Reproducible Research Course Project 1

Loading and preprocessing the data

Load and process the dataset

setwd("~/repos/coursera/data-science-specialization-github-assignments/reproducible-research-course-project-1")
activityDataFile <- "data/activity.csv"
if (!file.exists(activityDataFile)) {
    tempFile <- tempfile()
    download.file("https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip", tempFile)
    unzip(tempFile, exdir = "data")
    unlink(tempFile)
}
activityData <- read.csv(activityDataFile, sep = ",")
activityData$date <- as.POSIXct(activityData$date, format = "%Y-%m-%d", tz = "")

Display dataset summary

names(activityData)

## [1] "steps"    "date"     "interval"

str(activityData)

## 'data.frame':    17568 obs. of  3 variables:
##  $ steps   : int  NA NA NA NA NA NA NA NA NA NA ...
##  $ date    : POSIXct, format: "2012-10-01" "2012-10-01" ...
##  $ interval: int  0 5 10 15 20 25 30 35 40 45 ...

summary(activityData)

##      steps             date                        interval     
##  Min.   :  0.00   Min.   :2012-10-01 00:00:00   Min.   :   0.0  
##  1st Qu.:  0.00   1st Qu.:2012-10-16 00:00:00   1st Qu.: 588.8  
##  Median :  0.00   Median :2012-10-31 00:00:00   Median :1177.5  
##  Mean   : 37.38   Mean   :2012-10-31 00:25:34   Mean   :1177.5  
##  3rd Qu.: 12.00   3rd Qu.:2012-11-15 00:00:00   3rd Qu.:1766.2  
##  Max.   :806.00   Max.   :2012-11-30 00:00:00   Max.   :2355.0  
##  NA's   :2304

head(activityData)

What is mean total number of steps taken per day?

Calculate steps per day (excluding missing values)

stepsPerDay <- with(activityData, aggregate(steps, list(date), FUN = sum, na.rm = TRUE))
colnames(stepsPerDay) <- c("date", "steps")
head(stepsPerDay)

Display histogram

g <- ggplot(stepsPerDay, aes(stepsPerDay$steps))
g + geom_histogram(boundary = 0, binwidth = 2500, col = "wheat4", fill = "wheat2") + ggtitle("Histogram of steps per day") + xlab("Total number of steps") + ylab("Frequency") + theme(plot.title = element_text(face = "bold", size = 12, hjust = 0.5)) + scale_x_continuous(breaks = seq(0, 25000, 2500)) + scale_y_continuous(breaks = seq(0, 18, 2))

Mean and median number of steps taken each day

Mean (excluding missing values)

mean(stepsPerDay$steps, na.rm = TRUE)

## [1] 9354.23

Median (excluding missing values)

median(stepsPerDay$steps, na.rm = TRUE)

## [1] 10395

Note: adding na.rm = TRUE was not necessary since the stepsPerDay data frame already excluded NA values.

What is the average daily activity pattern?

Calculate steps per time interval

stepsPerIntervalAvg <- aggregate(steps ~ interval, data = activityData, FUN = mean, na.action = na.omit)
colnames(stepsPerIntervalAvg) <- c("interval", "steps")

Make a time series plot (i.e. type = “l”) of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all days (y-axis)

g <- ggplot(stepsPerIntervalAvg, aes(stepsPerIntervalAvg$interval, stepsPerIntervalAvg$steps))
g + geom_line(col = "wheat4") + ggtitle("Average steps per 5-minute time interval") + xlab("Interval") + ylab("Average number of steps") + theme(plot.title = element_text(face = "bold", size = 12, hjust = 0.5))

Which 5-minute interval, on average across all the days in the dataset, contains the maximum number of steps?

activityData %>% group_by(interval) %>%
    summarize(meanByInterval = mean(steps, na.rm = TRUE)) %>%
    filter(meanByInterval == max(meanByInterval))

Imputing missing values

There are a number of days/intervals where there are missing values (coded as NA). The presence of missing days may introduce bias into some calculations or summaries of the data.

Calculate and report the total number of missing values in the dataset (i.e. the total number of rows with NAs)

sum(is.na(activityData$steps) == TRUE)

## [1] 2304

Devise a strategy for filling in all of the missing values in the dataset. The strategy does not need to be sophisticated. For example, you could use the mean/median for that day, or the mean for that 5-minute interval, etc.

