Logistic regression: Why protective factor become risk factor? - statistics

I used logistic analysis to build a predictive model for liver toxicity after radiotherapy. I got:
(1+e^(-S))^(-1)
S = 0.00337 ∙ normal liver volume - 0.00579 ∙ VS10 + 2.88 ∙ preCP - 2.51; normal liver volume, OR = 1.003 (95% CI, 1.000-1.007), p = 0.047; VS10, OR = 0.994 (95% CI, 0.991-0.998), p = 0.002; preCP, OR = 17.883 (95% CI, 3.235-98.848), p = 0.001.
The normal liver tissue should be the protective factor in clinical practice. Here its OR is greater than 1. Why? (The outcome setting was right.)

Related

Linear decay as learning rate scheduler (pytorch)

I have read about LinearLR and ConstantLR in the Pytorch docs but I can't figure out, how to get a linear decay of my learning rate. Say I have epochs = 10 and lr=0.1 then I want to linearly reduce my learning-rate from 0.1 to 0 (or any other number) in 10 steps i.e by 0.01 in each step.
The two constraints you have are: lr(step=0)=0.1 and lr(step=10)=0. So naturally, lr(step) = -0.1*step/10 + 0.1 = 0.1*(1 - step/10).
This is known as the polynomial learning rate scheduler. Its general form is:
def polynomial(base_lr, iter, max_iter, power):
return base_lr * ((1 - float(iter) / max_iter) ** power)
Which in your case would be called with polynomial(base_lr=0.1, max_iter=10, power=1).

Bayesian Linear Regression with PyMC3 and a large dataset - bracket nesting level exceeded maximum and slow performance

I would like to use a Bayesian multivariate linear regression to estimate the strength of players in team sports (e.g. ice hockey, basketball or soccer). For that purpose, I create a matrix, X, containing the players as columns and the matches as rows. For each match the player entry is either 1 (player plays in the home team), -1 (player plays in the away team) or 0 (player does not take part in this game). The dependent variable Y is defined as the scoring differences for both teams in each match (Score_home_team - Score_away_team).
Thus, the number of parameters will be quite large for one season (e.g. X is defined by 300 rows x 450 columns; i.e. 450 player coefficients + y-intercept). When running the fit I came across a compilation error:
('Compilation failed (return status=1): /Users/me/.theano/compiledir_Darwin-17.7.0-x86_64-i386-64bit-i386-3.6.5-64/tmpdxxc2379/mod.cpp:27598:32: fatal error: bracket nesting level exceeded maximum of 256.
I tried to handle this error by setting:
theano.config.gcc.cxxflags = "-fbracket-depth=1024"
Now, the sampling is running. However, it is so slow that even if I take only 35 of 300 rows the sampling is not completed within 20 minutes.
This is my basic code:
import pymc3 as pm
basic_model = pm.Model()
with basic_model:
# Priors for beta coefficients - these are the coefficients of the players
dict_betas = {}
for col in X.columns:
dict_betas[col] = pm.Normal(col, mu=0, sd=10)
# Priors for unknown model parameters
alpha = pm.Normal('alpha', mu=0, sd=10) # alpha is the y-intercept
sigma = pm.HalfNormal('sigma', sd=1) # standard deviation of the observations
# Expected value of outcome
mu = alpha
for col in X.columns:
mu = mu + dict_betas[col] * X[col] # mu = alpha + beta_1 * Player_1 + beta_2 * Player_2 + ...
# Likelihood (sampling distribution) of observations
Y_obs = pm.Normal('Y_obs', mu=mu, sd=sigma, observed=Y)
The instantiation of the model runs within one minute for the large dataset. I do the sampling using:
with basic_model:
# draw 500 posterior samples
trace = pm.sample(500)
The sampling is completed for small sample sizes (e.g. 9 rows, 80 columns) within 7 minutes. However, the time is increasing substantially with increasing sample size.
Any suggestions how I can get this Bayesian linear regression to run in a feasible amount of time? Are these kind of problems doable using PyMC3 (remember I came across a bracket nesting error)? I saw in a recent publication that this kind of analysis is doable in R (https://arxiv.org/pdf/1810.08032.pdf). Therefore, I guess it should also somehow work with Python 3.
Any help is appreciated!
Eliminating the for loops should improve performance and might also take care of the nesting issue you are reporting. Theano TensorVariables and the PyMC3 random variables that derive from them are already multidimensional and support linear algebra operations. Try changing your code to something along the lines of
beta = pm.Normal('beta', mu=0, sd=10, shape=X.shape[1])
...
mu = alpha + pm.math.dot(X, beta)
...
If you need specify different prior values for mu and/or sd, those arguments accept anything that theano.tensor.as_tensor_variable() accepts, so you can pass a list or numpy array.
I highly recommend getting familiar with the theano.tensor and pymc3.math operations since sometimes you must use these to properly manipulate random variables, and in general it should lead to more efficient code.

