I am training a XGBoost model and doing hyper-parameter tuning using randomizedSearchCV. I specify the parameter distribution as:
from xgboost import XGBRegressor
# Define a xgboost regression model
model = XGBRegressor()
params = {
"colsample_bytree": uniform(0.1, 0.2), # fraction of cols to sample
"gamma": uniform(0, 0.3), # min loss reduction required for next split
"learning_rate": uniform(0.02, 0.3), # default 0.1
"n_estimators": randint(100, 150), # default 100
"subsample": uniform(0.8, 0.75) # % of rows to use in training sample
}
r = RandomizedSearchCV(model, param_distributions=params, n_iter=100,
scoring="neg_mean_absolute_error", cv=3, n_jobs=1)
I get the following error when even though the range I have specified for subsample is lower the bound [0,1].
raise XGBoostError(py_str(_LIB.XGBGetLastError()))
xgboost.core.XGBoostError: value 1.10671 for Parameter subsample exceed bound [0,1]
warnings.warn("Estimator fit failed. The score on this train-test"
Any ideas why this could be happening?
I think the issue comes from:
uniform(0.8, 0.75)
For numpy and random the first value of the function defines the lower limit and the second value the upper limit. Hence, for numpy and random you want:
uniform(0.75, 0.8)
This applies for numpy.random.uniform and random.uniform:
https://numpy.org/doc/stable/reference/random/generated/numpy.random.uniform.html
https://docs.python.org/3/library/random.html
However, for scipy.stats.uniform the definition is slightly different. Which is "Using the parameters loc and scale, one obtains the uniform distribution on [loc, loc + scale]." So for scipy you want:
uniform(0.75, 0.05)
Related
can I use the y-hat variance, the bounds, and the point estimate from the forecast data frame to calculate the confidence level that would contain a given value?
I've seen that I can change my interval level prior to fitting but programmatically that feels like a LOT of expensive trial and error.
Is there a way to estimate the confidence bound using only the information from the model parameters and the forecast data frame?
Something like:
for level in [.05, .1, .15, ... , .95]:
if value_in_question in (yhat - Z_{level}*yhat_variance/N, yhat + Z_{level}*yhat_variance/N):
print 'im in the bound level {level}'
# This is sudo code not meant to run in console
EDIT: working prophet example
# csv from fbprohets working examples https://github.com/facebook/prophet/blob/master/examples/example_wp_log_peyton_manning.csv
import pandas as pd
from fbprophet import Prophet
import os
df = pd.read_csv('example_wp_log_peyton_manning.csv')
m = Prophet()
m.fit(df)
future = m.make_future_dataframe(periods=30)
forecast = m.predict(future)
# the smallest confidence level s.t. the confidence interval of the 30th prediction contains 9
## My current approach
def __probability_calculation(estimate, forecast, j = 30):
sd_residuals = (forecast.yhat_lower[j] - forecast.yhat[j])/(-1.28)
for alpha in np.arange(.5, .95, .01):
z_val = st.norm.ppf(alpha)
if (forecast.yhat[j]-z_val*sd_residuals < estimate < forecast.yhat[j]+z_val*sd_residuals):
return alpha
prob = __probability_calculation(9, forecast)
fbprophet uses the numpy.percentile method to estimate the percentiles as you can see here in the source code:
https://github.com/facebook/prophet/blob/0616bfb5daa6888e9665bba1f95d9d67e91fed66/python/prophet/forecaster.py#L1448
How to inverse calculate percentiles for values is already answered here:
Map each list value to its corresponding percentile
Combining everything based on your code example:
import pandas as pd
import numpy as np
import scipy.stats as st
from fbprophet import Prophet
url = 'https://raw.githubusercontent.com/facebook/prophet/master/examples/example_wp_log_peyton_manning.csv'
df = pd.read_csv(url)
# put the amount of uncertainty samples in a variable so we can use it later.
uncertainty_samples = 1000 # 1000 is the default
m = Prophet(uncertainty_samples=uncertainty_samples)
m.fit(df)
future = m.make_future_dataframe(periods=30)
# You need to replicate some of the preparation steps which are part of the predict() call internals
tmpdf = m.setup_dataframe(future)
tmpdf['trend'] = m.predict_trend(tmpdf)
sim_values = m.sample_posterior_predictive(tmpdf)
The sim_values object contains for every datapoint 1000 simulations on which the confidence interval is based.
Now you can call the scipy.stats.percentileofscore method with any target value
target_value = 8
st.percentileofscore(sim_values['yhat'], target_value, 'weak') / uncertainty_samples
# returns 44.26
To prove this works backwards and forwards you can get the output of the np.percentile method and put it in the scipy.stats.percentileofscore method.