Strategy: Add a new column to the origin dataset named stepsCompleted that replaces missing values with the rounded average of the 5-minute interval.

activityData$stepsCompleted <- ifelse(is.na(activityData$steps), round(stepsPerIntervalAvg$steps[match(activityData$interval, stepsPerIntervalAvg$interval)], 0), activityData$steps)

Create a new dataset that is equal to the original dataset but with the missing data filled in.

activityDataNoNA <- data.frame(steps = activityData$stepsCompleted, interval = activityData$interval, date = activityData$date)
head(activityDataNoNA, n = 10)

Make a histogram of the total number of steps taken each day and calculate and report the mean and median total number of steps taken per day. Do these values differ from the estimates from the first part of the assignment? What is the impact of imputing missing data on the estimates of the total daily number of steps?

stepsPerDayCompleted <- aggregate(activityDataNoNA$steps, list(activityDataNoNA$date), FUN = sum)
colnames(stepsPerDayCompleted) <- c("date", "steps")
g <- ggplot(stepsPerDayCompleted, aes(stepsPerDayCompleted$steps))
g + geom_histogram(boundary = 0, binwidth = 2500, col = "wheat4", fill = "wheat2") + ggtitle("Histogram of steps per day\n(missing values replaced by mean)") + xlab("Total number of steps") + ylab("Frequency") + theme(plot.title = element_text(face = "bold", size = 12, hjust = 0.5)) + scale_x_continuous(breaks = seq(0, 25000, 2500)) + scale_y_continuous(breaks=seq(0, 26, 2))

Calculate and report the mean and median total number of steps taken per day.

Mean

mean(stepsPerDayCompleted$steps)

## [1] 10765.64

Median

median(stepsPerDayCompleted$steps)

## [1] 10762

1. Do these values differ from the estimates from the first part of the assignment?

   They do differ, but not significantly when looking at the mean and the median of the total daily number of steps.

2. What is the impact of imputing missing data on the estimates of the total daily number of steps?

   Reviewing the histogram, the only two bins that were impacted are the intervals 0 - 2500 and 10000 - 12500 steps; the latter of which grew from a frequency of 18 to a frequency of 26 (44%). Instead of replacing missing values with the mean, such as the mode or median, may have produced different results.

   Statistic	Original Dataset	Imputed Dataset	Difference
   mean	9,354.23	10,765.64	1,411.41 (15%)
   median	10,395	10,762	367 (3%)

Are there differences in activity patterns between weekdays and weekends?

For this part, the weekdays() function may be of some help here. Use the dataset with the filled-in missing values for this part.

In this example, I created the following two factor variables:

• weekDay: indicate the day of the week (non-abbreviated)
• dayType: indicate whether the date is a weekday or a weekend

The current date variable in the dataset has already been formatted to represent a date value using the as.POSIXct() function.

# create a factor variable indicating the day of the week
weekDay <- weekdays(activityDataNoNA$date, abbreviate = FALSE)
activityDataNoNA <- cbind(activityDataNoNA, weekDay)
names(activityDataNoNA)[4] <- "weekDay"

# create a factor variable indicating weekday or weekend
dayType <- ifelse(activityDataNoNA$weekDay == 'Saturday' | activityDataNoNA$weekDay == 'Sunday', 'weekend', 'weekday')
activityDataNoNA <- cbind(activityDataNoNA, dayType)
names(activityDataNoNA)[5] <- "dayType"

# let's see the first 10 observations
head(activityDataNoNA, n = 10)

Prepare data:

• create a data frame stepsPerTime that represents average steps per time interval by weekday/weekend
• add a time variable to display the time interval average in hours

stepsPerTime <- aggregate(steps ~ interval + dayType, data = activityDataNoNA, FUN = mean, na.action = na.omit)
stepsPerTime$time <- stepsPerTime$interval/100
head(stepsPerTime, n = 10)

Make a panel plot containing a time series plot (i.e. type="l") of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all weekday days or weekend days (y-axis).

g <- ggplot(stepsPerTime, aes(time, steps))
g + geom_line(col = "wheat4") + ggtitle("Average steps per time interval\n(weekdays vs. weekends)") + xlab("Time (in hours)") + ylab("Steps") + theme(plot.title = element_text(face = "bold", size = 12, hjust = 0.5)) + facet_grid(dayType ~ .)