04: Linear Regression with Multiple Variables

[ 04: Linear Regression with Multiple Variables ]

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Octave Programming: Linear Regression


Video Lectures


Linear regression with multiple features

New version of linear regression with multiple features
  • Multiple variables = multiple features
  • In original version we had
    • X = house size, use this to predict
    • y = house price
  • If in a new scheme we have more variables (such as number of bedrooms, number floors, age of the home)
    • x1, x2, x3, xare the four features
      • x1 – size (feet squared)
      • x2 – Number of bedrooms
      • x3 – Number of floors
      • x4 – Age of home (years)
    • y is the output variable (price)
  • More notation
      • number of features (n = 4)
    • m 
      • number of examples (i.e. number of rows in a table)
    • xi 
      • vector of the input for an example (so a vector of the four parameters for the ith input example)
      • i is an index into the training set
      • So
        • x is an n-dimensional feature vector
        • x3 is, for example, the 3rd house, and contains the four features associated with that house
    • xj
      • The value of feature j in the ith training example
      • So
        • x2is, for example, the number of bedrooms in the third house
  • Now we have multiple features
    • What is the form of our hypothesis?
    • Previously our hypothesis took the form;
      • hθ(x) = θ0 + θ1x
        • Here we have two parameters (theta 1 and theta 2) determined by our cost function
        • One variable x
    • Now we have multiple features
      • hθ(x) = θ0 + θ1x1 + θ2x2 + θ3x3 + θ4x4
    • For example
      • hθ(x) = 80 + 0.1x1 + 0.01x2 + 3x3 – 2x4
        • An example of a hypothesis which is trying to predict the price of a house
        • Parameters are still determined through a cost function
    • For convenience of notation, x0 = 1
      • For every example i you have an additional 0th feature for each example
      • So now your feature vector is n + 1 dimensional feature vector indexed from 0
        • This is a column vector called x
        • Each example has a column vector associated with it
        • So let’s say we have a new example called “X”
      • Parameters are also in a 0 indexed n+1 dimensional vector
        • This is also a column vector called θ
        • This vector is the same for each example
    • Considering this, hypothesis can be written
      • hθ(x) = θ0x0 + θ1x1 + θ2x2 + θ3x3 + θ4x4
    • If we do
      • hθ(x) =θT X
        • θis an [1 x n+1] matrix
        • In other words, because θ is a column vector, the transposition operation transforms it into a row vector
        • So before
          • θ was a matrix [n + 1 x 1]
        • Now
          • θis a matrix [1 x n+1]
        • Which means the inner dimensions of θT and X match, so they can be multiplied together as
          • [1 x n+1] * [n+1 x 1]
          • hθ(x)
          • So, in other words, the transpose of our parameter vector * an input example X gives you a predicted hypothesis which is [1 x 1] dimensions (i.e. a single value)
      • This x0 = 1 lets us write this like this
    • This is an example of multivariate linear regression
Gradient descent for multiple variables
  • Fitting parameters for the hypothesis with gradient descent
    • Parameters are θ0 to θn
    • Instead of thinking about this as n separate values, think about the parameters as a single vector (θ)
      • Where θ is n+1 dimensional
  • Our cost function is
  • Similarly, instead of thinking of J as a function of the n+1 numbers, J() is just a function of the parameter vector
    • J(θ)
  • Gradient descent
  • Once again, this is
    • θj = θj – learning rate (α) times the partial derivative of J(θ) with respect to θJ(…)
    • We do this through a simultaneous update of every θj value
  • Implementing this algorithm
    • When n = 1

  • Above, we have slightly different update rules for θ0 and θ1
    • Actually they’re the same, except the end has a previously undefined x0(i) as 1, so wasn’t shown
  • We now have an almost identical rule for multivariate gradient descent