This works for an accuracy of 4 decimals:
ACCURACY = 4
for test_percentile in np.arange(0, 100, 0.5):
target_value = np.percentile(sim_values['yhat'], test_percentile)
if not np.round(st.percentileofscore(sim_values['yhat'], target_value, 'weak') / uncertainty_samples, ACCURACY) == np.round(test_percentile, ACCURACY):
print(test_percentile)
raise ValueError('This doesnt work')
I wrote a vanilla autoencoder using only Dense layer.
Below is my code:
iLayer = Input ((784,))
layer1 = Dense(128, activation='relu' ) (iLayer)
layer2 = Dense(64, activation='relu') (layer1)
layer3 = Dense(28, activation ='relu') (layer2)
layer4 = Dense(64, activation='relu') (layer3)
layer5 = Dense(128, activation='relu' ) (layer4)
layer6 = Dense(784, activation='softmax' ) (layer5)
model = Model (iLayer, layer6)
model.compile(loss='binary_crossentropy', optimizer='adam')
(trainX, trainY), (testX, testY) = mnist.load_data()
print ("shape of the trainX", trainX.shape)
trainX = trainX.reshape(trainX.shape[0], trainX.shape[1]* trainX.shape[2])
print ("shape of the trainX", trainX.shape)
model.fit (trainX, trainX, epochs=5, batch_size=100)
Questions:
1) softmax provides probability distribution. Understood. This means, I would have a vector of 784 values with probability between 0 and 1. For example [ 0.02, 0.03..... upto 784 items], summing all 784 elements provides 1.
2) I don't understand how the binary crossentropy works with these values. Binary cross entropy is for two values of output, right?
In the context of autoencoders the input and output of the model is the same. So, if the input values are in the range [0,1] then it is acceptable to use sigmoid as the activation function of last layer. Otherwise, you need to use an appropriate activation function for the last layer (e.g. linear which is the default one).
As for the loss function, it comes back to the values of input data again. If the input data are only between zeros and ones (and not the values between them), then binary_crossentropy is acceptable as the loss function. Otherwise, you need to use other loss functions such as 'mse' (i.e. mean squared error) or 'mae' (i.e. mean absolute error). Note that in the case of input values in range [0,1] you can use binary_crossentropy, as it is usually used (e.g. Keras autoencoder tutorial and this paper). However, don't expect that the loss value becomes zero since binary_crossentropy does not return zero when both prediction and label are not either zero or one (no matter they are equal or not). Here is a video from Hugo Larochelle where he explains the loss functions used in autoencoders (the part about using binary_crossentropy with inputs in range [0,1] starts at 5:30)
Concretely, in your example, you are using the MNIST dataset. So by default the values of MNIST are integers in the range [0, 255]. Usually you need to normalize them first:
trainX = trainX.astype('float32')
trainX /= 255.
Now the values would be in range [0,1]. So sigmoid can be used as the activation function and either of binary_crossentropy or mse as the loss function.
Why binary_crossentropy can be used even when the true label values (i.e. ground-truth) are in the range [0,1]?
Note that we are trying to minimize the loss function in training. So if the loss function we have used reaches its minimum value (which may not be necessarily equal to zero) when prediction is equal to true label, then it is an acceptable choice. Let's verify this is the case for binray cross-entropy which is defined as follows:
bce_loss = -y*log(p) - (1-y)*log(1-p)
where y is the true label and p is the predicted value. Let's consider y as fixed and see what value of p minimizes this function: we need to take the derivative with respect to p (I have assumed the log is the natural logarithm function for simplicity of calculations):
bce_loss_derivative = -y*(1/p) - (1-y)*(-1/(1-p)) = 0 =>
-y/p + (1-y)/(1-p) = 0 =>
-y*(1-p) + (1-y)*p = 0 =>
-y + y*p + p - y*p = 0 =>
p - y = 0 => y = p
As you can see binary cross-entropy have the minimum value when y=p, i.e. when the true label is equal to predicted label and this is exactly what we are looking for.
I'm in a situation where I need to train a model to predict a scalar value, and it's important to have the predicted value be in the same direction as the true value, while the squared error being minimum.
What would be a good choice of loss function for that?
For example:
Let's say the predicted value is -1 and the true value is 1. The loss between the two should be a lot greater than the loss between 3 and 1, even though the squared error of (3, 1) and (-1, 1) is equal.
Thanks a lot!
