Learning Objectives
- To become familiar with common statistical functions available in R
- Linear Regression
- Logistic Regression
- K-Means Clustering
Linear regression is one of the most commonly used methods in all of statistics. It is used for a large variety of applications and offers highly interpretable results. It was the first regression method discovered and belongs to one of the most important families of models, generalized linear models.
Simple linear regression estimates the linear relationship between two variables, an outcome variable y, and an explanatory variable x.
To fit a linear regression in R, we can use the lm()
function (think linear model). We use the formula notation, y~x
where y
is the name of your outcome variable, and x
is the name of your explanatory variable, both are unquoted. The easiest way to view the results interactively is with the summary()
function.
weight_fit <- lm(formula = wander_dist ~ weight, data = cats)
summary(weight_fit)
Call:
lm(formula = wander_dist ~ weight, data = cats)
Residuals:
Min 1Q Median 3Q Max
-0.076139 -0.019173 -0.004538 0.018240 0.094993
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.0765553 0.0042990 17.808 <2e-16 ***
weight -0.0043866 0.0006988 -6.278 9e-10 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.0316 on 398 degrees of freedom
Multiple R-squared: 0.09009, Adjusted R-squared: 0.08781
F-statistic: 39.41 on 1 and 398 DF, p-value: 9.003e-10
In this case, the summary
function returns an object that provides a lot of interesting information when printed out. It also stores that information as part of the object, things like the terms used in the model, the coefficients of the model estimates, and the residuals of the model. This is nice if we want to do something programmatic with the results.
We aren’t restricted to just one explanatory variable in linear regression. We can test the effect of a linear relationship between multiple explanatory variables simultaneously. In the lm
function, we just add extra variable names in the formula separated by +
’s.
wander_fit <- lm(formula = wander_dist ~ weight + age , data = cats)
summary(wander_fit)
Call:
lm(formula = wander_dist ~ weight + age, data = cats)
Residuals:
Min 1Q Median 3Q Max
-0.083498 -0.019287 -0.002461 0.017039 0.068823
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.046857 0.004724 9.918 <2e-16 ***
weight -0.014260 0.001119 -12.747 <2e-16 ***
age 0.017581 0.001661 10.587 <2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.02794 on 397 degrees of freedom
Multiple R-squared: 0.2904, Adjusted R-squared: 0.2868
F-statistic: 81.24 on 2 and 397 DF, p-value: < 2.2e-16
Challenge
Fit a model predicting wander_dist and include weight, age, and fixed as predictors. What is the estimate for the effect of being fixed on the wandering distance?
wander_fit <- lm(formula = wander_dist ~ weight + age + fixed, data = cats)
summary(wander_fit)
Call:
lm(formula = wander_dist ~ weight + age + fixed, data = cats)
Residuals:
Min 1Q Median 3Q Max
-0.091273 -0.018117 -0.001609 0.017690 0.071914
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.054592 0.005074 10.760 < 2e-16 ***
weight -0.014130 0.001101 -12.836 < 2e-16 ***
age 0.017582 0.001633 10.764 < 2e-16 ***
fixed -0.011700 0.003083 -3.796 0.00017 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.02748 on 396 degrees of freedom
Multiple R-squared: 0.3153, Adjusted R-squared: 0.3101
F-statistic: 60.79 on 3 and 396 DF, p-value: < 2.2e-16
If an explanatory variable is not binary (coded as 0s or 1s), we can still include it in the model. The lm
function understands factors to be categorical variables automatically and will output the estimates with a reference category.
wander_fit <- lm(formula = wander_dist ~ weight + age + factor(coat) + sex, data = cats)
summary_fit <- summary(wander_fit)
summary_fit['coefficients']
$coefficients
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.043847799 0.005622582 7.7985172 5.719688e-14
weight -0.014196808 0.001125488 -12.6139100 7.470796e-31
age 0.017760029 0.001675466 10.6000533 2.978738e-23
factor(coat)brown 0.006257836 0.004003785 1.5629800 1.188642e-01
factor(coat)calico -0.001345917 0.005120491 -0.2628492 7.928049e-01
factor(coat)maltese 0.005543267 0.005907650 0.9383201 3.486577e-01
factor(coat)tabby 0.001579503 0.004265994 0.3702545 7.113928e-01
sexmale -0.002616410 0.002870695 -0.9114206 3.626340e-01
Challenge
What command will return the r-squared value from the summary_lm_fit object after running these commands:
wander_fit <- lm(formula = wander_dist ~ weight + age + fixed, data = cats) summary_lm_fit <- summary(wander_fit)
The lm
function also can estimate interactions between explanatory variables. This is useful if we think that the linear relationship between our outcome y and a variable x1 is different depending on the variable x2. This can be accomplished by connecting two variables in the formula with a *
instead of a +
.
