• NumericAndScientificRecipes

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""" sinc(x) = sin(\pi x) / (\pi x), if x != 0

• = 1, if x = 0

""" def sinc(x):

• from math import pi, sin try:
  • x = pi * x return sin(x) / x

  except ZeroDivisionError: # sinc(0) = 1

  • return 1.0

""" cbrt(x) = x^{1/3}, if x >= 0

• = -|x|^{1/3}, if x < 0

""" def cbrt(x):

• from math import pow

  if x >= 0:

  • return pow(x, 1.0/3.0)
  else:
  • return -pow(abs(x), 1.0/3.0)

""" Convert from polar (r,w) to rectangular (x,y)

• x = r cos(w) y = r sin(w)

""" def rect(r, w, deg=0): # radian if deg=0; degree if deg=1

• from math import cos, sin, pi if deg:
  • w = pi * w / 180.0
  return r * cos(w), r * sin(w)

""" Convert from rectangular (x,y) to polar (r,w)

• r = sqrt(x^{2 + y}2) w = arctan(y/x) = [-\pi,\pi] = [-180,180]

""" def polar(x, y, deg=0): # radian if deg=0; degree if deg=1

• from math import hypot, atan2, pi if deg:
  • return hypot(x, y), 180.0 * atan2(y, x) / pi
  else:
  • return hypot(x, y), atan2(y, x)

• Normally, rect() and polar() uses radian for angle; but, if deg=1 specified (any non-zero will do), degree is used instead.

""" p(x) = polyeval(a, x)

• = a[0] + a[1]x + a[2]x^{2 +...+ a[n-1]x}{n-1} + a[n]x^n = a[0] + x(a[1] + x(a[2] +...+ x(a[n-1] + a[n]x)...)

""" def polyeval(a, x):

• p = 0 a.reverse() for coef in a:
  • p = p * x + coef
  a.reverse() return p

""" p'(x) = polyderiv(a)

• = b[0] + b[1]x + b[2]x^{2 +...+ b[n-2]x}{n-2} + b[n-1]x^{n-1}

where

• b[i] = (i+1)a[i+1]

""" def polyderiv(a):

• b = [] for i in range(1, len(a)):
  • b.append(i * a[i])
  return b

""" Given x = r is a root of n'th degree polynomial p(x) = (x-r)q(x), divide p(x) by linear factor (x-r) using the same algorithm as polynomial evaluation. Then, return the (n-1)'th degree quotient q(x) = polyreduce(a, r)

• = c[0] + c[1]x + c[2]x^{2 +...+ c[n-2]x}{n-2} + c[n-1]x^{n-1}

""" def polyreduce(a, root):

• c, p = [], 0 a.reverse() for coef in a:
  • p = p * root + coef c.append(p)
  a.reverse() c.reverse() return c[1:]

""" x^{2 + ax + b = 0 (or ax}2 + bx + c = 0) By substituting x = y-t and t = a/2, the equation reduces to

• y^{2 + (b-t}2) = 0

which has easy solution

• y = +/- sqrt(t^2-b)

""" def quadratic(a, b, c=None):

• import math, cmath if c: # (ax^2 + bx + c = 0)
  • a, b = b / float(a), c / float(a)
  t = a / 2.0 r = t**2 - b

  if r >= 0: # real roots

  • y1 = math.sqrt(r)
  else: # complex roots
  • y1 = cmath.sqrt(r)
  y2 = -y1 return y1 - t, y2 - t

""" x^{3 + ax}2 + bx + c = 0 (or ax^{3 + bx}2 + cx + d = 0) With substitution x = y-t and t = a/3, the cubic equation reduces to

• y^3 + py + q = 0,

where p = b-3t^{2 and q = c-bt+2t}3. Then, one real root y1 = u+v can be determined by solving

• w^{2 + qw - (p/3)}3 = 0

where w = u^{3, v}3. From Vieta's theorem,

• y1 + y2 + y3 = 0 y1 y2 + y1 y3 + y2 y3 = p y1 y2 y3 = -q,

the other two (real or complex) roots can be obtained by solving

• y^{2 + (y1)y + (p+y1}2) = 0

""" def cubic(a, b, c, d=None):

