin = Z.max(), Z.min() Z = (Z  Zmin)/(Zmax  Zmin)
Hint
See the linear algebra documentation
Multiply a 5x3 matrix by a 3x2 matrix (real matrix product)
Z = np.dot(np.ones((5,3)), np.ones((3,2)))
Create a 10x10 matrix with row values ranging from 0 to 9
Z = np.zeros((10,10)) Z += np.arange(10)
Create a vector of size 1000 with values ranging from 0 to 1, both excluded
Z = np.random.linspace(0,1,1002,endpoint=True)[1:1]
Create a random vector of size 100 and sort it
Z = np.random.random(100) Z.sort()
Consider two random matrices A anb B, check if they are equal.
A = np.random.randint(0,2,(2,2)) B = np.random.randint(0,2,(2,2)) equal = np.allclose(A,B)
Create a random vector of size 1000 and find the mean value
Z = np.random.random(1000) m = Z.mean()
Consider a random 100x2 matrix representing cartesian coordinates, convert them to polar coordinates
Z = np.random.random((100,2)) X,Y = Z[:,0], Z[:,1] R = np.sqrt(X**2+Y**2) T = np.arctan2(Y,X)
Create random vector of size 100 and replace the maximum value by 0
Z = np.random.random(100) Z[Z.argmax()] = 0
Hint
See the documentation on Structured arrays
Declare a structured array with x and y coordinates covering the [0,1]x[0,1] area.
Z = np.zeros((10,10), [('x',float),('y',float)]) Z['x'], Z['y'] = np.meshgrid(np.linspace(0,1,10), np.linspace(0,1,10))
Hint
Have a look at Data type routines
Print the minimum and maximum representable value for each numpy scalar type
for dtype in [np.int8, np.int32, np.int64]: print np.iinfo(dtype).min print np.iinfo(dtype).max for dtype in [np.float32, np.float64]: print np.finfo(dtype).min print np.finfo(dtype).max print np.finfo(dtype).eps
Create a structured array representing a position (x,y) and a color (r,g,b)
Z = np.zeros(10, [ ('position', [ ('x', float, 1), ('y', float, 1)]), ('color', [ ('r', float, 1), ('g', float, 1), ('b', float, 1)])])
Consider a random vector with shape (100,2) representing coordinates, find point by point distances
Z = np.random.random((10,2)) X,Y = np.atleast_2d(Z[:,0]), np.atleast_2d(Z[:,1]) D = np.sqrt( (XX.T)**2 + (YY.T)**2)
Generate a generic 2D Gaussianlike array
X, Y = np.meshgrid(np.linspace(1,1,100), np.linspace(1,1,100)) D = np.sqrt(X*X+Y*Y) sigma, mu = 1.0, 0.0 G = np.exp(( (Dmu)**2 / ( 2.0 * sigma**2 ) ) )
Consider the vector [1, 2, 3, 4, 5], how to build a new vector with 3 consecutive zeros interleaved between each value ?
Z = np.array([1,2,3,4,5]) nz = 3 Z0 = np.zeros(len(Z) + (len(Z)1)*(nz)) Z0[::nz+1] = Z
Numpy benefits from extensive documentation as well as a large community of users and developpers. Here are some links of interest:
The SciPy Lecture notes offers a teaching material on the scientific Python ecosystem as well as quick introduction to central tools and techniques. The different chapters each correspond to a 1 to 2 hours course with increasing level of expertise, from beginner to expert.
A firstaid kit for the numerically adventurous by Stéfan van der Walt.
An introduction to Numpy and Scipy
A short introduction to Numpy and Scipy by M. Scott Shell.
This guide is intended as an introductory overview of NumPy and explains how to install and make use of the most important features of NumPy.
This reference manual details functions, modules, and objects included in Numpy, describing what they are and what they do.
 General questions about numpy
 General questions about SciPy
 Basic SciPy/numpy usage
 Advanced NumPy/SciPy usage
 NumPy/SciPy installation
 Miscellaneous Issues
The code is fairly well documented and you can quickly access a specific command from within a python session:
>>> import numpy as np >>> help(np.ones) Help on function ones in module numpy.core.numeric: ones(shape, dtype=None, order='C') Return a new array of given shape and type, filled with ones. Please refer to the documentation for `zeros` for further details. See Also  zeros, ones_like Examples  >>> np.ones(5) array([ 1., 1., 1., 1., 1.]) >>> np.ones((5,), dtype=np.int) array([1, 1, 1, 1, 1]) >>> np.ones((2, 1)) array([[ 1.], [ 1.]]) >>> s = (2,2) >>> np.ones(s) array([[ 1., 1.], [ 1., 1.]])
Finally, there is a mailing list where you can ask for help.
Data type  Description 

bool  Boolean (True or False) stored as a byte 
int  Platform integer (normally either int32 or int64) 
int8  Byte (128 to 127) 
int16  Integer (32768 to 32767) 
int32  Integer (2147483648 to 2147483647) 
int64  Integer (9223372036854775808 to 9223372036854775807) 
uint8  Unsigned integer (0 to 255) 
uint16  Unsigned integer (0 to 65535) 
uint32  Unsigned integer (0 to 4294967295) 
uint64  Unsigned integer (0 to 18446744073709551615) 
float  Shorthand for float64. 
float16  Half precision float: sign bit, 5 bits exponent, 10 bits mantissa 
float32  Single precision float: sign bit, 8 bits exponent, 23 bits mantissa 
float64  Double precision float: sign bit, 11 bits exponent, 52 bits mantissa 
complex  Shorthand for complex128. 
complex64  Complex number, represented by two 32bit floats 
complex128  Complex number, represented by two 64bit floats 
Code  Result  Code  Result  

Z = zeros(9) 
Z = zeros((5,9)) 

Z = ones(9) 
Z = ones((5,9)) 

Z = array( [0,0,0,0,0,0,0,0,0] ) 
Z = array( [[0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,0]]) 

Z = arange(9) 
Z = arange(5*9).reshape(5,9) 

Z = random.uniform(0,1,9) 
Z = random.uniform(0,1,(5,9)) 
Code  Result  Code  Result 

Z[2,2] = 1 
Z = Z.reshape(1,12) 

Z = Z.reshape(4,3) 
Z = Z.reshape(12,1) 

Z = Z.reshape(6,2) 

Z = Z.reshape(2,6) 
Code  Result  Code  Result  

Z 
Z[...] = 1 

Z[1,1] = 1 
Z[:,0] = 1 

Z[0,:] = 1 
Z[2:,2:] = 1 

Z[:,::2] = 1 
Z[::2,:] = 1 

Z[:2,:2] = 1 
Z[2:4,2:4] = 1 

Z[::2,::2] = 1 
Z[3::2,3::2] = 1 
+  →  +  =  
+  →  +  =  
+  →  +  =  
+  →  +  = 
Code  Before  After 

Z = np.where(Z > 0.5, 0, 1) 

Z = np.maximum(Z, 0.5) 

Z = np.minimum(Z, 0.5) 

Z = np.sum(Z, axis=0) 