# answers 08: the cycle of 12¶

The observed data are downloadable here, and my notebook page assumes this file is in the current directory.

In [1]:
import numpy as np
import math
import re
import sys
import scipy.misc        as misc
import scipy.stats       as stats
import scipy.optimize    as optimize
import matplotlib.pyplot as plt
%matplotlib notebook

datafile = 'w08-data.tbl'


This parsing code is taking from moriarty.py, and slightly modified, adding a reliability[] flag that I'll use to color points in my plots at the end.

In [2]:
# Parse the data file.
# This leaves us with
#    N              : number of experiments (columns in the table)
#    G              : number of genes (rows in the table)
#    genenames[i]   : names of the genes
#    X[i]           : array of time points, in hrs, for the N experiments
#    S_true[i]      : array of sigmas for the experiments
#    Y[i][t]        : GxN: observed tpm for gene i, time point t
#    reliability[i] : 'H', 'M', 'L' for sigma = 2, 5, 20
#
with open(datafile) as f:
# First header line gives us the time points
X = []
for s in fields:
match = re.search(r'^(\d+)hr', s)
X.append(int(match.group(1)))
X = np.array(X)
N = len(X)

# Second header line gives us "gene" followed by +=SD's
S_true = np.zeros(N)
reliability = []
for i,s in enumerate(fields[1:]):
match = re.search(r'^\+-(\d+)', s)
if   match.group(1) == '2': reliability.append('H')
elif match.group(1) == '5': reliability.append('M')
else:                       reliability.append('L')
S_true[i] = float(match.group(1))

# Third header line is just ------ stuff

# Remaining lines are data
genenames = []
Y = []
fields = line.split()
genenames.append(fields[0])
Y.append( np.array( [ float(s) for s in fields[1:]] ))
G = len(Y)


### Moriarty's solution¶

Moriarty solved with least squares. Each solution gives us three variables per gene: baseline $b$, amplitude $a$, and phase $\phi$ that we'll call p in our python variables. We'll call the least squares solutions a_fit, etc.

You may have noticed that there's a degenerate solution: for every solution with positive amplitude $a$ and phase $\phi$, there's an equivalent solution with negative amplitude $-a$ and phase $\phi + 12$hr. Here, and in the ML solution below, I check for that and flip a $-a,\phi$ solution to $+a, \phi +12$.

In [3]:
b_fit = np.zeros(G)   # least squares baseline fit, for each gene
a_fit = np.zeros(G)   # amplitude fit, for each gene
p_fit = np.zeros(G)   # phase \phi fit, for each gene

for g in range(G):
# This solves for baseline, amplitude, phase by least squares fitting:
#    y_t = b + (a cos p) sin t + (a sin p) cos t
#
A = np.zeros((N, 3))  # observations x coefficients
for i in range(N):
A[i][0] = 1.
A[i][1] = np.sin(2. * math.pi * X[i] / 24)
A[i][2] = np.cos(2. * math.pi * X[i] / 24)

try:
result    = np.linalg.lstsq(A, Y[g], rcond=-1)[0]    # numpy's ordinary least squares fitter
except:
sys.exit("Linear least square fit failed")

p_fit[g]  = np.arctan(result[2] / result[1])   # phase in radians
b_fit[g]  = result[0]
a_fit[g]  = result[1] / np.cos(p_fit[g])

p_fit[g] = 24 * p_fit[g] / (2 * math.pi)       # convert phase to hours
if a_fit[g] < 0:                               # there's a degeneracy in the solution: negative amplitude & phase offset by 12hr
a_fit[g]  = -a_fit[g]
p_fit[g] += 12
while p_fit[g] < 0:  p_fit[g] += 24
while p_fit[g] > 24: p_fit[g] -= 24


## 1. solve by maximum likelihood¶

Now the maximum likelihood solutions, which we'll call a_opt, etc.

To solve, we'll just define a negative log likelihood function nll(), in the format that the SciPy optimizer needs, and hand it to the SciPy optimizer. This isn't the most efficient way of solving it (in terms of computation time)! But it works fine, and it's pretty efficient in terms of our time.