• from math import cos

  if d: # (ax^{3 + bx}2 + cx + d = 0)

  • a, b, c = b / float(a), c / float(a), d / float(a)
  t = a / 3.0 p, q = b - 3 * t**2, c - b * t + 2 * t**3 u, v = quadratic(q, -(p/3.0)**3) if type(u) == type(0j): # complex cubic root
  • r, w = polar(u.real, u.imag) y1 = 2 * cbrt(r) * cos(w / 3.0)
  else: # real root
  • y1 = cbrt(u) + cbrt(v)
  y2, y3 = quadratic(y1, p + y1**2) return y1 - t, y2 - t, y3 - t

""" Ubiquitous Newton-Raphson algorithm for solving

• f(x) = 0

where a root is repeatedly estimated by

• x = x - f(x)/f'(x)

until |dx|/(1+|x|) < TOL is achieved. This termination condition is a compromise between

• |dx| < TOL, if x is small |dx|/|x| < TOL, if x is large

""" def newton(func, funcd, x, TOL=1e-6): # f(x)=func(x), f'(x)=funcd(x)

• f, fd = func(x), funcd(x) count = 0 while 1:
  • dx = f / float(fd)

    if abs(dx) < TOL * (1 + abs(x)): return x - dx x = x - dx f, fd = func(x), funcd(x) count = count + 1 print "newton(%d): x=%s, f(x)=%s" % (count, x, f)

""" Similar to Newton's method, but the derivative is estimated by divided difference using only function calls. A root is estimated by

• x = x - f(x) (x - oldx)/(f(x) - f(oldx))

where oldx = x[i-1] and x = x[i]. """ def secant(func, oldx, x, TOL=1e-6): # f(x)=func(x)

• oldf, f = func(oldx), func(x)

  if (abs(f) > abs(oldf)): # swap so that f(x) is closer to 0

  • oldx, x = x, oldx oldf, f = f, oldf
  count = 0 while 1:
  • dx = f * (x - oldx) / float(f - oldf)

    if abs(dx) < TOL * (1 + abs(x)): return x - dx oldx, x = x, x - dx oldf, f = f, func(x) count = count + 1 print "secant(%d): x=%s, f(x)=%s" % (count, x, f)

""" Closed Simpson's rule for

• \int_a^b f(x) dx

Divide [a,b] iteratively into h, h/2, h/4, h/8, ... step sizes; and, for each step size, evaluate f(x) at a+h, a+3h, a+5h, a+7h, ..., b-3h, b-h, noting that other points have already been sampled.

At each iteration step, data are sampled only where necessary so that the total data is represented by adding sampled points from all previous steps:

• step 1: h a

b

• step 2: h/2 a

^

b

• step 3: h/4 a---^----------b step 4: h/8 a-^------^----b total: a-^--^--^--^-b

So, for step size of h/n, there are n intervals, and the data are sampled at the boundaries including the 2 end points.

If old = Trapezoid formula for an old step size 2h, then Trapezoid formula for the new step size h is obtained by

• new = old/2 + h{f(a+h) + f(a+3h) + f(a+5h) + f(a+7h) +...+ f(b-3h)
  • + f(b-h)}

Also, Simpson formula for the new step size h is given by

• simpson = (4 new - old)/3

""" def closedpoints(func, a, b, TOL=1e-6): # f(x)=func(x)

• h = b - a old2 = old = h * (func(a) + func(b)) / 2.0 count = 0 while 1:
  • h = h / 2.0 x, sum = a + h, 0

    while x < b:

    • sum = sum + func(x) x = x + 2 * h
    new = old / 2.0 + h * sum new2 = (4 * new - old) / 3.0

    if abs(new2 - old2) < TOL * (1 + abs(old2)): return new2 old = new # Trapezoid old2 = new2 # Simpson count = count + 1 print 'closedpoints(%d): Trapezoid=%s, Simpson=%s' % (count, new, new2)

""" Open Simpson's rule (excluding end points) for

• \int_a^b f(x) dx

Divide [a,b] iteratively into h, h/3, h/9, h/27, ... step sizes; and, for each step size, evaluate f(x) at a+h/2, a+2h+h/2, a+3h+h/2, a+5h+h/2, a+6h+h/2, ... , b-3h-h/2, b-2h-h/2, b-h/2, noting that other points have already been sampled.

