Develop Reference

How can I improve performance in my forward substitution method for lower triangle matrices?

I tried implementing the forward substitution method, a solving process to solve the problem Lx = b with L being a lower triangle matrix and x,b as vectors.
This was an easy task:
def tri_solve(L,b):
n = len(b)
x = np.zeros(n)
x[0] = b[0]/L[0,0];
for i in range(1,n):
comp = 0;
for k in range(0,i):
index = L[i,k]
preSolution = x[k]
comp = comp + index * preSolution
x[i] = 1/L[i,i] * (b[i] - comp)
return x;
Now I compared my calculation times for different sized matrices several times with linalg.solve from the scipy module and it turns out that it is much faster. This makes sense in some points, since SciPy is written in C and C++, but I still expected similar or better calculation times for matrices up to 10x10 dimension. Beginning with 6x6 matrices, linalg.solves becomes slightly faster on average.
Is there a way to improve my rather simple solution?

You could try solve_triangular
If you want to accelerate your code, what you could do is to vectorize the inner loop.
def tri_solve(L,b):
n = len(b)
x = np.zeros(n)
x[0] = b[0]/L[0,0];
for i in range(1,n):
comp = np.sum(L[i,:i] * x[:i])
x[i] = 1/L[i,i] * (b[i] - comp)
return x;
Edit: How to use it
You have to pass as first argument a square lower triangular matrix and as second argument you can pass a 1D array
N = 20
A = np.tril(np.random.randn(N, N))
b = np.random.randn(N)
assert np.allclose(np.linalg.solve(A, b), tri_solve(A, b))
Of course this is a naive implementation and is not stable, you can't use it to solve very large or ill conditioned systems.

Finding a way to replace a column in an np.ogrid with a different formula for a specific value of the iterable

a,b=np.ogrid[0:n:1,0:n:1]
A=np.exp(1j*(np.pi/3)*np.abs(a-b))
a,b=np.diag_indices_from(A)
A[a,b]=1-1j/np.sqrt(3)
is my basis. it produces a grid which acts as an n*n matrix.
My issue is I need to replace a column in the grid, say for example where b=17.
I need for this column to be:
A=np.exp(1j*(np.pi/3)*np.abs(a-17+geo_mean(x)))
except for where a=b where it needs to stay as:
A[a,b]=1-1j/np.sqrt(3)
geo_mean(x) is just a geometric average of 50 values determined from a pseudo random number generator, defined in my code as:
x=[random.uniform(0,0.5) for p in range(0,50)]
def geo_mean(iterable):
a = np.array(iterable)
return a.prod()**(1.0/len(a))
So how do i go about replacing a column to include the geo_mean in the exponent formula and do it without changing the diagonal value?

Let's start by saying that diag_indices_from() is kind of useless here since we already know that diagonal elements are those that have equal indices i and j and run up to value n. Therefore, let's simplify the code a little bit at the beginning:
a, b = np.ogrid[0:n:1, 0:n:1]
A = np.exp(1j * (np.pi / 3) * np.abs(a - b))
diag = np.arange(n)
A[diag, diag] = 1 - 1j / np.sqrt(3)
Now, let's say you would like to set the column k values, except for the diagonal element, to
np.exp(1j * (np.pi/3) * np.abs(a - 17 + geo_mean(x)))
(I guess a in the above formula is row index).
This can be done using integer indices, especially that they are almost computed: we already have diag and we just need to remove from it the index of the diagonal element that needs to be kept unchanged:
r = np.delete(diag, k)
Then
x = np.random.uniform(0, 0.5, (r.size, 50))
A[r, k] = np.exp(1j * (np.pi/3) * np.abs(r - k + geo_mean(x)))
However, for the above to work, you need to rewrite your geo_mean() function in a such a way that it will work with 2D input arrays (I will also add some checks and conversions to make it backward compatible):
def geo_mean(x):
x = np.asarray(x)
dim = len(x.shape)
x = np.atleast_2d(x)
v = np.prod(x, axis=1) ** (1.0 / x.shape[1])
return v[0] if dim == 1 else v

python multiprocessing for euclidean distance loop [duplicate]