This turned out to be a really interesting question - thanks for asking it! First, remember that you want your loss functions to be defined entirely of differential operations, so that you can back-propagation though it. This means that any old arbitrary logic won't necessarily do. To restate your problem: you want to find a differentiable function of two variables that increases sharply when the two variables take on values of different signs, and more slowly when they share the same sign. Additionally, you want some control over how sharply these values increase, relative to one another. Thus, we want something with two configurable constants. I started constructing a function that met these needs, but then remembered one you can find in any high school geometry text book: the elliptic paraboloid!
The standard formulation doesn't meet the requirement of sign agreement symmetry, so I had to introduce a rotation. The plot above is the result. Note that it increases more sharply when the signs don't agree, and less sharply when they do, and that the input constants controlling this behaviour are configurable. The code below is all that was needed to define and plot the loss function. I don't think I've ever used a geometric form as a loss function before - really neat.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
def elliptic_paraboloid_loss(x, y, c_diff_sign, c_same_sign):
# Compute a rotated elliptic parabaloid.
t = np.pi / 4
x_rot = (x * np.cos(t)) + (y * np.sin(t))
y_rot = (x * -np.sin(t)) + (y * np.cos(t))
z = ((x_rot**2) / c_diff_sign) + ((y_rot**2) / c_same_sign)
return(z)
c_diff_sign = 4
c_same_sign = 2
a = np.arange(-5, 5, 0.1)
b = np.arange(-5, 5, 0.1)
loss_map = np.zeros((len(a), len(b)))
for i, a_i in enumerate(a):
for j, b_j in enumerate(b):
loss_map[i, j] = elliptic_paraboloid_loss(a_i, b_j, c_diff_sign, c_same_sign)
fig = plt.figure()
ax = fig.gca(projection='3d')
X, Y = np.meshgrid(a, b)
surf = ax.plot_surface(X, Y, loss_map, cmap=cm.coolwarm,
linewidth=0, antialiased=False)
plt.show()
From what I understand, your current loss function is something like:
loss = mean_square_error(y, y_pred)
What you could do, is to add one other component to your loss, being this a component that penalizes negative numbers and does nothing with positive numbers. And you can choose a coefficient for how much you want to penalize it. For that, we can use like a negative shaped ReLU. Something like this:
Let's call "Neg_ReLU" to this component. Then, your loss function will be:
loss = mean_squared_error(y, y_pred) + Neg_ReLU(y_pred)
So for example, if your result is -1, then the total error would be:
mean_squared_error(1, -1) + 1
And if your result is 3, then the total error would be:
mean_squared_error(1, -1) + 0
(See in the above function how Neg_ReLU(3) = 0, and Neg_ReLU(-1) = 1.
If you want to penalize more the negative values, then you can add a coefficient:
coeff_negative_value = 2
loss = mean_squared_error(y, y_pred) + coeff_negative_value * Neg_ReLU
Now the negative values are more penalized.
The ReLU negative function we can build it like this:
tf.nn.relu(tf.math.negative(value))
So summarizing, in the end your total loss will be:
coeff = 1
Neg_ReLU = tf.nn.relu(tf.math.negative(y))
total_loss = mean_squared_error(y, y_pred) + coeff * Neg_ReLU
I read this thread about the difference between SVC() and LinearSVC() in scikit-learn.
Now I have a data set of binary classification problem(For such a problem, the one-to-one/one-to-rest strategy difference between both functions could be ignore.)
I want to try under what parameters would these 2 functions give me the same result. First of all, of course, we should set kernel='linear' for SVC()
However, I just could not get the same result from both functions. I could not find the answer from the documents, could anybody help me to find the equivalent parameter set I am looking for?
Updated:
I modified the following code from an example of the scikit-learn website, and apparently they are not the same:
import numpy as np
import matplotlib.pyplot as plt
from sklearn import svm, datasets
# import some data to play with
iris = datasets.load_iris()
X = iris.data[:, :2] # we only take the first two features. We could
# avoid this ugly slicing by using a two-dim dataset
y = iris.target
for i in range(len(y)):
if (y[i]==2):
y[i] = 1
h = .02 # step size in the mesh
# we create an instance of SVM and fit out data. We do not scale our
# data since we want to plot the support vectors
C = 1.0 # SVM regularization parameter
svc = svm.SVC(kernel='linear', C=C).fit(X, y)
lin_svc = svm.LinearSVC(C=C, dual = True, loss = 'hinge').fit(X, y)
# create a mesh to plot in
x_min, x_max = X[:, 0].min() - 1, X[:, 0].max() + 1
y_min, y_max = X[:, 1].min() - 1, X[:, 1].max() + 1
xx, yy = np.meshgrid(np.arange(x_min, x_max, h),
np.arange(y_min, y_max, h))
# title for the plots
titles = ['SVC with linear kernel',
'LinearSVC (linear kernel)']
for i, clf in enumerate((svc, lin_svc)):
