Homework 4
Moiya Josephs
2024-02-25
library(mlbench)
library(fpp3)## ── Attaching packages ────────────────────────────────────────────── fpp3 0.5 ──## ✔ tibble 3.2.1 ✔ tsibble 1.1.4
## ✔ dplyr 1.1.2 ✔ tsibbledata 0.4.1
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## ✔ lubridate 1.9.2 ✔ fable 0.3.3
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## ✖ tsibble::union() masks base::union()library(e1071)##
## Attaching package: 'e1071'## The following object is masked from 'package:fabletools':
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## interpolatelibrary(caret)## Loading required package: lattice##
## Attaching package: 'caret'## The following objects are masked from 'package:fabletools':
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## MAE, RMSEAsk
Do problems 3.1 and 3.2 in the Kuhn and Johnson book Applied Predictive Modeling. Please submit your Rpubs link along with your .pdf for your run code.
3.1
data(Glass)head(Glass )## RI Na Mg Al Si K Ca Ba Fe Type
## 1 1.52101 13.64 4.49 1.10 71.78 0.06 8.75 0 0.00 1
## 2 1.51761 13.89 3.60 1.36 72.73 0.48 7.83 0 0.00 1
## 3 1.51618 13.53 3.55 1.54 72.99 0.39 7.78 0 0.00 1
## 4 1.51766 13.21 3.69 1.29 72.61 0.57 8.22 0 0.00 1
## 5 1.51742 13.27 3.62 1.24 73.08 0.55 8.07 0 0.00 1
## 6 1.51596 12.79 3.61 1.62 72.97 0.64 8.07 0 0.26 1A
Glass %>%
dplyr::select(where(is.numeric)) %>%
gather() %>%
ggplot(aes(value)) +
geom_histogram(bins = 15) +
facet_wrap(~key, scales = 'free')
The variables Al, Ca, K,
Na, and RI are right skewed variables.
Fe has a large number of 0s.
B
Fe, K appears right skewed due to the large
number of 0 values. While Mg appears left skewed, and also has outliers
at 0.
The skewness function can also help determine the skew
of the values.
Glass %>%
dplyr::select(where(is.numeric)) %>%
apply(., 2, skewness)## RI Na Mg Al Si K Ca
## 1.6027151 0.4478343 -1.1364523 0.8946104 -0.7202392 6.4600889 2.0184463
## Ba Fe
## 3.3686800 1.7298107B
We are able to observe the skew in a data using the
skewness function. There seems to be outliers in Ba, K, RI,
Ca, Fe, and possibly Na.
Glass %>%
dplyr::select(where(is.numeric)) %>%
apply(., 2, skewness)## RI Na Mg Al Si K Ca
## 1.6027151 0.4478343 -1.1364523 0.8946104 -0.7202392 6.4600889 2.0184463
## Ba Fe
## 3.3686800 1.7298107C
The categorical variables that are not centered would benefit from a
series of transformation. With the use of caret function
PreProcess we can apply multiple transformations.
caret::preProcess(Glass, method = c("BoxCox", "center", "scale", "pca"))## Created from 214 samples and 10 variables
##
## Pre-processing:
## - Box-Cox transformation (5)
## - centered (9)
## - ignored (1)
## - principal component signal extraction (9)
## - scaled (9)
##
## Lambda estimates for Box-Cox transformation:
## -2, -0.1, 0.5, 2, -1.1
## PCA needed 7 components to capture 95 percent of the varianceThe lambda estimates show the optimal lambda values for the transformation.
3.2
A
data(Soybean)
columns <- colnames(Soybean)
lapply(columns,
function(col) {
ggplot(Soybean,
aes_string(col)) + geom_bar() + coord_flip() + ggtitle(col)})## Warning: `aes_string()` was deprecated in ggplot2 3.0.0.
## ℹ Please use tidy evaluation idioms with `aes()`.
## ℹ See also `vignette("ggplot2-in-packages")` for more information.
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_lifecycle_warnings()` to see where this warning was
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B
Values like lodging, germ and
mold are more likely to have N/A because there is a chance
none of these things were observed on the plants.
colSums(is.na(Soybean))## Class date plant.stand precip temp
## 0 1 36 38 30
## hail crop.hist area.dam sever seed.tmt
## 121 16 1 121 121
## germ plant.growth leaves leaf.halo leaf.marg
## 112 16 0 84 84
## leaf.size leaf.shread leaf.malf leaf.mild stem
## 84 100 84 108 16
## lodging stem.cankers canker.lesion fruiting.bodies ext.decay
## 121 38 38 106 38
## mycelium int.discolor sclerotia fruit.pods fruit.spots
## 38 38 38 84 106
## seed mold.growth seed.discolor seed.size shriveling
## 92 92 106 92 106
## roots
## 31C
Depending on the shape of the variable different inferencing techniques should be used. If a missing value makes sense for that variable, then using 0 would be optimal. If the N/A does not make sense we can either use the mean or the median. For a centered variable, the mean would be a good value to use. For a skewed data, the median is optimal. Additionally, removing variables with a high number of NAs from the predictor list, may be useful to optimize the model.