I have two points in 3D space:
a = (ax, ay, az)
b = (bx, by, bz)
I want to calculate the distance between them:
dist = sqrt((ax-bx)^2 + (ay-by)^2 + (az-bz)^2)
How do I do this with NumPy? I have:
import numpy
a = numpy.array((ax, ay, az))
b = numpy.array((bx, by, bz))

Use numpy.linalg.norm:
dist = numpy.linalg.norm(a-b)
This works because the Euclidean distance is the l2 norm, and the default value of the ord parameter in numpy.linalg.norm is 2.
For more theory, see Introduction to Data Mining:

Use scipy.spatial.distance.euclidean:
from scipy.spatial import distance
a = (1, 2, 3)
b = (4, 5, 6)
dst = distance.euclidean(a, b)

For anyone interested in computing multiple distances at once, I've done a little comparison using perfplot (a small project of mine).
The first advice is to organize your data such that the arrays have dimension (3, n) (and are C-contiguous obviously). If adding happens in the contiguous first dimension, things are faster, and it doesn't matter too much if you use sqrt-sum with axis=0, linalg.norm with axis=0, or
a_min_b = a - b
numpy.sqrt(numpy.einsum('ij,ij->j', a_min_b, a_min_b))
which is, by a slight margin, the fastest variant. (That actually holds true for just one row as well.)
The variants where you sum up over the second axis, axis=1, are all substantially slower.
Code to reproduce the plot:
import numpy
import perfplot
from scipy.spatial import distance
def linalg_norm(data):
a, b = data[0]
return numpy.linalg.norm(a - b, axis=1)
def linalg_norm_T(data):
a, b = data[1]
return numpy.linalg.norm(a - b, axis=0)
def sqrt_sum(data):
a, b = data[0]
return numpy.sqrt(numpy.sum((a - b) ** 2, axis=1))
def sqrt_sum_T(data):
a, b = data[1]
return numpy.sqrt(numpy.sum((a - b) ** 2, axis=0))
def scipy_distance(data):
a, b = data[0]
return list(map(distance.euclidean, a, b))
def sqrt_einsum(data):
a, b = data[0]
a_min_b = a - b
return numpy.sqrt(numpy.einsum("ij,ij->i", a_min_b, a_min_b))
def sqrt_einsum_T(data):
a, b = data[1]
a_min_b = a - b
return numpy.sqrt(numpy.einsum("ij,ij->j", a_min_b, a_min_b))
def setup(n):
a = numpy.random.rand(n, 3)
b = numpy.random.rand(n, 3)
out0 = numpy.array([a, b])
out1 = numpy.array([a.T, b.T])
return out0, out1
b = perfplot.bench(
setup=setup,
n_range=[2 ** k for k in range(22)],
kernels=[
linalg_norm,
linalg_norm_T,
scipy_distance,
sqrt_sum,
sqrt_sum_T,
sqrt_einsum,
sqrt_einsum_T,
],
xlabel="len(x), len(y)",
)
b.save("norm.png")