# Plot the decision boundary. For that, we will assign a color to each
# point in the mesh [x_min, m_max]x[y_min, y_max].
plt.subplot(1, 2, i + 1)
plt.subplots_adjust(wspace=0.4, hspace=0.4)
Z = clf.predict(np.c_[xx.ravel(), yy.ravel()])
# Put the result into a color plot
Z = Z.reshape(xx.shape)
plt.contourf(xx, yy, Z, cmap=plt.cm.Paired, alpha=0.8)
# Plot also the training points
plt.scatter(X[:, 0], X[:, 1], c=y, cmap=plt.cm.Paired)
plt.xlabel('Sepal length')
plt.ylabel('Sepal width')
plt.xlim(xx.min(), xx.max())
plt.ylim(yy.min(), yy.max())
plt.xticks(())
plt.yticks(())
plt.title(titles[i])
plt.show()
Result:
Output Figure from previous code
In mathematical sense you need to set:
SVC(kernel='linear', **kwargs) # by default it uses RBF kernel
and
LinearSVC(loss='hinge', **kwargs) # by default it uses squared hinge loss
Another element, which cannot be easily fixed is increasing intercept_scaling in LinearSVC, as in this implementation bias is regularized (which is not true in SVC nor should be true in SVM - thus this is not SVM) - consequently they will never be exactly equal (unless bias=0 for your problem), as they assume two different models
SVC : 1/2||w||^2 + C SUM xi_i
LinearSVC: 1/2||[w b]||^2 + C SUM xi_i
Personally I consider LinearSVC one of the mistakes of sklearn developers - this class is simply not a linear SVM.
After increasing intercept scaling (to 10.0)
However, if you scale it up too much - it will also fail, as now tolerance and number of iterations are crucial.
To sum up: LinearSVC is not linear SVM, do not use it if do not have to.
I understand the concept behind Stacked/ Deep Autoencoders and therefore want to implement it with the following code of a single-layer de-noising Autoencoder. Theano also provides a tutorial for a Stacked Autoencoder but this is trained in a supervised fashion - I need to stack it to establish unsupervised (hierarchical) feature learning.
Any idea how to get this working with the following code?
import os
import sys
import timeit
import numpy
import theano
import theano.tensor as T
from theano.tensor.shared_randomstreams import RandomStreams
from logistic_sgd import load_data
from utils import tile_raster_images
try:
import PIL.Image as Image
except ImportError:
import Image
class dA(object):
"""Denoising Auto-Encoder class (dA)
A denoising autoencoders tries to reconstruct the input from a corrupted
version of it by projecting it first in a latent space and reprojecting
it afterwards back in the input space. Please refer to Vincent et al.,2008
for more details. If x is the input then equation (1) computes a partially
destroyed version of x by means of a stochastic mapping q_D. Equation (2)
computes the projection of the input into the latent space. Equation (3)
computes the reconstruction of the input, while equation (4) computes the
reconstruction error.
.. math::
\tilde{x} ~ q_D(\tilde{x}|x) (1)
y = s(W \tilde{x} + b) (2)
x = s(W' y + b') (3)
L(x,z) = -sum_{k=1}^d [x_k \log z_k + (1-x_k) \log( 1-z_k)] (4)
"""
def __init__(
self,
numpy_rng,
theano_rng=None,
input=None,
n_visible=784,
n_hidden=500,
W=None,
bhid=None,
bvis=None
):
"""
Initialize the dA class by specifying the number of visible units (the
dimension d of the input ), the number of hidden units ( the dimension
d' of the latent or hidden space ) and the corruption level. The
constructor also receives symbolic variables for the input, weights and
bias. Such a symbolic variables are useful when, for example the input
is the result of some computations, or when weights are shared between
the dA and an MLP layer. When dealing with SdAs this always happens,
the dA on layer 2 gets as input the output of the dA on layer 1,
and the weights of the dA are used in the second stage of training
to construct an MLP.