  • What’s going on here?
    • We’re doing this for each j (0 until n) as a simultaneous update (like when n = 1)
    • So, we re-set θj to
      • θj minus the learning rate (α) times the partial derivative of of the θ vector with respect to θj
      • In non-calculus words, this means that we do
        • Learning rate
        • Times 1/m (makes the maths easier)
        • Times the sum of
          • The hypothesis taking in the variable vector, minus the actual value, times the j-th value in that variable vector for EACH example
    • It’s important to remember that
  • These algorithm are highly similar
Gradient Decent in practice: 1 Feature Scaling
  • Having covered the theory, we now move on to learn about some of the practical tricks
  • Feature scaling
    • If you have a problem with multiple features
    • You should make sure those features have a similar scale
      • Means gradient descent will converge more quickly
    • e.g.
      • x1 = size (0 – 2000 feet)
      • x2 = number of bedrooms (1-5)
      • Means the contours generated if we plot θ1 vs. θ2 give a very tall and thin shape due to the huge range difference
    • Running gradient descent on this kind of cost function can take a long time to find the global minimum
  • Pathological input to gradient descent
    • So we need to rescale this input so it’s more effective
    • So, if you define each value from x1 and x2 by dividing by the max for each feature
    • Contours become more like circles (as scaled between 0 and 1)
  • May want to get everything into -1 to +1 range (approximately)
    • Want to avoid large ranges, small ranges or very different ranges from one another
    • Rule a thumb regarding acceptable ranges
      • -3 to +3 is generally fine – any bigger bad
      • -1/3 to +1/3 is ok – any smaller bad
  • Can do mean normalization
    • Take a feature xi
      • Replace it by (xi – mean)/max
      • So your values all have an average of about 0
  • Instead of max can also use standard deviation
Learning Rate α 
  • Focus on the learning rate (α)
  • Topics
    • Update rule
    • Debugging
    • How to chose α

Make sure gradient descent is working

  • Plot min J(θ) vs. no of iterations
    • (i.e. plotting J(θ) over the course of gradient descent
  • If gradient descent is working then J(θ) should decrease after every iteration
  • Can also show if you’re not making huge gains after a certain number
    • Can apply heuristics to reduce number of iterations if need be
    • If, for example, after 1000 iterations you reduce the parameters by nearly nothing you could chose to only run 1000 iterations in the future
    • Make sure you don’t accidentally hard-code thresholds like this in and then forget about why they’re their though!
    • Number of iterations varies a lot
      • 30 iterations
      • 3000 iterations
      • 3000 000 iterations
      • Very hard to tel in advance how many iterations will be needed
      • Can often make a guess based a plot like this after the first 100 or so iterations
    • Automatic convergence tests
      • Check if J(θ) changes by a small threshold or less
        • Choosing this threshold is hard
        • So often easier to check for a straight line
          • Why? – Because we’re seeing the straightness in the context of the whole algorithm
          • Could you design an automatic checker which calculates a threshold based on the systems preceding progress?
    • Checking its working
      • If you plot J(θ) vs iterations and see the value is increasing – means you probably need a smaller α
        • Cause is because your minimizing a function which looks like this

    • But you overshoot, so reduce learning rate so you actually reach the minimum (green line)

    • So, use a smaller α
  • Another problem might be if J(θ) looks like a series of waves
    • Here again, you need a smaller α
  • However
    • If α is small enough, J(θ) will decrease on every iteration
    • BUT, if α is too small then rate is too slow
      • A less steep incline is indicative of a slow convergence, because we’re decreasing by less on each iteration than a steeper slope
  • Typically
    • Try a range of alpha values
    • Plot J(θ) vs number of iterations for each version of alpha
    • Go for roughly threefold increases
      • 0.001, 0.003, 0.01, 0.03. 0.1, 0.3
Features and polynomial regression
  • Choice of features and how you can get different learning algorithms by choosing appropriate features
  • Polynomial regression for non-linear function
  • Example
    • House price prediction
      • Two features
        • Frontage – width of the plot of land along road (x1)
        • Depth – depth away from road (x2)
    • You don’t have to use just two features
      • Can create new features
    • Might decide that an important feature is the land area
      • So, create a new feature = frontage * depth (x3)
      • h(x) = θ0 + θ1x3
        • Area is a better indicator
    • Often, by defining new features you may get a better model
  • Polynomial regression
    • May fit the data better
    • θ0 + θ1x + θ2x2 e.g. here we have a quadratic function
    • For housing data could use a quadratic function
      • But may not fit the data so well – inflection point means housing prices decrease when size gets really big
      • So instead must use a cubic function
  • How do we fit the model to this data
    • To map our old linear hypothesis and cost functions to these polynomial descriptions the easy thing to do is set
      • x1 = x
      • x2 = x2
      • x3 = x3
    • By selecting the features like this and applying the linear regression algorithms you can do polynomial linear regression
    • Remember, feature scaling becomes even more important here
  • Instead of a conventional polynomial you could do variable ^(1/something) – i.e. square root, cubed root etc
  • Lots of features – later look at developing an algorithm to chose the best features
Normal equation
  • For some linear regression problems the normal equation provides a better solution
  • So far we’ve been using gradient descent
    • Iterative algorithm which takes steps to converse
  • Normal equation solves θ analytically
    • Solve for the optimum value of theta
  • Has some advantages and disadvantages

How does it work?