How can I interpret the P-value of nonlinear term under rms: anova?

Here I demonstrated a survival model with rcs term. I was wondering whether the anova()under rms package is the way to test the linearity association? And How can I interpret the P-value of the Nonlinear term (see 0.094 here), does that support adding a rcs() term in the cox model?
library(rms)
data(pbc)
d <- pbc
rm(pbc, pbcseq)
d$status <- ifelse(d$status != 0, 1, 0)
dd = datadist(d)
options(datadist='dd')
# rcs model
m2 <- cph(Surv(time, status) ~ rcs(albumin, 4), data=d)
anova(m2)
Wald Statistics Response: Surv(time, status)
Factor Chi-Square d.f. P
albumin 82.80 3 <.0001
Nonlinear 4.73 2 0.094
TOTAL 82.80 3 <.0001
The proper way to test is with model comparison of the log-likelihood (aka deviance) across two models: a full and reduced:
m2 <- cph(Surv(time, status) ~ rcs(albumin, 4), data=d)
anova(m2)
m <- cph(Surv(time, status) ~ albumin, data=d)
p.val <- 1- pchisq( (m2$loglik[2]- m$loglik[2]), 2 )
You can see the difference in the inference using the less accurate Wald statistic (which in your case was not significant anyway since the p-value was > 0.05) versus this more accurate method in the example that Harrell used in his ?cph help page. Using his example:
> anova(f)
Wald Statistics Response: S
Factor Chi-Square d.f. P
age 57.75 3 <.0001
Nonlinear 8.17 2 0.0168
sex 18.75 1 <.0001
TOTAL 75.63 4 <.0001
You would incorrectly conclude that the nonlinear term was "significant" at conventional 0.05 level. This despite the fact that code creating the model was constructed as entirely linear in age (on the log-hazard scale):
h <- .02*exp(.04*(age-50)+.8*(sex=='Female'))
Create a reduced mode and compare:
f0 <- cph(S ~ age + sex, x=TRUE, y=TRUE)
anova(f0)
#-------------
Wald Statistics Response: S
Factor Chi-Square d.f. P
age 56.64 1 <.0001
sex 16.26 1 1e-04
TOTAL 75.85 2 <.0001
The difference in deviance is not significant with 2 degrees of freedom difference:
1-pchisq((f$loglik[2]- f0$loglik[2]),2)
[1] 0.1243212
I don't know why Harrell leaves this example in, because I've taken his RMS course and know that he endorses the cross-model comparison of deviance as the more accurate approach.