:type numpy_rng: numpy.random.RandomState
:param numpy_rng: number random generator used to generate weights
:type theano_rng: theano.tensor.shared_randomstreams.RandomStreams
:param theano_rng: Theano random generator; if None is given one is
generated based on a seed drawn from `rng`
:type input: theano.tensor.TensorType
:param input: a symbolic description of the input or None for
standalone dA
:type n_visible: int
:param n_visible: number of visible units
:type n_hidden: int
:param n_hidden: number of hidden units
:type W: theano.tensor.TensorType
:param W: Theano variable pointing to a set of weights that should be
shared belong the dA and another architecture; if dA should
be standalone set this to None
:type bhid: theano.tensor.TensorType
:param bhid: Theano variable pointing to a set of biases values (for
hidden units) that should be shared belong dA and another
architecture; if dA should be standalone set this to None
:type bvis: theano.tensor.TensorType
:param bvis: Theano variable pointing to a set of biases values (for
visible units) that should be shared belong dA and another
architecture; if dA should be standalone set this to None
"""
self.n_visible = n_visible
self.n_hidden = n_hidden
# create a Theano random generator that gives symbolic random values
if not theano_rng:
theano_rng = RandomStreams(numpy_rng.randint(2 ** 30))
# note : W' was written as `W_prime` and b' as `b_prime`
if not W:
# W is initialized with `initial_W` which is uniformely sampled
# from -4*sqrt(6./(n_visible+n_hidden)) and
# 4*sqrt(6./(n_hidden+n_visible))the output of uniform if
# converted using asarray to dtype
# theano.config.floatX so that the code is runable on GPU
initial_W = numpy.asarray(
numpy_rng.uniform(
low=-4 * numpy.sqrt(6. / (n_hidden + n_visible)),
high=4 * numpy.sqrt(6. / (n_hidden + n_visible)),
size=(n_visible, n_hidden)
),
dtype=theano.config.floatX
)
W = theano.shared(value=initial_W, name='W', borrow=True)
if not bvis:
bvis = theano.shared(
value=numpy.zeros(
n_visible,
dtype=theano.config.floatX
),
borrow=True
)
if not bhid:
bhid = theano.shared(
value=numpy.zeros(
n_hidden,
dtype=theano.config.floatX
),
name='b',
borrow=True
)
self.W = W
# b corresponds to the bias of the hidden
self.b = bhid
# b_prime corresponds to the bias of the visible
self.b_prime = bvis
# tied weights, therefore W_prime is W transpose
self.W_prime = self.W.T
self.theano_rng = theano_rng
# if no input is given, generate a variable representing the input
if input is None:
# we use a matrix because we expect a minibatch of several
# examples, each example being a row
self.x = T.dmatrix(name='input')
else:
self.x = input
self.params = [self.W, self.b, self.b_prime]
def get_corrupted_input(self, input, corruption_level):
"""This function keeps ``1-corruption_level`` entries of the inputs the
same and zero-out randomly selected subset of size ``coruption_level``
Note : first argument of theano.rng.binomial is the shape(size) of
random numbers that it should produce
second argument is the number of trials
third argument is the probability of success of any trial
this will produce an array of 0s and 1s where 1 has a
probability of 1 - ``corruption_level`` and 0 with
``corruption_level``
The binomial function return int64 data type by
default. int64 multiplicated by the input
type(floatX) always return float64. To keep all data
in floatX when floatX is float32, we set the dtype of
the binomial to floatX. As in our case the value of
the binomial is always 0 or 1, this don't change the
result. This is needed to allow the gpu to work
correctly as it only support float32 for now.
"""
return self.theano_rng.binomial(size=input.shape, n=1,
p=1 - corruption_level,
dtype=theano.config.floatX) * input
def get_hidden_values(self, input):
""" Computes the values of the hidden layer """
return T.nnet.sigmoid(T.dot(input, self.W) + self.b)
def get_reconstructed_input(self, hidden):
"""Computes the reconstructed input given the values of the
hidden layer
"""
return T.nnet.sigmoid(T.dot(hidden, self.W_prime) + self.b_prime)
def get_cost_updates(self, corruption_level, learning_rate):
""" This function computes the cost and the updates for one trainng
step of the dA """
tilde_x = self.get_corrupted_input(self.x, corruption_level)
y = self.get_hidden_values(tilde_x)
z = self.get_reconstructed_input(y)
# note : we sum over the size of a datapoint; if we are using
# minibatches, L will be a vector, with one entry per
# example in minibatch
L = - T.sum(self.x * T.log(z) + (1 - self.x) * T.log(1 - z), axis=1)
# note : L is now a vector, where each element is the
# cross-entropy cost of the reconstruction of the
# corresponding example of the minibatch. We need to
# compute the average of all these to get the cost of
# the minibatch
cost = T.mean(L)
# compute the gradients of the cost of the `dA` with respect
# to its parameters
gparams = T.grad(cost, self.params)
# generate the list of updates
updates = [
(param, param - learning_rate * gparam)
for param, gparam in zip(self.params, gparams)
]
return (cost, updates)