I want to expound on the simple answer with various performance notes. np.linalg.norm will do perhaps more than you need:
dist = numpy.linalg.norm(a-b)
Firstly - this function is designed to work over a list and return all of the values, e.g. to compare the distance from pA to the set of points sP:
sP = set(points)
pA = point
distances = np.linalg.norm(sP - pA, ord=2, axis=1.) # 'distances' is a list
Remember several things:
Python function calls are expensive.
[Regular] Python doesn't cache name lookups.
So
def distance(pointA, pointB):
dist = np.linalg.norm(pointA - pointB)
return dist
isn't as innocent as it looks.
>>> dis.dis(distance)
2 0 LOAD_GLOBAL 0 (np)
2 LOAD_ATTR 1 (linalg)
4 LOAD_ATTR 2 (norm)
6 LOAD_FAST 0 (pointA)
8 LOAD_FAST 1 (pointB)
10 BINARY_SUBTRACT
12 CALL_FUNCTION 1
14 STORE_FAST 2 (dist)
3 16 LOAD_FAST 2 (dist)
18 RETURN_VALUE
Firstly - every time we call it, we have to do a global lookup for "np", a scoped lookup for "linalg" and a scoped lookup for "norm", and the overhead of merely calling the function can equate to dozens of python instructions.
Lastly, we wasted two operations on to store the result and reload it for return...
First pass at improvement: make the lookup faster, skip the store
def distance(pointA, pointB, _norm=np.linalg.norm):
return _norm(pointA - pointB)
We get the far more streamlined:
>>> dis.dis(distance)
2 0 LOAD_FAST 2 (_norm)
2 LOAD_FAST 0 (pointA)
4 LOAD_FAST 1 (pointB)
6 BINARY_SUBTRACT
8 CALL_FUNCTION 1
10 RETURN_VALUE
The function call overhead still amounts to some work, though. And you'll want to do benchmarks to determine whether you might be better doing the math yourself:
def distance(pointA, pointB):
return (
((pointA.x - pointB.x) ** 2) +
((pointA.y - pointB.y) ** 2) +
((pointA.z - pointB.z) ** 2)
) ** 0.5 # fast sqrt
On some platforms, **0.5 is faster than math.sqrt. Your mileage may vary.
**** Advanced performance notes.
Why are you calculating distance? If the sole purpose is to display it,
print("The target is %.2fm away" % (distance(a, b)))
move along. But if you're comparing distances, doing range checks, etc., I'd like to add some useful performance observations.
Let’s take two cases: sorting by distance or culling a list to items that meet a range constraint.
# Ultra naive implementations. Hold onto your hat.
def sort_things_by_distance(origin, things):
return things.sort(key=lambda thing: distance(origin, thing))
def in_range(origin, range, things):
things_in_range = []
for thing in things:
if distance(origin, thing) <= range:
things_in_range.append(thing)
The first thing we need to remember is that we are using Pythagoras to calculate the distance (dist = sqrt(x^2 + y^2 + z^2)) so we're making a lot of sqrt calls. Math 101:
dist = root ( x^2 + y^2 + z^2 )
:.
dist^2 = x^2 + y^2 + z^2
and
sq(N) < sq(M) iff M > N
and
sq(N) > sq(M) iff N > M
and
sq(N) = sq(M) iff N == M
In short: until we actually require the distance in a unit of X rather than X^2, we can eliminate the hardest part of the calculations.
# Still naive, but much faster.
def distance_sq(left, right):
""" Returns the square of the distance between left and right. """
return (
((left.x - right.x) ** 2) +
((left.y - right.y) ** 2) +
((left.z - right.z) ** 2)
)
def sort_things_by_distance(origin, things):
return things.sort(key=lambda thing: distance_sq(origin, thing))
def in_range(origin, range, things):
things_in_range = []
# Remember that sqrt(N)**2 == N, so if we square
# range, we don't need to root the distances.
range_sq = range**2
for thing in things:
if distance_sq(origin, thing) <= range_sq:
things_in_range.append(thing)
Great, both functions no-longer do any expensive square roots. That'll be much faster, but before you go further, check yourself: why did sort_things_by_distance need a "naive" disclaimer both times above? Answer at the very bottom (*a1).
We can improve in_range by converting it to a generator:
def in_range(origin, range, things):
range_sq = range**2
yield from (thing for thing in things
if distance_sq(origin, thing) <= range_sq)
This especially has benefits if you are doing something like:
if any(in_range(origin, max_dist, things)):
...
But if the very next thing you are going to do requires a distance,
for nearby in in_range(origin, walking_distance, hotdog_stands):
print("%s %.2fm" % (nearby.name, distance(origin, nearby)))
consider yielding tuples:
def in_range_with_dist_sq(origin, range, things):
range_sq = range**2
for thing in things:
dist_sq = distance_sq(origin, thing)
if dist_sq <= range_sq: yield (thing, dist_sq)
This can be especially useful if you might chain range checks ('find things that are near X and within Nm of Y', since you don't have to calculate the distance again).
But what about if we're searching a really large list of things and we anticipate a lot of them not being worth consideration?
There is actually a very simple optimization:
def in_range_all_the_things(origin, range, things):
range_sq = range**2
for thing in things:
dist_sq = (origin.x - thing.x) ** 2
if dist_sq <= range_sq:
dist_sq += (origin.y - thing.y) ** 2
if dist_sq <= range_sq:
dist_sq += (origin.z - thing.z) ** 2
if dist_sq <= range_sq:
yield thing
Whether this is useful will depend on the size of 'things'.
def in_range_all_the_things(origin, range, things):
range_sq = range**2
if len(things) >= 4096:
for thing in things:
dist_sq = (origin.x - thing.x) ** 2
if dist_sq <= range_sq:
dist_sq += (origin.y - thing.y) ** 2
if dist_sq <= range_sq:
dist_sq += (origin.z - thing.z) ** 2
if dist_sq <= range_sq:
yield thing
elif len(things) > 32:
for things in things:
dist_sq = (origin.x - thing.x) ** 2
if dist_sq <= range_sq:
dist_sq += (origin.y - thing.y) ** 2 + (origin.z - thing.z) ** 2
if dist_sq <= range_sq:
yield thing
else:
... just calculate distance and range-check it ...
And again, consider yielding the dist_sq. Our hotdog example then becomes:
# Chaining generators
info = in_range_with_dist_sq(origin, walking_distance, hotdog_stands)
info = (stand, dist_sq**0.5 for stand, dist_sq in info)
for stand, dist in info:
print("%s %.2fm" % (stand, dist))
(*a1: sort_things_by_distance's sort key calls distance_sq for every single item, and that innocent looking key is a lambda, which is a second function that has to be invoked...)