Incorporating uncertainty into a pymc3 model

I have a set of data for which I have the mean, standard deviation and number of observations for each point (i.e., I have knowledge regarding the accuracy of the measure). In a traditional pymc3 model where I look only at the means, I may do something along the lines of:
x = data['mean']
with pm.Model() as m:
a = pm.Normal('a', mu=0, sd=1)
b = pm.Normal('b', mu=1, sd=1)
y = a + b*x
eps= pm.HalfNormal('eps', sd=1)
likelihood = pm.Normal('likelihood', mu=y, sd=eps, observed=x)
What is the best way to incorporate the information regarding the variance of the observations into the model? Obviously the result should weight low-variance observations more heavily than high-variance (less certain) observations.
One approach a statistician suggested was to do the following:
x = data['mean'] # mean of observation
x_sd = data['sd'] # sd of observation
x_n = data['n'] # of measures for observation
x_sem = x_sd/np.sqrt(x_n)
with pm.Model() as m:
a = pm.Normal('a', mu=0, sd=1)
b = pm.Normal('b', mu=1, sd=1)
y = a + b*x
eps = pm.HalfNormal('eps', sd=1)
obs = mc.Normal('obs', mu=x, sd=x_sem, shape=len(x))
likelihood = pm.Normal('likelihood', mu=y, eps=eps, observed=obs)
However, when I run this I get:
TypeError: observed needs to be data but got: <class 'pymc3.model.FreeRV'>
I am running the master branch of pymc3 (3.0 has some performance issues resulting in very slow sample times).
You are close, you just need to make some small changes. The main reason is that for PyMC3 data is always constant. Check the following code:
with pm.Model() as m:
a = pm.Normal('a', mu=0, sd=1)
b = pm.Normal('b', mu=1, sd=1)
mu = a + b*x
mu_est = pm.Normal('mu_est', mu, x_sem, shape=len(x))
likelihood = pm.Normal('likelihood', mu=mu_est, sd=x_sd, observed=x)
Notice than I keep the data fixed and I introduce the observed uncertainty at two points: for the estimation of mu_est and for the likelihood. Of course you are free to do not use x_sem or x_sd and instead estimate them, like you did in your code with the variable eps.
On a historical note, code with "random data" used to work on PyMC3 (at least for some models), but given that it was not really designed to work that way, developers decided to prevent the user from using random data, and that explains the message you got.

How to find the fundamental frequency of a guitar string sound?

