10. 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
Tuesday, 16 November 2010
11. X is measured without error.
Relationship is linear.
Errors are independent.
Errors have normal distribution.
Errors have constant variance.
Assumptions
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18. qplot(x, y - x, data=dclean)
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19. 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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20. 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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22. Can we use a
linear model to
remove this trend?
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23. Can we use a
linear model to
remove this trend?
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24. 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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25. 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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26. 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)
Tuesday, 16 November 2010
27. 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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29. # Useful trick - close to 0, exp(x) ~ x + 1
x <- 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
# transformed
coef(modprice)
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31. 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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32. 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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34. # 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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35. 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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36. Plot the predictions from the two sets of
models. How are they different?
Your turn
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