Another instance of this problem solving method:
def dist(x,y):
return numpy.sqrt(numpy.sum((x-y)**2))
a = numpy.array((xa,ya,za))
b = numpy.array((xb,yb,zb))
dist_a_b = dist(a,b)

Starting Python 3.8, the math module directly provides the dist function, which returns the euclidean distance between two points (given as tuples or lists of coordinates):
from math import dist
dist((1, 2, 6), (-2, 3, 2)) # 5.0990195135927845
And if you're working with lists:
dist([1, 2, 6], [-2, 3, 2]) # 5.0990195135927845

It can be done like the following. I don't know how fast it is, but it's not using NumPy.
from math import sqrt
a = (1, 2, 3) # Data point 1
b = (4, 5, 6) # Data point 2
print sqrt(sum( (a - b)**2 for a, b in zip(a, b)))

A nice one-liner:
dist = numpy.linalg.norm(a-b)
However, if speed is a concern I would recommend experimenting on your machine. I've found that using math library's sqrt with the ** operator for the square is much faster on my machine than the one-liner NumPy solution.
I ran my tests using this simple program:
#!/usr/bin/python
import math
import numpy
from random import uniform
def fastest_calc_dist(p1,p2):
return math.sqrt((p2[0] - p1[0]) ** 2 +
(p2[1] - p1[1]) ** 2 +
(p2[2] - p1[2]) ** 2)
def math_calc_dist(p1,p2):
return math.sqrt(math.pow((p2[0] - p1[0]), 2) +
math.pow((p2[1] - p1[1]), 2) +
math.pow((p2[2] - p1[2]), 2))
def numpy_calc_dist(p1,p2):
return numpy.linalg.norm(numpy.array(p1)-numpy.array(p2))
TOTAL_LOCATIONS = 1000
p1 = dict()
p2 = dict()
for i in range(0, TOTAL_LOCATIONS):
p1[i] = (uniform(0,1000),uniform(0,1000),uniform(0,1000))
p2[i] = (uniform(0,1000),uniform(0,1000),uniform(0,1000))
total_dist = 0
for i in range(0, TOTAL_LOCATIONS):
for j in range(0, TOTAL_LOCATIONS):
dist = fastest_calc_dist(p1[i], p2[j]) #change this line for testing
total_dist += dist
print total_dist
On my machine, math_calc_dist runs much faster than numpy_calc_dist: 1.5 seconds versus 23.5 seconds.
To get a measurable difference between fastest_calc_dist and math_calc_dist I had to up TOTAL_LOCATIONS to 6000. Then fastest_calc_dist takes ~50 seconds while math_calc_dist takes ~60 seconds.
You can also experiment with numpy.sqrt and numpy.square though both were slower than the math alternatives on my machine.
My tests were run with Python 2.6.6.