At each iteration step, data are sampled only where necessary so that the total data is represented by adding sampled points from all previous steps:

• step 1: h a

^

b

• step 2: h/3 a

^{-----------------------}

b

• step 3: h/9 a-^----------^----------^--------b total: a-^------^------^------^------^-b

So, for step size of h/n, there are n intervals, and the data are sampled at the midpoints.

If old = Trapezoid formula for an old step size 3h, then Trapezoid formula for the new step size h is obtained by

• new = old/3 + h{f(a+h/2) + f(a+2h+h/2) + f(a+3h+h/2) + f(a+5h+h/2)
  • + f(a+6h+h/2) +...+ f(b-3h-h/2) + f(b-2h-h/2) + f(b-h/2)}

Also, Simpson formula for the new step size h is given by

• simpson = (9 new - old)/8

""" def openpoints(func, a, b, TOL=1e-6): # f(x)=func(x)

• h = b - a old2 = old = h * func((a + b) / 2.0) count = 0 while 1:
  • h = h / 3.0 x, sum = a + 0.5 * h, 0

    while x < b:

    • sum = sum + func(x) + func(x + 2 * h) x = x + 3 * h
    new = old / 3.0 + h * sum new2 = (9 * new - old) / 8.0

    if abs(new2 - old2) < TOL * (1 + abs(old2)): return new2 old = new # Trapezoid old2 = new2 # Simpson count = count + 1 print 'openpoints(%d): Trapezoid=%s, Simpson=%s' % (count, new, new2)

""" Find 2^n that is equal to or greater than. """ def nextpow2(i):

• n = 2

  while n < i: n = n * 2 return n

• This is internal function used by fft(), because the FFT routine requires that the data size be a power of 2.

""" Return bit-reversed list, whose length is assumed to be 2^n: eg. 0111 <--> 1110 for N=2^4. """ def bitrev(x):

• N, x = len(x), x[:]

  if N != nextpow2(N): raise ValueError, 'N is not power of 2' for i in range(N):

  • k, b, a = 0, N>>1, 1 while b >= a:

    • if b & i: k = k | a if a & i: k = k | b b, a = b>>1, a<<1

    if i < k: # important not to swap back

    • x[i], x[k] = x[k], x[i]
  return x

""" FFT using Cooley-Tukey algorithm where N = 2^n. The choice of e^{{-j2\pi/N} or e}{j2\pi/N} is made by 'sign=-1' or 'sign=1' respectively. Since I prefer Engineering convention, I chose 'sign=-1' as the default.

FFT is performed as follows: 1. bit-reverse the array. 2. partition the data into group of m = 2, 4, 8, ..., N data points. 3. for each group with m data points,

1. divide into upper half (section A) and lower half (section B),
   • each containing m/2 data points.
2. divide unit circle by m.
3. apply "butterfly" operation

   • |a| = |1 w||a| or a, b = a+w*b, a-w*b |b| |1 -w||b|

   • where a and b are data points of section A and B starting from the top of each section, and w is data points along the unit circle starting from z = 1+0j.

FFT ends after applying "butterfly" operation on the entire data array as whole, when m = N. """ def fft(x, sign=-1):

• from cmath import pi, exp N, W = len(x), [] for i in range(N): # exp(-j...) is default
  • W.append(exp(sign * 2j * pi * i / N))
  x = bitrev(x) m = 2

  while m <= N:

  • for s in range(0, N, m):
    • for i in range(m/2):
      • n = i * N / m a, b = s + i, s + i + m/2 x[a], x[b] = x[a] + W[n % N] * x[b], x[a] - W[n % N] * x[b]
    m = m * 2
  return x

""" Inverse FFT with normalization by N, so that x == ifft(fft(x)) within round-off errors. """ def ifft(X):

• N, x = len(X), fft(X, sign=1) # e^{j2\pi/N} for i in range(N):
  • x[i] = x[i] / float(N)
  return x

""" DFT using discrete summation

• X(n) = \sum_k W^{{nk} x(k), W = e}{-j2\pi/N}

where N need not be power of 2. The choice of e^{-j2\pi/N} or e^{j2\pi/N} is made by "sign=-1" or "sign=1" respectively. """ def dft(x, sign=-1):

