%matplotlib inline

from statsmodels.compat import lmap
import numpy as np
from scipy import stats
import matplotlib.pyplot as plt

import statsmodels.api as sm

"""
Explanation: M-Estimators for Robust Linear Modeling
End of explanation
"""


norms = sm.robust.norms

def plot_weights(support, weights_func, xlabels, xticks):
    fig = plt.figure(figsize=(12,8))
    ax = fig.add_subplot(111)
    ax.plot(support, weights_func(support))
    ax.set_xticks(xticks)
    ax.set_xticklabels(xlabels, fontsize=16)
    ax.set_ylim(-.1, 1.1)
    return ax

"""
Explanation: An M-estimator minimizes the function 

$$Q(e_i, \rho) = \sum_i~\rho \left (\frac{e_i}{s}\right )$$
where $\rho$ is a symmetric function of the residuals 

The effect of $\rho$ is to reduce the influence of outliers
$s$ is an estimate of scale. 

The robust estimates $\hat{\beta}$ are computed by the iteratively re-weighted least squares algorithm


We have several choices available for the weighting functions to be used
End of explanation
"""


help(norms.AndrewWave.weights)

a = 1.339
support = np.linspace(-np.pi*a, np.pi*a, 100)
andrew = norms.AndrewWave(a=a)
plot_weights(support, andrew.weights, ['$-\pi*a$', '0', '$\pi*a$'], [-np.pi*a, 0, np.pi*a]);

"""
Explanation: Andrew's Wave
End of explanation
"""


help(norms.Hampel.weights)

c = 8
support = np.linspace(-3*c, 3*c, 1000)
hampel = norms.Hampel(a=2., b=4., c=c)
plot_weights(support, hampel.weights, ['3*c', '0', '3*c'], [-3*c, 0, 3*c]);

"""
Explanation: Hampel's 17A
End of explanation
"""


help(norms.HuberT.weights)

t = 1.345
support = np.linspace(-3*t, 3*t, 1000)
huber = norms.HuberT(t=t)
plot_weights(support, huber.weights, ['-3*t', '0', '3*t'], [-3*t, 0, 3*t]);

"""
Explanation: Huber's t
End of explanation
"""


help(norms.LeastSquares.weights)

support = np.linspace(-3, 3, 1000)
lst_sq = norms.LeastSquares()
plot_weights(support, lst_sq.weights, ['-3', '0', '3'], [-3, 0, 3]);

"""
Explanation: Least Squares
End of explanation
"""


help(norms.RamsayE.weights)

a = .3
support = np.linspace(-3*a, 3*a, 1000)
ramsay = norms.RamsayE(a=a)
plot_weights(support, ramsay.weights, ['-3*a', '0', '3*a'], [-3*a, 0, 3*a]);

"""
Explanation: Ramsay's Ea
End of explanation
"""


help(norms.TrimmedMean.weights)

c = 2
support = np.linspace(-3*c, 3*c, 1000)
trimmed = norms.TrimmedMean(c=c)
plot_weights(support, trimmed.weights, ['-3*c', '0', '3*c'], [-3*c, 0, 3*c]);

"""
Explanation: Trimmed Mean
End of explanation
"""


help(norms.TukeyBiweight.weights)

c = 4.685
support = np.linspace(-3*c, 3*c, 1000)
tukey = norms.TukeyBiweight(c=c)
plot_weights(support, tukey.weights, ['-3*c', '0', '3*c'], [-3*c, 0, 3*c]);

"""
Explanation: Tukey's Biweight
End of explanation
"""


x = np.array([1, 2, 3, 4, 500])

"""
Explanation: Scale Estimators

Robust estimates of the location
End of explanation
"""


x.mean()

"""
Explanation: The mean is not a robust estimator of location
End of explanation
"""


np.median(x)

"""
Explanation: The median, on the other hand, is a robust estimator with a breakdown point of 50%
End of explanation
"""


x.std()

"""
Explanation: Analogously for the scale
The standard deviation is not robust
End of explanation
"""


stats.norm.ppf(.75)

print(x)

sm.robust.scale.mad(x)

np.array([1,2,3,4,5.]).std()

"""
Explanation: Median Absolute Deviation
$$ median_i |X_i - median_j(X_j)|) $$
Standardized Median Absolute Deviation is a consistent estimator for $\hat{\sigma}$
$$\hat{\sigma}=K \cdot MAD$$
where $K$ depends on the distribution. For the normal distribution for example,
$$K = \Phi^{-1}(.75)$$
End of explanation
"""


sm.robust.scale.iqr(x)