I find a 'dist' function in matplotlib.mlab, but I don't think it's handy enough.
I'm posting it here just for reference.
import numpy as np
import matplotlib as plt
a = np.array([1, 2, 3])
b = np.array([2, 3, 4])
# Distance between a and b
dis = plt.mlab.dist(a, b)

You can just subtract the vectors and then innerproduct.
Following your example,
a = numpy.array((xa, ya, za))
b = numpy.array((xb, yb, zb))
tmp = a - b
sum_squared = numpy.dot(tmp.T, tmp)
result = numpy.sqrt(sum_squared)

I like np.dot (dot product):
a = numpy.array((xa,ya,za))
b = numpy.array((xb,yb,zb))
distance = (np.dot(a-b,a-b))**.5

With Python 3.8, it's very easy.
https://docs.python.org/3/library/math.html#math.dist
math.dist(p, q)
Return the Euclidean distance between two points p and q, each given
as a sequence (or iterable) of coordinates. The two points must have
the same dimension.
Roughly equivalent to:
sqrt(sum((px - qx) ** 2.0 for px, qx in zip(p, q)))

Having a and b as you defined them, you can use also:
distance = np.sqrt(np.sum((a-b)**2))

Since Python 3.8
Since Python 3.8 the math module includes the function math.dist().
See here https://docs.python.org/3.8/library/math.html#math.dist.
math.dist(p1, p2)
Return the Euclidean distance between two points p1 and p2,
each given as a sequence (or iterable) of coordinates.
import math
print( math.dist( (0,0), (1,1) )) # sqrt(2) -> 1.4142
print( math.dist( (0,0,0), (1,1,1) )) # sqrt(3) -> 1.7321

Here's some concise code for Euclidean distance in Python given two points represented as lists in Python.
def distance(v1,v2):
return sum([(x-y)**2 for (x,y) in zip(v1,v2)])**(0.5)

import math
dist = math.hypot(math.hypot(xa-xb, ya-yb), za-zb)

Calculate the Euclidean distance for multidimensional space:
import math
x = [1, 2, 6]
y = [-2, 3, 2]
dist = math.sqrt(sum([(xi-yi)**2 for xi,yi in zip(x, y)]))
5.0990195135927845

import numpy as np
from scipy.spatial import distance
input_arr = np.array([[0,3,0],[2,0,0],[0,1,3],[0,1,2],[-1,0,1],[1,1,1]])
test_case = np.array([0,0,0])
dst=[]
for i in range(0,6):
temp = distance.euclidean(test_case,input_arr[i])
dst.append(temp)
print(dst)

You can easily use the formula
distance = np.sqrt(np.sum(np.square(a-b)))
which does actually nothing more than using Pythagoras' theorem to calculate the distance, by adding the squares of Δx, Δy and Δz and rooting the result.

import numpy as np
# any two python array as two points
a = [0, 0]
b = [3, 4]
You first change list to numpy array and do like this: print(np.linalg.norm(np.array(a) - np.array(b))). Second method directly from python list as: print(np.linalg.norm(np.subtract(a,b)))