I want to build a guitar tuner app for Iphone. My goal is to find the fundamental frequency of sound generated by a guitar string. I have used bits of code from aurioTouch sample provided by Apple to calculate frequency spectrum and I find the frequency with the highest amplitude . It works fine for pure sounds (the ones that have only one frequency) but for sounds from a guitar string it produces wrong results. I have read that this is because of the overtones generate by the guitar string that might have higher amplitudes than the fundamental one. How can I find the fundamental frequency so it works for guitar strings? Is there an open-source library in C/C++/Obj-C for sound analyzing (or signal processing)?
You can use the signal's autocorrelation, which is the inverse transform of the magnitude squared of the DFT. If you're sampling at 44100 samples/s, then a 82.4 Hz fundamental is about 535 samples, whereas 1479.98 Hz is about 30 samples. Look for the peak positive lag in that range (e.g. from 28 to 560). Make sure your window is at least two periods of the longest fundamental, which would be 1070 samples here. To the next power of two that's a 2048-sample buffer. For better frequency resolution and a less biased estimate, use a longer buffer, but not so long that the signal is no longer approximately stationary. Here's an example in Python:
from pylab import *
import wave
fs = 44100.0 # sample rate
K = 3 # number of windows
L = 8192 # 1st pass window overlap, 50%
M = 16384 # 1st pass window length
N = 32768 # 1st pass DFT lenth: acyclic correlation
# load a sample of guitar playing an open string 6
# with a fundamental frequency of 82.4 Hz (in theory),
# but this sample is actually at about 81.97 Hz
g = fromstring(wave.open('dist_gtr_6.wav').readframes(-1),
dtype='int16')
g = g / float64(max(abs(g))) # normalize to +/- 1.0
mi = len(g) / 4 # start index
def welch(x, w, L, N):
# Welch's method
M = len(w)
K = (len(x) - L) / (M - L)
Xsq = zeros(N/2+1) # len(N-point rfft) = N/2+1
for k in range(K):
m = k * ( M - L)
xt = w * x[m:m+M]
# use rfft for efficiency (assumes x is real-valued)
Xsq = Xsq + abs(rfft(xt, N)) ** 2
Xsq = Xsq / K
Wsq = abs(rfft(w, N)) ** 2
bias = irfft(Wsq) # for unbiasing Rxx and Sxx
p = dot(x,x) / len(x) # avg power, used as a check
return Xsq, bias, p
# first pass: acyclic autocorrelation
x = g[mi:mi + K*M - (K-1)*L] # len(x) = 32768
w = hamming(M) # hamming[m] = 0.54 - 0.46*cos(2*pi*m/M)
# reduces the side lobes in DFT
Xsq, bias, p = welch(x, w, L, N)
Rxx = irfft(Xsq) # acyclic autocorrelation
Rxx = Rxx / bias # unbias (bias is tapered)
mp = argmax(Rxx[28:561]) + 28 # index of 1st peak in 28 to 560
# 2nd pass: cyclic autocorrelation
N = M = L - (L % mp) # window an integer number of periods
# shortened to ~8192 for stationarity
x = g[mi:mi+K*M] # data for K windows
w = ones(M); L = 0 # rectangular, non-overlaping
Xsq, bias, p = welch(x, w, L, N)
Rxx = irfft(Xsq) # cyclic autocorrelation
Rxx = Rxx / bias # unbias (bias is constant)
mp = argmax(Rxx[28:561]) + 28 # index of 1st peak in 28 to 560
Sxx = Xsq / bias[0]
Sxx[1:-1] = 2 * Sxx[1:-1] # fold the freq axis
Sxx = Sxx / N # normalize S for avg power
n0 = N / mp
np = argmax(Sxx[n0-2:n0+3]) + n0-2 # bin of the nearest peak power
# check
print "\nAverage Power"
print " p:", p
print "Rxx:", Rxx[0] # should equal dot product, p
print "Sxx:", sum(Sxx), '\n' # should equal Rxx[0]
figure().subplots_adjust(hspace=0.5)
subplot2grid((2,1), (0,0))
title('Autocorrelation, R$_{xx}$'); xlabel('Lags')
mr = r_[:3 * mp]
plot(Rxx[mr]); plot(mp, Rxx[mp], 'ro')
xticks(mp/2 * r_[1:6])
grid(); axis('tight'); ylim(1.25*min(Rxx), 1.25*max(Rxx))
subplot2grid((2,1), (1,0))
title('Power Spectral Density, S$_{xx}$'); xlabel('Frequency (Hz)')
fr = r_[:5 * np]; f = fs * fr / N;
vlines(f, 0, Sxx[fr], colors='b', linewidth=2)
xticks((fs * np/N * r_[1:5]).round(3))
grid(); axis('tight'); ylim(0,1.25*max(Sxx[fr]))
show()
Output:
Average Power
p: 0.0410611012542
Rxx: 0.0410611012542
Sxx: 0.0410611012542
The peak lag is 538, which is 44100/538 = 81.97 Hz. The first-pass acyclic DFT shows the fundamental at bin 61, which is 82.10 +/- 0.67 Hz. The 2nd pass uses a window length of 538*15 = 8070, so the DFT frequencies include the fundamental period and harmonics of the string. This enables an ubiased cyclic autocorrelation for an improved PSD estimate with less harmonic spreading (i.e. the correlation can wrap around the window periodically).
Edit: Updated to use Welch's method to estimate the autocorrelation. Overlapping the windows compensates for the Hamming window. I also calculate the tapered bias of the hamming window to unbias the autocorrelation.
Edit: Added a 2nd pass with cyclic correlation to clean up the power spectral density. This pass uses 3 non-overlapping, rectangular windows length 538*15 = 8070 (short enough to be nearly stationary). The bias for cyclic correlation is a constant, instead of the Hamming window's tapered bias.
Finding the musical pitches in a chord is far more difficult than estimating the pitch of one single string or note played at a time. The overtones for the multiple notes in a chord might all be overlapping and interleaving. And all the notes in common chords may themselves be at overtone frequencies for one or more non-existent lower pitched notes.
For single notes, autocorrelation is a common technique used by some guitar tuners. But with autocorrelation, you have to be aware of some potential octave uncertainty, as guitars may produce inharmonic and decaying overtones which thus don't exactly match from pitch period to pitch period. Cepstrum and Harmonic Product Spectrum are two other pitch estimation methods which may or may not have different problems, depending on the guitar and the note.
RAPT appears to be one published algorithm for more robust pitch estimation. YIN is another.
Also Objective C is a superset of ANSI C. So you can use any C DSP routines you find for pitch estimation within an Objective C app.
Use libaubio (link) and be happy . It was one the biggest time lose for me to try to implement a fundemental frequency estimator. If you want to do it yourself I advise you follow to YINFFT method (link)

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