"""
Explanation: Another robust estimator of scale is the Interquartile Range (IQR)
$$\left(\hat{X}{0.75} - \hat{X}{0.25}\right),$$
where $\hat{X}_{p}$ is the sample p-th quantile and $K$ depends on the distribution. 
The standardized IQR, given by $K \cdot \text{IQR}$ for
$$K = \frac{1}{\Phi^{-1}(.75) - \Phi^{-1}(.25)} \approx 0.74,$$
is a consistent estimator of the standard deviation for normal data.
End of explanation
"""


sm.robust.scale.qn(x)

"""
Explanation: The IQR is less robust than the MAD in the sense that it has a lower breakdown point: it can withstand 25\% outlying observations before being completely ruined, whereas the MAD can withstand 50\% outlying observations. However, the IQR is better suited for asymmetric distributions.
Yet another robust estimator of scale is the $Q_n$ estimator, introduced in Rousseeuw & Croux (1993), 'Alternatives to the Median Absolute Deviation'. Then $Q_n$ estimator is given by
$$
Q_n = K \left\lbrace \vert X_{i} - X_{j}\vert : i<j\right\rbrace_{(h)}
$$
where $h\approx (1/4){{n}\choose{2}}$ and $K$ is a given constant. In words, the $Q_n$ estimator is the normalized $h$-th order statistic of the absolute differences of the data. The normalizing constant $K$ is usually chosen as 2.219144, to make the estimator consistent for the standard deviation in the case of normal data. The $Q_n$ estimator has a 50\% breakdown point and a 82\% asymptotic efficiency at the normal distribution, much higher than the 37\% efficiency of the MAD.
End of explanation
"""


np.random.seed(12345)
fat_tails = stats.t(6).rvs(40)

kde = sm.nonparametric.KDEUnivariate(fat_tails)
kde.fit()
fig = plt.figure(figsize=(12,8))
ax = fig.add_subplot(111)
ax.plot(kde.support, kde.density);

print(fat_tails.mean(), fat_tails.std())

print(stats.norm.fit(fat_tails))

print(stats.t.fit(fat_tails, f0=6))

huber = sm.robust.scale.Huber()
loc, scale = huber(fat_tails)
print(loc, scale)

sm.robust.mad(fat_tails)

sm.robust.mad(fat_tails, c=stats.t(6).ppf(.75))

sm.robust.scale.mad(fat_tails)

"""
Explanation: The default for Robust Linear Models is MAD
another popular choice is Huber's proposal 2
End of explanation
"""


from statsmodels.graphics.api import abline_plot
from statsmodels.formula.api import ols, rlm

prestige = sm.datasets.get_rdataset("Duncan", "carData", cache=True).data

print(prestige.head(10))

fig = plt.figure(figsize=(12,12))
ax1 = fig.add_subplot(211, xlabel='Income', ylabel='Prestige')
ax1.scatter(prestige.income, prestige.prestige)
xy_outlier = prestige.loc['minister', ['income','prestige']]
ax1.annotate('Minister', xy_outlier, xy_outlier+1, fontsize=16)
ax2 = fig.add_subplot(212, xlabel='Education',
                           ylabel='Prestige')
ax2.scatter(prestige.education, prestige.prestige);

ols_model = ols('prestige ~ income + education', prestige).fit()
print(ols_model.summary())

infl = ols_model.get_influence()
student = infl.summary_frame()['student_resid']
print(student)

print(student.loc[np.abs(student) > 2])

print(infl.summary_frame().loc['minister'])

sidak = ols_model.outlier_test('sidak')
sidak.sort_values('unadj_p', inplace=True)
print(sidak)

fdr = ols_model.outlier_test('fdr_bh')
fdr.sort_values('unadj_p', inplace=True)
print(fdr)

rlm_model = rlm('prestige ~ income + education', prestige).fit()
print(rlm_model.summary())

print(rlm_model.weights)

