Reproducible Research: Peer Assessment 1

rcquan 06-07-2014

Loading and preprocessing the data

After setting our working directory, we can begin to load our data into R. If the file activity.zip is not in the current working directory, this function will download and read in the required dataset. We also change the column classes to prepare data for analysis in the following steps.

read_data <- function(){
        fileName <- "activity.zip"
        url <- "https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip"
        if (!file.exists(fileName)){
                download.file(url, fileName, method = "curl")
        }
        unzip(fileName, "activity.csv")
        data <- read.csv("activity.csv", colClasses = c("numeric", "character", "numeric"))
        data$date <- as.Date(data$date)
        return(data)
}
data <- read_data()

What is mean total number of steps taken per day?

Histogram

By using the aggregate function, we can subset by date and apply the sum function to each day to obtain the number of steps taken per day. We first visualize this using barplot to look at the total number of steps taken per day.

barplot_StepsPerDay <- function(data){
        df <- aggregate(steps ~ date, data = data, sum)
        df$date <- as.Date(df$date)
        bplot <- barplot(df$steps, 
             xlab = "Date by Day (2012-10-02 to 2012-11-29)",
             ylab = "Total Steps",
             main = "Total Steps Taken Per Day")
        bplot
        axis(1, at = bplot, labels = FALSE)
}
barplot_StepsPerDay(data)

Mean

Let’s calculate the mean by using the aggregate function once again. Since aggregate returns a data.frame object, we can simply call mean on the steps column.

mean_StepsPerDay <- function(data){
        df <- aggregate(steps ~ date, data = data, sum)
        mean(df$steps)
}
mean_StepsPerDay(data)

## [1] 10766

Median

We do the same to calculate the median number of steps taken each day.

median_StepsPerDay <- function(data){
        df <- aggregate(steps ~ date, data = data, sum)
        median(df$steps)
}
median_StepsPerDay(data)

## [1] 10765

What is the average daily activity pattern?

Time-Series Plot

We can use the aggregate function once again to split the data into subsets, this time into each 5-minute interval, and call the mean function on each to get the average number of steps taken, averaged across all days.

ts_plot <- function(data){
        df <- aggregate(steps ~ interval, data = data, mean)
        plot(df$interval, df$steps, 
                        type = "l", 
                        xlab = "5-Minute Intervals", 
                        ylab = "Average Steps Taken Across All Days",
                        col = "blue")
}
ts_plot(data)

5-Minute Interval with Maximum Number of Steps on Average

Since the aggregate function returns a data.frame, we can return the row with the maximum number of steps averaged across all days by subsetting.

max_interval <- function(){
        df <- aggregate(steps ~ interval, data = data, mean)
        df[which.max(df$steps),]
}
max_interval()

##     interval steps
## 104      835 206.2

Imputing missing values

Total number of missing values in the dataset

For the missing values, we will be imputing the mean number of steps taken for the 5-minute intervals on the assumption that the average daily activity pattern is relatively consistent.

impute_mean <- function(){
        # create data frame with column of indices for NA values
        na_indices <- which(is.na(data$steps))
        df_na <- cbind(data[na_indices,], na_indices)
        
        # using indices of NA values, impute average steps per interval 
        df_mean <- aggregate(steps ~ interval, data = data, mean)
        df_merged <- merge(df_na, df_mean, by.x = "interval", by.y = "interval")
        df_merged <- df_merged[order(df_merged$na_indices),]
        
        imputed_data <- data
        imputed_data$steps[na_indices] <- df_merged$steps.y
        return(imputed_data)
}
imputed_data <- impute_mean()

Histogram

barplot_StepsPerDay(imputed_data)

Mean

mean_StepsPerDay(imputed_data)

## [1] 10766

Median

median_StepsPerDay(imputed_data)

## [1] 10766

Comparing the imputed dataset to the dataset with missing values, the mean total number of steps taken per day stays the same while the median total number of steps taken per day shifts up towards the mean.

Are there differences in activity patterns between weekdays and weekends?

Adding New Variable

To approach this problem, we first add a new variable to the imputed dataset that indicates whether the given date is a weekend or a weekday.

# categorizes days of week into weekdays or weekends
weekend_or_weekday <- function(var){
     if (var %in% c("Sunday", "Saturday")){
                var <- as.factor("Weekend")
        }else{
                var <- as.factor("Weekday")
        }   
}
# adds factor variable WeekEndDay
add_WeekEndDay <- function(df){
        list_DaysOfTheWeek <- weekdays(df$date)
        WeekEndDay <- sapply(list_DaysOfTheWeek, weekend_or_weekday)
        day_of_week_data <- cbind(df, WeekEndDay)
        return(day_of_week_data)
}
day_of_week_data <- add_WeekEndDay(imputed_data)

Side-By-Side Time-Series Plots

With the WeekEndDay variable now indicating whether or not the given date is a weekend or a weekday, we can subset the dataset and call our previously made ts_plot function to create time-series plots for each.

library(lattice)
plot_WeekendVsWeekday <- function(){
        df <- aggregate(steps ~ interval + WeekEndDay, data = day_of_week_data, mean)
        with(df, xyplot(steps~interval|WeekEndDay, 
                        type = "l", 
                        layout = c(1,2)))
}
plot_WeekendVsWeekday()

Comparing the two time-series plots, we can see that the total steps taken during weekends are, on average, greater than that taken during weekday, particularly around midday. However, the weekday still has the highest peak around 800 minutes, perhaps due to the subject’s daily commute to work.