Reproducible Research: Peer Assessment 1
Loading libraries
library(ggplot2)
library(gridExtra)
library(plyr)
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:plyr':
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## arrange, count, desc, failwith, id, mutate, rename, summarise,
## summarize## The following object is masked from 'package:gridExtra':
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## combine## The following objects are masked from 'package:stats':
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## filter, lag## The following objects are masked from 'package:base':
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## intersect, setdiff, setequal, unionLoading and preprocessing the data
1. Set up directory for downloading data:
dir.create("./data", showWarnings = FALSE)
downloadURL <- "https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip"
downloadedFile <- "./data/activity.zip"2. Download and unzip data:
if(!file.exists(downloadedFile)) {
download.file(downloadURL, downloadedFile, method = "curl")
unzip(downloadedFile, exdir = "./data")
}3. Load data to environment:
df <- read.csv(file = "./data/activity.csv", header = TRUE)4. Convert date feature to DateTime format:
df$date <- as.Date(df$date, "%Y-%m-%d")5. Take a quick look of our data:
head(df)## steps date interval
## 1 NA 2012-10-01 0
## 2 NA 2012-10-01 5
## 3 NA 2012-10-01 10
## 4 NA 2012-10-01 15
## 5 NA 2012-10-01 20
## 6 NA 2012-10-01 25What is mean total number of steps taken per day?
1. Calculate the total number of steps taken per day:
total.steps <- tapply(df$step, df$date, sum)2. Histogram of the total number of steps taken per day:
histogram.steps <- qplot(total.steps,
geom = "histogram",
bins = 10,
main = "Total steps taken per day",
xlab = "Steps")
histogram.steps## Warning: Removed 8 rows containing non-finite values (stat_bin).
3. The mean and median of the total number of steps taken per day:
Mean total number of steps taken per day:
mean.steps <- tapply(df$steps, df$date, mean)
barplot(mean.steps, main = "Mean total number of steps taken per day")
→ The minimum value for the mean of the total number steps is very low with 0.14. Otherwise, the maximum value is 73.59 which is very high value for this figure.
Median total number of steps taken per day:
median.steps <- tapply(df$steps, df$date, median)
boxplot(median.steps, main = "Mean total number of steps taken per day")
→ The median values for all of the dates are the same with 0. It means, for each date, there are more than 50% of the time that the enthusiasts did not take any steps.
What is the average daily activity pattern?
Calculate average number of steps taken per day:
avarage.steps <- df %>%
group_by(date) %>%
summarise(mean(steps)) %>%
rename("steps" = "mean(steps)")Time series plot:
ggplot(data = avarage.steps, aes(x = steps, y = date)) +
geom_line() +
xlab("avarage steps") +
ggtitle("Time series plot") +
annotate("segment",
x = avarage.steps[which(avarage.steps$steps == max(avarage.steps$steps, na.rm = TRUE)), "steps"][[1]] + 1,
xend = avarage.steps[which(avarage.steps$steps == max(avarage.steps$steps, na.rm = TRUE)), "steps"][[1]] + 1,
y = avarage.steps[which(avarage.steps$steps == max(avarage.steps$steps, na.rm = TRUE)), "date"][[1]],
yend = avarage.steps[which(avarage.steps$steps == max(avarage.steps$steps, na.rm = TRUE)), "date"][[1]],
arrow = arrow(),
color = "blue")## Warning: Removed 8 row(s) containing missing values (geom_path).
→ From the plot, the day Dec 23 has the highest average number of steps within 5 minutes.
Imputing missing values
1. Calculate the total of missing values:
summary(df)## steps date interval
## Min. : 0.00 Min. :2012-10-01 Min. : 0.0
## 1st Qu.: 0.00 1st Qu.:2012-10-16 1st Qu.: 588.8
## Median : 0.00 Median :2012-10-31 Median :1177.5
## Mean : 37.38 Mean :2012-10-31 Mean :1177.5
## 3rd Qu.: 12.00 3rd Qu.:2012-11-15 3rd Qu.:1766.2
## Max. :806.00 Max. :2012-11-30 Max. :2355.0
## NA's :2304→ In the data set, only the column of steps contains missing values with 2304 number. In other words, there are more than 13% NAs in this column.