In [4]:
def to_radians(h):
return (h * 2 * math.pi / 24)

# nll():
# negative log likelihood objective function, compatible
# with the interface to scipy's optimize.
#   v = current parameter values (b,a,p)
#   X = time points, hrs
#   Y = TPM measurements
#   s = std dev for each data point
#
def nll(v, X, Y, s):
b     = v[0]
a     = v[1]
p     = v[2]
LL    = 0.
for i,t in enumerate(X):
yhat     = b +  a * np.sin( to_radians(t + p))
residual = Y[i] - yhat
LL      += stats.norm.logpdf(residual, loc=0, scale=s[i])
return -LL

b_opt = np.zeros(G)
a_opt = np.zeros(G)
p_opt = np.zeros(G)

for g in range(G):
guess     = np.array([10., 10., 10.])                         # any initial guess will do.
result    = optimize.minimize(nll, guess, (X, Y[g], S_true))  # hand our objective function to the SciPy minimizer
if result.success != True:
sys.exit("Maximum likelihood fit failed")

b_opt[g] = result.x[0]
a_opt[g] = result.x[1]
p_opt[g] = result.x[2]

if a_opt[g] < 0:                       # Rectify a degenerate solution w/ a negative amplitude and a 12-hr offset in phase
a_opt[g] = -a_opt[g]
p_opt[g] += 12
while p_opt[g] < 0:  p_opt[g] += 24
while p_opt[g] > 24: p_opt[g] -= 24


## 2. compare solutions¶

We're asked to calculate the total log likelihood of the two solutions. Let's also print a table of the results.

In [5]:
# we're going to sort genes according to their maximum likelihood phase.
# a concise trick for that: create a sorted list of indices 0..G-1
# Sort from high to low on phase, so genes come out in order that their
# expression rises.
#
ranked_at = sorted(list(range(G)), key=lambda g: p_opt[g], reverse=True)

# we're asked to calculate the total log likelihood of Moriarty's solution,
# versus our maximum likelihood solution
ll_opt = 0
ll_fit = 0
for g in range(G):
ll_opt -= nll( [b_opt[g], a_opt[g], p_opt[g]], X, Y[g], S_true)
ll_fit -= nll( [b_fit[g], a_fit[g], p_fit[g]], X, Y[g], S_true)

print("{0:12s} {1:>20s} {2:>20s}".format('', 'least squares', 'maximum likelihood'))
print("{0:12s} {1:20s} {2:20s}".format('', '-'*20, '-'*20))
print("{0:12s} {1:>6s} {2:>6s} {3:>6s} {4:>6s} {5:>6s} {6:>6s}".format('genename', 'b', 'a', 'p', 'b', 'a', 'p'))
print("{0:12s} {1:6s} {2:6s} {3:6s} {4:6s} {5:6s} {6:6s}".format('-'*12, '-'*6,'-'*6,'-'*6,'-'*6,'-'*6,'-'*6))
for k in range(G):
g = ranked_at[k]
print("{0:12s} {1:6.1f} {2:6.1f} {3:6.1f} {4:6.1f} {5:6.1f} {6:6.1f}".format(genenames[g], b_fit[g], a_fit[g], p_fit[g], b_opt[g], a_opt[g], p_opt[g]))

print('')
print("total log likelihood of Moriarty's solution:         {0:.1f}".format(ll_fit))
print("total log likelihood of maximum likelihood solution: {0:.1f}".format(ll_opt))