The other answers work for floating point numbers, but do not correctly compute the distance for integer dtypes which are subject to overflow and underflow. Note that even scipy.distance.euclidean has this issue:
>>> a1 = np.array([1], dtype='uint8')
>>> a2 = np.array([2], dtype='uint8')
>>> a1 - a2
array([255], dtype=uint8)
>>> np.linalg.norm(a1 - a2)
255.0
>>> from scipy.spatial import distance
>>> distance.euclidean(a1, a2)
255.0
This is common, since many image libraries represent an image as an ndarray with dtype="uint8". This means that if you have a greyscale image which consists of very dark grey pixels (say all the pixels have color #000001) and you're diffing it against black image (#000000), you can end up with x-y consisting of 255 in all cells, which registers as the two images being very far apart from each other. For unsigned integer types (e.g. uint8), you can safely compute the distance in numpy as:
np.linalg.norm(np.maximum(x, y) - np.minimum(x, y))
For signed integer types, you can cast to a float first:
np.linalg.norm(x.astype("float") - y.astype("float"))
For image data specifically, you can use opencv's norm method:
import cv2
cv2.norm(x, y, cv2.NORM_L2)

Find difference of two matrices first. Then, apply element wise multiplication with numpy's multiply command. After then, find summation of the element wise multiplied new matrix. Finally, find square root of the summation.
def findEuclideanDistance(a, b):
euclidean_distance = a - b
euclidean_distance = np.sum(np.multiply(euclidean_distance, euclidean_distance))
euclidean_distance = np.sqrt(euclidean_distance)
return euclidean_distance

What's the best way to do this with NumPy, or with Python in general? I have:
Well best way would be safest and also the fastest
I would suggest hypot usage for reliable results for chances of underflow and overflow are very little compared to writing own sqroot calculator
Lets see math.hypot, np.hypot vs vanilla np.sqrt(np.sum((np.array([i, j, k])) ** 2, axis=1))
i, j, k = 1e+200, 1e+200, 1e+200
math.hypot(i, j, k)
# 1.7320508075688773e+200
np.sqrt(np.sum((np.array([i, j, k])) ** 2))
# RuntimeWarning: overflow encountered in square
Speed wise math.hypot look better
%%timeit
math.hypot(i, j, k)
# 100 ns ± 1.05 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)
%%timeit
np.sqrt(np.sum((np.array([i, j, k])) ** 2))
# 6.41 µs ± 33.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
Underflow
i, j = 1e-200, 1e-200
np.sqrt(i**2+j**2)
# 0.0
Overflow
i, j = 1e+200, 1e+200
np.sqrt(i**2+j**2)
# inf
No Underflow
i, j = 1e-200, 1e-200
np.hypot(i, j)
# 1.414213562373095e-200
No Overflow
i, j = 1e+200, 1e+200
np.hypot(i, j)
# 1.414213562373095e+200
Refer

The fastest solution I could come up with for large number of distances is using numexpr. On my machine it is faster than using numpy einsum:
import numexpr as ne
import numpy as np
np.sqrt(ne.evaluate("sum((a_min_b)**2,axis=1)"))

If you want something more explicit you can easily write the formula like this:
np.sqrt(np.sum((a-b)**2))
Even with arrays of 10_000_000 elements this still runs at 0.1s on my machine.

Tips on how to optimize the inner function with a large for loop

def testing(min_quadReq, stepsize, max_quadReq, S):
y = np.arange(min_quadReq, max_quadReq, stepsize)
print("Y", y)
I_avg = np.zeros(len(y))
Q_avg = np.zeros(len(y))
x = np.arange(0, (len(S)))
debugger = 0
for i in range(0, len(y)):
I = np.array(S * (np.cos(2 * np.pi * y[i] * x)))
Q = np.array(S * (np.sin(2 * np.pi * y[i] * x)))
I_avg[i] = np.sum(I, 0)
Q_avg[i] = np.sum(Q, 0)
debugger += 1
D = [I_avg**2 + Q_avg**2]
maxIndex = np.argmax(D)
#maxValue = D.max()
# in python is arctan2(b,a) compared to matlab's atan2(a,b)
phaseOut = np.arctan2(Q_avg[maxIndex], I_avg[maxIndex])
# returns the out value and the phase
out = min_quadReq + ((maxIndex + 1) - 1) * stepsize
return out, phaseOut
I'm working on a project where uses DSP to process a signal at get out the relevant data. The code above is from the inner function of a quadrature modulation. From what I have seen this is the part of the code that have the biggest potential to be optimized. For example the two sum function is called about 92k times each and the quadrature function itself 2696 times. I'm not that familiar with python so if any have any suggestion to how a more efficient way of writing it or some good documentation it would be lovely.
The signal S is the input source and it's a array of [481][251]. The outer shell of the quadrature is called on by quadReq(cavSig[j, :]) just some extra information to show how it's called and how many times.
def randomnumber():
s = np.random.random_sample((1, 251))
print(s)
return s
randomnumber()
Edit: Added some more information