"""
Explanation: Duncan's Occupational Prestige data - M-estimation for outliers
End of explanation
"""


dta = sm.datasets.get_rdataset("starsCYG", "robustbase", cache=True).data

from matplotlib.patches import Ellipse
fig = plt.figure(figsize=(12,8))
ax = fig.add_subplot(111, xlabel='log(Temp)', ylabel='log(Light)', title='Hertzsprung-Russell Diagram of Star Cluster CYG OB1')
ax.scatter(*dta.values.T)
# highlight outliers
e = Ellipse((3.5, 6), .2, 1, alpha=.25, color='r')
ax.add_patch(e);
ax.annotate('Red giants', xy=(3.6, 6), xytext=(3.8, 6),
            arrowprops=dict(facecolor='black', shrink=0.05, width=2),
            horizontalalignment='left', verticalalignment='bottom',
            clip_on=True, # clip to the axes bounding box
            fontsize=16,
     )
# annotate these with their index
for i,row in dta.loc[dta['log.Te'] < 3.8].iterrows():
    ax.annotate(i, row, row + .01, fontsize=14)
xlim, ylim = ax.get_xlim(), ax.get_ylim()

from IPython.display import Image
Image(filename='star_diagram.png')

y = dta['log.light']
X = sm.add_constant(dta['log.Te'], prepend=True)
ols_model = sm.OLS(y, X).fit()
abline_plot(model_results=ols_model, ax=ax)

rlm_mod = sm.RLM(y, X, sm.robust.norms.TrimmedMean(.5)).fit()
abline_plot(model_results=rlm_mod, ax=ax, color='red')

"""
Explanation: Hertzprung Russell data for Star Cluster CYG 0B1 - Leverage Points

Data is on the luminosity and temperature of 47 stars in the direction of Cygnus.
End of explanation
"""


infl = ols_model.get_influence()

h_bar = 2*(ols_model.df_model + 1 )/ols_model.nobs
hat_diag = infl.summary_frame()['hat_diag']
hat_diag.loc[hat_diag > h_bar]

sidak2 = ols_model.outlier_test('sidak')
sidak2.sort_values('unadj_p', inplace=True)
print(sidak2)

fdr2 = ols_model.outlier_test('fdr_bh')
fdr2.sort_values('unadj_p', inplace=True)
print(fdr2)

"""
Explanation: Why? Because M-estimators are not robust to leverage points.
End of explanation
"""


l =  ax.lines[-1]
l.remove()
del l

weights = np.ones(len(X))
weights[X[X['log.Te'] < 3.8].index.values - 1] = 0
wls_model = sm.WLS(y, X, weights=weights).fit()
abline_plot(model_results=wls_model, ax=ax, color='green')

"""
Explanation: Let's delete that line
End of explanation
"""


yy = y.values[:,None]
xx = X['log.Te'].values[:,None]

"""
Explanation: MM estimators are good for this type of problem, unfortunately, we do not yet have these yet. 
It's being worked on, but it gives a good excuse to look at the R cell magics in the notebook.
End of explanation
"""


params = [-4.969387980288108, 2.2531613477892365]  # Computed using R
print(params[0], params[1])

abline_plot(intercept=params[0], slope=params[1], ax=ax, color='red')

"""
Explanation: Note: The R code and the results in this notebook has been converted to markdown so that R is not required to build the documents. The R results in the notebook were computed using R 3.5.1 and robustbase 0.93.
```ipython
%load_ext rpy2.ipython
%R library(robustbase)
%Rpush yy xx
%R mod <- lmrob(yy ~ xx);
%R params <- mod$coefficients;
%Rpull params
```
ipython
%R print(mod)
Call:
lmrob(formula = yy ~ xx)
 \--&gt; method = "MM"
Coefficients:
(Intercept)           xx  
     -4.969        2.253
End of explanation
"""


np.random.seed(12345)
nobs = 200
beta_true = np.array([3, 1, 2.5, 3, -4])
X = np.random.uniform(-20,20, size=(nobs, len(beta_true)-1))
# stack a constant in front
X = sm.add_constant(X, prepend=True) # np.c_[np.ones(nobs), X]
mc_iter = 500
contaminate = .25 # percentage of response variables to contaminate

all_betas = []
for i in range(mc_iter):
    y = np.dot(X, beta_true) + np.random.normal(size=200)
    random_idx = np.random.randint(0, nobs, size=int(contaminate * nobs))
    y[random_idx] = np.random.uniform(-750, 750)
    beta_hat = sm.RLM(y, X).fit().params
    all_betas.append(beta_hat)

all_betas = np.asarray(all_betas)
se_loss = lambda x : np.linalg.norm(x, ord=2)**2
se_beta = lmap(se_loss, all_betas - beta_true)

"""
Explanation: Exercise: Breakdown points of M-estimator
End of explanation
"""


np.array(se_beta).mean()

all_betas.mean(0)

beta_true

se_loss(all_betas.mean(0) - beta_true)

"""
Explanation: Squared error loss
End of explanation
"""