2. Strategy for filling the gaps:
Beside the simple methods like filling the gaps with the mean or median value of steps column, I will be replacing missing values in the column of steps with the Median per group of interval.
3. Conduct the strategy:
Steps to fill the gap:
+ Using the ddply() function;
+ Specify the data frame that containg the missing values. In this case, the data frame is df;
+ Specify the column that define the groups. In this case, the column is interval;
+ Use the transform function;
+ Specify the column that contains the missing values. In this case, the column is steps;
+ Use the ifelse function to identify and replace the NAs with the median;
+ Finish the ddply() function.new.df <- ddply(df,
~ interval,
transform,
steps = ifelse(is.na(steps),
median(steps, na.rm = TRUE),
steps))Take a look at the tidy data set:
head(new.df)## steps date interval
## 1 0 2012-10-01 0
## 2 0 2012-10-02 0
## 3 0 2012-10-03 0
## 4 47 2012-10-04 0
## 5 0 2012-10-05 0
## 6 0 2012-10-06 0Summary after filling the NAs:
summary(new.df)## steps date interval
## Min. : 0 Min. :2012-10-01 Min. : 0.0
## 1st Qu.: 0 1st Qu.:2012-10-16 1st Qu.: 588.8
## Median : 0 Median :2012-10-31 Median :1177.5
## Mean : 33 Mean :2012-10-31 Mean :1177.5
## 3rd Qu.: 8 3rd Qu.:2012-11-15 3rd Qu.:1766.2
## Max. :806 Max. :2012-11-30 Max. :2355.0→ There is no missing values at all.
4. Re-make histogram and re-calculate the mean vs median total number of steps taken per day:
Histogram based on the total number of steps each day:
Calculate the new histogram based on the new data sets:
new.total.steps <- tapply(new.df$step, new.df$date, sum)
new.histogram.steps <- qplot(new.total.steps,
geom = "histogram",
bins = 10,
main = "New total steps taken per day",
xlab = "Steps")Plot the old and new version of histogram:
grid.arrange(histogram.steps, new.histogram.steps, nrow = 2)## Warning: Removed 8 rows containing non-finite values (stat_bin).
→ The two histograms look almost the same except for the steps with value 0. The steps 0 go up from 2.5 to 10.
Mean total number of steps taken per day:
Calculate the new mean:
new.mean.steps <- tapply(new.df$steps, new.df$date, mean)Plot the barplots based on the old and new data sets:
par(mfrow = c(2, 1))
barplot(mean.steps, main = "Mean total number of steps taken per day")
barplot(new.mean.steps, main = "New mean total number of steps taken per day")
→ Overall, the mean total number of steps before and after filling the NAs do not change much.
Median total number of steps taken per day:
Calculate the new median:
new.median.steps <- tapply(new.df$steps, new.df$date, median)Plot the boxplots based on the old and new data sets:
par(mfrow = c(2, 1))
boxplot(median.steps, main = "Median total number of steps taken per day")
boxplot(new.median.steps, main = "New median total number of steps taken per day")
→ The two plots are identical.
Are there differences in activity patterns between weekdays and weekends?
1. Create a new factor indicating whether a given date is a weekday:
Create a function indicating a given date is a weekdays:
check.weekday <- function(x) !(weekdays(x, abbreviate = TRUE) %in% c("Sat", "Sun"))Create weekday feature:
new.df$weekday <- sapply(new.df$date, check.weekday)Summary on the new weekday column:
summary(new.df$weekday)## Mode FALSE TRUE
## logical 4608 129602. Make the panel plot:
p <- ggplot(new.df, aes(x = interval, y = steps, fill = weekday))
p <- p + geom_line() +
facet_grid(weekday ~ .,
labeller = labeller(weekday = c("TRUE" = "Weekday", "FALSE" = "Weekend"))) +
xlab("Interval") +
ylab("Number of steps") +
ggtitle("Time series plot")
p