Challenge
Fit a linear regression model estimating the relationship between the outcome, wandering distance (
wander_dist
) and explanatory variables age (age
), weight (weight
), with an interaction between age and weight. What is the coefficient associated with the interaction between age and weight?
wander_fit <- lm(formula = wander_dist ~ weight * age, data = cats)
summary(wander_fit)
Call:
lm(formula = wander_dist ~ weight * age, data = cats)
Residuals:
Min 1Q Median 3Q Max
-0.070356 -0.019174 -0.002753 0.019351 0.066527
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.0176044 0.0096232 1.829 0.068094 .
weight -0.0086451 0.0019568 -4.418 1.29e-05 ***
age 0.0241292 0.0024971 9.663 < 2e-16 ***
weight:age -0.0011310 0.0003255 -3.474 0.000569 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.02756 on 396 degrees of freedom
Multiple R-squared: 0.3114, Adjusted R-squared: 0.3062
F-statistic: 59.69 on 3 and 396 DF, p-value: < 2.2e-16
If we are analyzing a binary outcome, we can use logistic regression. Logistic regression uses the linear model framework, but makes different assumptions about the distribution of the outcome. So we can look for associations between binary outcome variables and multiple explanatory variables.
ggplot(cats, aes(x = roamer, y = wander_dist)) +
geom_boxplot(width = 0.5)
For logistic regression, we use the glm
function. It takes formula and data arguments like the lm
function, but we also need to specify a family. For logistic, we pass binomial as the family, which tells the glm
function that we have a binary outcome, and we want to use the logit link function.
roamer_fit <- glm(formula = roamer ~ wander_dist, data = cats, family = binomial )
We can use the summary
function to extract important information from the object that glm
returns, just like with the lm
function
glm_summary <- summary(roamer_fit)
glm_summary
Call:
glm(formula = roamer ~ wander_dist, family = binomial, data = cats)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.9068 -0.3812 -0.2234 -0.0767 3.4678
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -6.0103 0.5634 -10.668 <2e-16 ***
wander_dist 72.8672 7.4867 9.733 <2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 421.53 on 399 degrees of freedom
Residual deviance: 212.89 on 398 degrees of freedom
AIC: 216.89
Number of Fisher Scoring iterations: 6
Challenge
roamer_fit <- glm(formula = roamer ~ wander_dist + weight, data = cats, family = binomial )
summary(roamer_fit)
Call:
glm(formula = roamer ~ wander_dist + weight, family = binomial,
data = cats)
Deviance Residuals:
Min 1Q Median 3Q Max
-2.2674 -0.3752 -0.1984 -0.0728 3.6661
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -4.44909 0.75765 -5.872 4.3e-09 ***
wander_dist 68.99577 7.57569 9.108 < 2e-16 ***
weight -0.25096 0.09484 -2.646 0.00814 **
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 421.53 on 399 degrees of freedom
Residual deviance: 205.44 on 397 degrees of freedom
AIC: 211.44
Number of Fisher Scoring iterations: 6
We can look at the effects of multiple covariates on our binary outcome with logistic regression, just like with linear regression. We just add as many variable names as we’d like to the right side of the formula argument, separated by +
symbols.
ggplot(cats, aes(x = weight, y = wander_dist, color = roamer)) +
geom_point(size = 2) +
geom_smooth(method = 'lm', se = FALSE, size = 3) +
theme_bw(base_size = 18)
# cats$roamer <- relevel(cats$roamer, ref = 'yes')
roamer_fit <- glm(formula = roamer ~ wander_dist + weight, data = cats, family = binomial )
glm_summary <- summary(roamer_fit)
glm_summary
Call:
glm(formula = roamer ~ wander_dist + weight, family = binomial,
data = cats)
Deviance Residuals:
Min 1Q Median 3Q Max
-2.2674 -0.3752 -0.1984 -0.0728 3.6661
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -4.44909 0.75765 -5.872 4.3e-09 ***
wander_dist 68.99577 7.57569 9.108 < 2e-16 ***
weight -0.25096 0.09484 -2.646 0.00814 **
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 421.53 on 399 degrees of freedom
Residual deviance: 205.44 on 397 degrees of freedom
AIC: 211.44
Number of Fisher Scoring iterations: 6
names(glm_summary)
[1] "call" "terms" "family" "deviance"
[5] "aic" "contrasts" "df.residual" "null.deviance"
[9] "df.null" "iter" "deviance.resid" "coefficients"
[13] "aliased" "dispersion" "df" "cov.unscaled"
[17] "cov.scaled"
glm_summary$coefficients
Estimate Std. Error z value Pr(>|z|)
(Intercept) -4.4490855 0.75764856 -5.872229 4.299754e-09
wander_dist 68.9957683 7.57569390 9.107518 8.428814e-20
weight -0.2509641 0.09484151 -2.646141 8.141576e-03
glm_summary$null.deviance
[1] 421.5264
glm_summary$deviance
[1] 205.4406
glm_summary$aic
[1] 211.4406
We can also use the model objects to predict on unobserved values. We just need to pass a data frame with all of the terms used in the original model to the predict
function. The predict
function will return values in a few different ways. The default value of the type
argument is “link” and will return things on the same scale as the linear predictors. This is often not what we want. If we pass “response” to the type
argument, we’ll get predicted values on the same scale as the response. In the logistic case, this is the predicted probability.