                    least squares   maximum likelihood
-------------------- --------------------
genename          b      a      p      b      a      p
------------ ------ ------ ------ ------ ------ ------
lentil         38.1   42.3   22.4   44.5   30.9   23.7
clementine     41.9   27.8   21.4   44.3   25.3   22.0
grape          46.1   21.3   19.8   42.0   27.7   20.2
melon          44.0   28.9   18.2   44.8   27.9   17.6
tangerine      39.5   19.8   13.4   40.9   26.2   16.0
anise          38.8   16.6   14.4   40.2   17.2   13.7
spinach        43.3   22.7   11.7   45.4   20.3   11.7
huckleberry    48.4   34.8   10.3   42.5   22.2    9.9
beet           40.6   17.2    7.3   41.6   21.4    7.5
kiwi           36.5   21.5    2.1   41.5   21.2    5.7
carrot         44.8   37.2    4.8   44.7   34.7    4.2
cauliflower    40.6   25.7   23.3   39.7   27.0    2.1

total log likelihood of Moriarty's solution:         -556.3
total log likelihood of maximum likelihood solution: -317.9


summary: The two solutions order tangerine versus anise and kiwi versus carrot differently. Moriarty has (anise, tangerine, ..., carrot, kiwi) whereas the ML solution has (tangerine, anise, ..., kiwi, carrot).

Also, Moriarty's solutions for the phases of several genes differ from the ML solution by an hour or more.

We'd happily take pretty much any bet Moriarty wants to offer us; the odds are massively in favor of the ML solution. (If Moriarty proposed a specific bet -- say, that his order is correct as opposed to ours -- and if we wanted to estimate fair betting odds precisely, we'd have to do more calculation and marginalization, since we wouldn't care about the exact value of $\phi$, only the order, and we wouldn't care about $a$ or $b$ at all.)

It turns out that the ML solution is the correct order. The true phases $\phi$ are evenly distributed, just 2, 4, ... 24 hr for the 12 genes, as you might have guessed.

## 3. plot the fits¶

First some bonus plotting: scatter plots of the least squares fits versus the ML estimates, for all 12 genes.

In [6]:
f, (ax1, ax2, ax3) = plt.subplots(1, 3)
ax1.scatter(b_opt, b_fit)
ax1.axis('equal')
ax1.set_title('baseline b')
ax2.scatter(a_opt, a_fit)
ax2.axis('equal')
ax2.set_title('amplitude a')
ax3.scatter(p_opt, p_fit)
ax3.set_title('phase $\phi$')

Out[6]:
Text(0.5, 1.0, 'phase $\\phi$')

We haven't been paying a lot of attention to the baseline and amplitude, but we can see there's a lot of difference between us and Moriarty there too.

Now let's look at plots of the data versus the two fits, for each gene, ordered by the ML phases. Our ML estimate will be a thick green line. Moriarty's fit is a thin grey line.

In [7]:
def make_plot_line(xpoints, b, a, p):
N = len(xpoints)
Y = np.zeros(N)
for i,x in enumerate(xpoints):
Y[i] = b + a * np.sin( to_radians(x + p) )
return Y

f, axarr = plt.subplots(6, 2, figsize=(10,20))
xpoints  = np.linspace(0., 48., 200)
for k in range(G):
g = ranked_at[k]    # plots are sorted by their maximum likelihood phase
row = k // 2
col = k % 2
axarr[row,col].plot(xpoints, make_plot_line(xpoints, b_fit[g], a_fit[g], p_fit[g]), color='#aaaaaa')
axarr[row,col].plot(xpoints, make_plot_line(xpoints, b_opt[g], a_opt[g], p_opt[g]), color='g', linewidth=2)
for i in range(N):
if reliability[i] == 'L':
axarr[row,col].plot(X[i], Y[g][i], 'ro', markersize=5, mfc='w', markeredgewidth=1)
elif reliability[i] == 'M':
axarr[row,col].plot(X[i], Y[g][i], 'yo', markersize=5, mfc='w', markeredgewidth=1)
else:
axarr[row,col].plot(X[i], Y[g][i], 'go', markersize=5, mfc='w', markeredgewidth=1)

axarr[row,col].set_xticks( list(range(0,49,4)) )
axarr[row,col].set_ylim(ymin=0)
axarr[row,col].set_title(genenames[g])
axarr[row,col].set_xlabel('time (hours)')
axarr[row,col].set_ylabel('expression (TPM)')

plt.tight_layout()