Your loop produces one I_avg value for each element of y. For compactness I could write it as a list comprehension.
In [61]: x=np.arange(4)
In [62]: y=np.arange(0,1,.2)
In [63]: [np.cos(2*np.pi*y[i]*x).sum() for i in range(y.shape[0])]
Out[63]:
[4.0,
-0.30901699437494745,
0.80901699437494767,
0.80901699437494834,
-0.30901699437494756]
But that y[i]*x part is just an outer product of y and x, which can be written with np.outter, or just as easily with broadcasting, y[:,None]*x.
In [64]: np.cos(2*np.pi*y[:,None]*x).sum(axis=1)
Out[64]: array([ 4. , -0.30901699, 0.80901699, 0.80901699, -0.30901699])
Get on to an interactive Python session, and play around with expressions like this. The best way to learn is by doing and seeing immediate results. Keep the test expressions and arrays small so you can see immediately what is happening.

What is the most pythonic way to conditionally compute?

I'm implementing Bayesian Changepoint Detection in Python/NumPy (if you are interested have a look at the paper). I need to compute likelihoods for data in ranges [a, b], where a and b can have all values from 1 to n. However I can prune the computation at some points, so that I don't have to compute every likelihood. On the other hand some likelihoods are used more than once, so that I can save time by saving the values in a matrix P[a, b]. Right now I check whether the value is already computed, whenever I use it, but I find that a bit of a hassle. It looks like this:
# ...
P = np.ones((n, n)) * np.inf # a likelihood can't get inf, so I use it
# as pseudo value
for a in range(n):
for b in range(a, n):
# The following two lines get annoying and error prone if you
# use P more than once
if P[a, b] == np.inf:
P[a, b] = likelihood(data, a, b)
Q[a] += P[a, b] * g[a] * Q[a - 1] # some computation using P[a, b]
# ...
I wonder, whether there is a more intuitive and pythonic way to achieve this, without having the if ... statement before every use of a P[a, b]. Something like an automagical function call if some condition is not met. I could of course make the likelihood function aware of the fact that it could save values, but then it needs some kind of state (e.g. becomes an object). I want to avoid that.
The likelihood function
Since it was asked for in a comment, I add the likelihood function. It actually computes the conjugate prior and then the likelihood. And all in log representation... So it is quite complicated.
from scipy.special import gammaln
def gaussian_obs_log_likelihood(data, t, s):
n = s - t
mean = data[t:s].sum() / n
muT = (n * mean) / (1 + n)
nuT = 1 + n
alphaT = 1 + n / 2
betaT = 1 + 0.5 * ((data[t:s] - mean) ** 2).sum() + ((n)/(1 + n)) * (mean**2 / 2)
scale = (betaT*(nuT + 1))/(alphaT * nuT)
# splitting the PDF of the student distribution up is /much/ faster. (~ factor 20)
prob = 1
for yi in data[t:s]:
prob += np.log(1 + (yi - muT)**2/(nuT * scale))
lgA = gammaln((nuT + 1) / 2) - np.log(np.sqrt(np.pi * nuT * scale)) - gammaln(nuT/2)
return n * lgA - (nuT + 1)/2 * prob
Although I work with Python 2.7, both answers for 2.7 and 3.x are appreciated.

I would use a sibling of defaultdict for this (you can't use defaultdict directly since it won't tell you the key that is missing):
class Cache(object):
def __init__(self):
self.cache = {}
def get(self, a, b):
key = (a,b)
result = self.cache.get(key, None)
if result is None:
result = likelihood(data, a, b)
self.cache[key] = result
return result
Another approach would be using a cache decorator on likelihood as described here.