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Hadley Wickham
Stat405Intro to modelling
Tuesday, 16 November 2010
1. What is a linear model?
2. Removing trends
3. Transformations
4. Categorical data
5. Visualising models
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What is a linear
model?
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observed value
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observed value
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predicted value
observed value
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predicted value
observed value
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predicted value
observed value
residual
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y ~ x# yhat = b1x + b0
# Want to find b's that minimise distance # between y and yhat
z ~ x + y # zhat = b2x + b1y + b0
# Want to find b's that minimise distance # between z and zhat
z ~ x * y # zhat = b3(x⋅y) + b2x + b1y + b0
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X is measured without error.
Relationship is linear.
Errors are independent.
Errors have normal distribution.
Errors have constant variance.
Assumptions
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Removing trends
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library(ggplot2)
diamonds$x[diamonds$x == 0] <- NA
diamonds$y[diamonds$y == 0] <- NA
diamonds$y[diamonds$y > 30] <- NA
diamonds$z[diamonds$z == 0] <- NA
diamonds$z[diamonds$z > 30] <- NA
diamonds <- subset(diamonds, carat < 2)
qplot(x, y, data = diamonds)
qplot(x, z, data = diamonds)
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mody <- lm(y ~ x, data = diamonds, na = na.exclude)
coef(mody)
# yhat = 0.05 + 0.99⋅x
# Plot x vs yhat
qplot(x, predict(mody), data = diamonds)
# Plot x vs (y - yhat) = residual
qplot(x, resid(mody), data = diamonds)
# Standardised residual:
qplot(x, rstandard(mody), data = diamonds)
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qplot(x, resid(mody), data=dclean)Tuesday, 16 November 2010
qplot(x, y - x, data=dclean)Tuesday, 16 November 2010
Your turn
Do the same thing for z and x. What threshold might you use to remove outlying values?
Are the errors from predicting z and y from x related?
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modz <- lm(z ~ x, data = diamonds, na = na.exclude)
coef(modz)# zhat = 0.03 + 0.61x
qplot(x, rstandard(modz), data = diamonds)last_plot() + ylim(-10, 10)
qplot(rstandard(mody), rstandard(modz))
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Transformations
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Can we use a linear model to remove this trend?
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Can we use a linear model to remove this trend?
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Can we use a linear model to remove this trend?
Linear models are linear in their parameters which can be any transformation of the data
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Your turn
Use a linear model to remove the effect of carat on price. Confirm that this worked by plotting model residuals vs. color.
How can you interpret the model coefficients and residuals?
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modprice <- lm(log(price) ~ log(carat), data = diamonds, na = na.exclude)
diamonds$relprice <- exp(resid(modprice))
qplot(carat, relprice, data = diamonds)diamonds <- subset(diamonds, carat < 2)qplot(carat, relprice, data = diamonds)
qplot(carat, relprice, data = diamonds) + facet_wrap(~ color)qplot(relprice, ..density.., data = diamonds, colour = color, geom = "freqpoly", binwidth = 0.2)qplot(relprice, ..density.., data = diamonds, colour = cut, geom = "freqpoly", binwidth = 0.2)
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log(Y) = a * log(X) + b
Y = c . dX
An additive model becomes a multiplicative model.
Intercept becomes starting point, slope becomes geometric growth.
Multiplicative model
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Residuals
resid(mod) = log(Y) - log(Yhat)
exp(resid(mod)) = Y / (Yhat)
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# Useful trick - close to 0, exp(x) ~ x + 1x <- seq(-0.2, 0.2, length = 100)qplot(x, exp(x)) + geom_abline(intercept = 1)
qplot(x, x / exp(x)) + scale_y_continuous("Percent error", formatter = percent)
# Not so useful here because the x is also # transformedcoef(modprice)
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Categorical data
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Compare the results of the following two functions. What can you say about the model?
ddply(diamonds, "color", summarise, mean = mean(price))
coef(lm(price ~ color, data = diamonds))
Your turn
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Categorical data
Converted into a numeric matrix, with one column for each level. Contains 1 if that observation has that level, 0 otherwise.
However, if we just do that naively, we end up with too many columns (because we have one extra column for the intercept)
So everything is relative to the first level.
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Visualising models
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# What do you think this model does?lm(log(price) ~ log(carat) + color, data = diamonds)
# What about this one?lm(log(price) ~ log(carat) * color, data = diamonds)
# Or this one?lm(log(price) ~ cut * color, data = diamonds)
# How can we interpret the results?
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mod1 <- lm(log(price) ~ log(carat) + cut, data = diamonds)mod2 <- lm(log(price) ~ log(carat) * cut, data = diamonds)
# One way is to explore predictions from the model# over an evenly spaced grid. expand.grid makes # this easy
grid <- expand.grid( carat = seq(0.2, 2, length = 20), cut = levels(diamonds$cut), KEEP.OUT.ATTRS = FALSE)str(grid)grid
grid$p1 <- exp(predict(mod1, grid))grid$p2 <- exp(predict(mod2, grid))
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Plot the predictions from the two sets of models. How are they different?
Your turn
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qplot(carat, p1, data = grid, colour = cut, geom = "line")qplot(carat, p2, data = grid, colour = cut, geom = "line")
qplot(log(carat), log(p1), data = grid, colour = cut, geom = "line")qplot(log(carat), log(p2), data = grid, colour = cut, geom = "line")
qplot(carat, p1 / p2, data = grid, colour = cut, geom = "line")
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# Another approach is the effects package# install.packages("effects")library(effects)effect("cut", mod1)
cut <- as.data.frame(effect("cut", mod1))qplot(fit, reorder(cut, fit), data = cut)qplot(fit, reorder(cut, fit), data = cut) + geom_errorbarh(aes(xmin = lower, xmax = upper), height = 0.1)
qplot(exp(fit), reorder(cut, fit), data = cut) + geom_errorbarh(aes(xmin = exp(lower), xmax = exp(upper)), height = 0.1)
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