new_cats <- data.frame(wander_dist = c(0.15, 0.10, 0.05),
weight = c(2, 6, 12))
new_cats
wander_dist weight
1 0.15 2
2 0.10 6
3 0.05 12
predicted_logit <- predict(object = roamer_fit, newdata = new_cats)
predicted_logit
1 2 3
5.3983516 0.9447069 -4.0108658
predicted_probs <- predict(object = roamer_fit, newdata = new_cats, type = 'response')
predicted_probs
1 2 3
0.99549634 0.72004946 0.01779529
We can then predict whether each cat is a roamer or not based on the predicted probabilty from our model. We need to assign a cut-off probability.
new_cats$predicted_prob <- predicted_probs
new_cats <- new_cats %>% mutate(roamer = ifelse(predicted_prob > 0.5, 'yes', 'no'))
ggplot(cats, aes(x = weight, y = wander_dist, color = roamer, group = roamer)) +
geom_point(size = 2) +
geom_point(data = new_cats, aes(x = weight, y = wander_dist, color = roamer), size = 8) +
theme_bw(base_size = 18)
Another popular machine learning technique is k-means clustering. It seeks to group your data into a fixed number of clusters based on a measure of distance. An important pre-processing step to clustering is the center and scale your data.
cats <- cats %>%
mutate(scale_weight = scale(weight),
scale_wander = scale(wander_dist),
scale_age = scale(age))
We use the kmeans
function to perform the clustering, and need to pass it a data frame, the number of centers we want, and another argument nstart
, which helps get around some numerical issues if the algorithm gets stuck in a local extrema.
cats_cluster <-
kmeans(x = cats %>% select(scale_weight, scale_wander, scale_age),
centers = 3,
nstart = 20)
str(cats_cluster)
List of 9
$ cluster : int [1:400] 1 1 1 1 2 3 1 2 2 1 ...
$ centers : num [1:3, 1:3] -0.621 0.902 -0.654 1.416 -0.455 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : chr [1:3] "1" "2" "3"
.. ..$ : chr [1:3] "scale_weight" "scale_wander" "scale_age"
$ totss : num 1197
$ withinss : num [1:3] 144 187 163
$ tot.withinss: num 494
$ betweenss : num 703
$ size : int [1:3] 99 166 135
$ iter : int 3
$ ifault : int 0
- attr(*, "class")= chr "kmeans"
cats_cluster$centers
scale_weight scale_wander scale_age
1 -0.6208415 1.4159815 -0.1528717
2 0.9019205 -0.4551119 0.7941463
3 -0.6537444 -0.4787674 -0.8643999
We can use ggplot
to visualize the clusters.
cats$cluster <- factor(cats_cluster$cluster)
cluster_centers <- as.data.frame(cats_cluster$centers)
ggplot(data = cats) +
geom_point(aes(x = scale_age, y = scale_weight, color = cluster), size = 4) +
geom_point(data = cluster_centers, aes(x = scale_age, y = scale_weight), color = 'black', size = 8) +
theme_bw(base_size = 18)
ggplot(data = cats) +
geom_point(aes(x = scale_age, y = scale_wander, color = cluster), size = 4) +
geom_point(data = cluster_centers, aes(x = scale_age, y = scale_wander), color = 'black', size = 8) +
theme_bw(base_size = 18)
ggplot(data = cats) +
geom_point(aes(x = scale_weight, y = scale_wander, color = cluster), size = 4) +
geom_point(data = cluster_centers, aes(x = scale_weight, y = scale_wander), color = 'black', size = 8) +
theme_bw(base_size = 18)