• from cmath import pi, exp N, W = len(x), [] for i in range(N): # exp(-j...) is default
  • W.append(exp(sign * 2j * pi * i / N))
  X = [] for n in range(N):
  • sum = 0 for k in range(N):
    • sum = sum + W[n * k % N] * x[k]
    X.append(sum)
  return X

""" Inverse DFT with normalization by N, so that x == idft(dft(x)) within round-off errors. """ def idft(X):

• N, x = len(X), dft(X, sign=1) # e^{j2\pi/N} for i in range(N):
  • x[i] = x[i] / float(N)
  return x

""" Convolution of 2 casual signals, x(t<0) = y(t<0) = 0, using discrete summation.

• x*y(t) = \int_{u=0}^t x(u) y(t-u) du = y*x(t)

where the size of x[], y[], x*y[] are P, Q, N=P+Q-1 respectively. """ def conv(x, y):

• P, Q, N = len(x), len(y), len(x)+len(y)-1 z = [] for k in range(N):
  • t, lower, upper = 0, max(0, k-(Q-1)), min(P-1, k) for i in range(lower, upper+1):
    • t = t + x[i] * y[k-i]
    z.append(t)
  return z

""" Correlation of 2 casual signals, x(t<0) = y(t<0) = 0, using discrete summation.

• Rxy(t) = \int_{u=0}^{\infty} x(u) y(t+u) du = Ryx(-t)

where the size of x[], y[], Rxy[] are P, Q, N=P+Q-1 respectively.

The Rxy[i] data is not shifted, so relationship with the continuous Rxy(t) is preserved. For example, Rxy(0) = Rxy[0], Rxy(t) = Rxy[i], and Rxy(-t) = Rxy[-i]. The data are ordered as follows:

• t: -(P-1), -(P-2), ..., -3, -2, -1, 0, 1, 2, 3, ..., Q-2, Q-1 i: N-(P-1), N-(P-2), ..., N-3, N-2, N-1, 0, 1, 2, 3, ..., Q-2, Q-1

""" def corr(x, y):

• P, Q, N = len(x), len(y), len(x)+len(y)-1 z1=[] for k in range(Q):
  • t, lower, upper = 0, 0, min(P-1, Q-1-k) for i in range(lower, upper+1):
    • t = t + x[i] * y[i+k]
    z1.append(t) # 0, 1, 2, ..., Q-1
  z2=[] for k in range(1,P):
  • t, lower, upper = 0, k, min(P-1, Q-1+k) for i in range(lower, upper+1):
    • t = t + x[i] * y[i-k]
    z2.append(t) # N-1, N-2, ..., N-(P-2), N-(P-1)
  z2.reverse() return z1 + z2

""" FFT convolution using relation

• x*y <==> XY

where x[] and y[] have been zero-padded to length N, such that N >= P+Q-1 and N = 2^n. """ def fftconv(x, y):

• N, X, Y, x_y = len(x), fft(x), fft(y), [] for i in range(N):
  • x_y.append(X[i] * Y[i])
  return ifft(x_y)

""" FFT correlation using relation

• Rxy <==> X'Y

where x[] and y[] have been zero-padded to length N, such that N >= P+Q-1 and N = 2^n. """ def fftcorr(x, y):

• N, X, Y, Rxy = len(x), fft(x), fft(y), [] for i in range(N):
  • Rxy.append(X[i].conjugate() * Y[i])
  return ifft(Rxy)

""" vdot(x, y) = <x|y> = \sum_i x_i y_i """ def vdot(x, y):

• if len(x) != len(y): raise ValueError, 'unequal length' sum = 0 for a, b in map(None, x, y):

  • sum = sum + a * b
  return sum

""" Various vector norms of x[]:

• ||

  x||1 = \sum_i |x_i| ||x||2 = sqrt(\sum_i x_i^2) = sqrt(<x|x>) ||x||oo = \max |x_i|

""" def vnorm(x, normtype=2):

• from math import sqrt

  if normtype == 1: # ||x||1

  • sum = 0 for a in x:
    • sum = sum + abs(a)
    return sum

  elif normtype == 2: # ||x||2 is default

  • return sqrt(vdot(x, x))

  elif normtype == 'oo': # ||x||oo

  • return max( abs(min(x)), abs(max(x)) )
  else:

  • raise ValueError, 'unknown option'

""" Calculate mean and standard deviation of data x[]:

• mean = {\sum_i x_i \over n} std = sqrt(\sum_i (x_i - mean)^2 \over n-1)

""" def meanstdv(x):

• from math import sqrt n, mean, std = len(x), 0, 0 for a in x:
  • mean = mean + a
  mean = mean / float(n) for a in x:
  • std = std + (a - mean)**2
  std = sqrt(std / float(n-1)) return mean, std

""" Read 1 number/line using eval() from <stdin> or 'file' if specified. If the first non-whitespace is not valid number characters '(+-.0123456789', then the line will be skipped. """ def getv(s=):

• import sys, string

  if s == :

  • f = sys.stdin
  else:
  • f = open(s, 'r')
  x = [] for a in f.readlines():
  • a = string.strip(a)

    if a != and a[0] in '(+-.0123456789':

    • x.append(eval(a))
  return x

""" Read 2 numbers/line using eval() from <stdin> or 'file' if specified. If the first non-whitespace is not valid number characters '(+-.0123456789', then the line will be skipped. """ def getv2(s=):

• import sys, string

  if s == :

  • f = sys.stdin
  else:
  • f = open(s, 'r')
  x, y = [], [] for a in f.readlines():
  • a = string.strip(a)

    if a != and a[0] in '(+-.0123456789':

    • b = string.split(a) x.append(eval(b[0])) y.append(eval(b[1]))
  return x, y

""" Write 1 number/line to <stdout> or 'file' if specified. """ def printv(x, s=):

• import sys

  out = for a in x:

  • out = out + a + '\n'

  if s == :

  • sys.stdout.write(out)
  else:
  • open(s, 'w').write(out)

""" Returns coefficients to the regression line "y=ax+b" from x[] and y[]. Basically, it solves

• Sxx a + Sx b = Sxy
  • Sx a + N b = Sy

where Sxy = \sum_i x_i y_i, Sx = \sum_i x_i, and Sy = \sum_i y_i. The solution is

• a = (Sxy N - Sy Sx)/det b = (Sxx Sy - Sx Sxy)/det

where det = Sxx N - Sx^2. In addition,

• Var|a| = s^{2 |Sxx Sx|}-1 = s^2 | N -Sx| / det

  • |b| |Sx N | |-Sx Sxx|

  s^{2 = {\sum_i (y_i - \hat{y_i})}2 \over N-2}

  • = {\sum_i (y_i - ax_i - b)^2 \over N-2} = residual / (N-2)

  R^{2 = 1 - {\sum_i (y_i - \hat{y_i})}2 \over \sum_i (y_i - \mean{y})^2}

  • = 1 - residual/meanerror

It also prints to <stdout> few other data,

• N, a, b, R^{2, s}2,

which are useful in assessing the confidence of estimation. """ def linreg(X, Y):

• from math import sqrt

  if len(X) != len(Y): raise ValueError, 'unequal length' N = len(X) Sx = Sy = Sxx = Syy = Sxy = 0.0 for x, y in map(None, X, Y):

  • Sx = Sx + x Sy = Sy + y Sxx = Sxx + x*x Syy = Syy + y*y Sxy = Sxy + x*y
  det = Sxx * N - Sx * Sx a, b = (Sxy * N - Sy * Sx)/det, (Sxx * Sy - Sx * Sxy)/det meanerror = residual = 0.0 for x, y in map(None, X, Y):
  • meanerror = meanerror + (y - Sy/N)**2 residual = residual + (y - a * x - b)**2
  RR = 1 - residual/meanerror ss = residual / (N-2) Var_a, Var_b = ss * N / det, ss * Sxx / det print "y=ax+b" print "N= %d" % N print "a= %g \\pm t_{%d;\\alpha/2} %g" % (a, N-2, sqrt(Var_a)) print "b= %g \\pm t_{%d;\\alpha/2} %g" % (b, N-2, sqrt(Var_b)) print "R^2= %g" % RR print "s^2= %g" % ss return a, b

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