  • Simplified cost function
    • J(θ) = aθ+ bθ + c
      • θ is just a real number, not a vector
    • Cost function is a quadratic function
    • How do you minimize this?
      • Do
          • Take derivative of J(θ) with respect to θ
          • Set that derivative equal to 0
          • Allows you to solve for the value of θ which minimizes J(θ)
  • In our more complex problems;
    • Here θ is an n+1 dimensional vector of real numbers
    • Cost function is a function of the vector value
      • How do we minimize this function
        • Take the partial derivative of J(θ) with respect θand set to 0 for every j
        • Do that and solve for θ0 to θn
        • This would give the values of θ which minimize J(θ)
    • If you work through the calculus and the solution, the derivation is pretty complex
      • Not going to go through here
      • Instead, what do you need to know to implement this process
Example of normal equation
  • Here
    • m = 4
    • n = 4
  • To implement the normal equation
    • Take examples
    • Add an extra column (x0 feature)
    • Construct a matrix (X – the design matrix) which contains all the training data features in an [m x n+1] matrix
    • Do something similar for y
      • Construct a column vector y vector [m x 1] matrix
    • Using the following equation (X transpose * X) inverse times X transpose y
  • If you compute this, you get the value of theta which minimize the cost function

General case

  • Have m training examples and n features
    • The design matrix (X)
      • Each training example is a n+1 dimensional feature column vector
      • X is constructed by taking each training example, determining its transpose (i.e. column -> row) and using it for a row in the design A
      • This creates an [m x (n+1)] matrix
    • Vector y
      • Used by taking all the y values into a column vector
  • What is this equation?!
    • (XT * X)-1
      • What is this –> the inverse of the matrix (XT * X)
        • i.e. A = XT X
        • A-1 = (XT X)-1
  • In octave and MATLAB you could do;        pinv(X’*x)*x’*y
      • X’ is the notation for X transpose
      • pinv is a function for the inverse of a matrix
  • In a previous lecture discussed feature scaling
    • If you’re using the normal equation then no need for feature scaling

When should you use gradient descent and when should you use feature scaling?

    • Gradient descent
      • Need to chose learning rate
      • Needs many iterations – could make it slower
      • Works well even when n is massive (millions)
        • Better suited to big data
        • What is a big n though
          • 100 or even a 1000 is still (relativity) small
          • If n is 10 000 then look at using gradient descent
    • Normal equation
      • No need to chose a learning rate
      • No need to iterate, check for convergence etc.
      • Normal equation needs to compute (XT X)-1
        • This is the inverse of an n x n matrix
        • With most implementations computing a matrix inverse grows by O(n)
          • So not great
      • Slow of n is large
        • Can be much slower
Normal equation and non-invertibility
  • Advanced concept
    • Often asked about, but quite advanced, perhaps optional material
    • Phenomenon worth understanding, but not probably necessary
  • When computing (XT X)-1 * XT * y)
    • What if (XT X) is non-invertible (singular/degenerate)
      • Only some matrices are invertible
      • This should be quite a rare problem
        • Octave can invert matrices using
          • pinv (pseudo inverse)
            • This gets the right value even if (XT X) is non-invertible
          • inv (inverse)
    • What does it mean for (XT X) to be non-invertible
      • Normally two common causes
        • Redundant features in learning model
          • e.g.
            • x1 = size in feet
            • x2 = size in meters squared
        • Too many features
          • e.g. m <= n (m is much larger than n)
            • m = 10
            • n = 100
          • Trying to fit 101 parameters from 10 training examples
          • Sometimes work, but not always a good idea
          • Not enough data
          • Later look at why this may be too little data
          • To solve this we
            • Delete features
            • Use regularization (let’s you use lots of features for a small training set)
    • If you find (XT X) to be non-invertible
      • Look at features –> are features linearly dependent?
        • So just delete one, will solve problem

Author: iotmaker

I am interested in IoT, robot, figures & leadership. Also, I have spent almost every day of the past 15 years making robots or electronic inventions or computer programs.

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