Reproducible Research: Peer Assessment 1

Loading and preprocessing the data

if (!file.exists("activity.csv")) {
  unzip("activity.zip")
}
activity <- read.csv("activity.csv", colClass=c('integer', 'Date', 'integer'))

What is mean total number of steps taken per day?

The total number of steps taken per day

steps.date <- aggregate(steps ~ date, activity, sum)
head(steps.date)

##         date steps
## 1 2012-10-02   126
## 2 2012-10-03 11352
## 3 2012-10-04 12116
## 4 2012-10-05 13294
## 5 2012-10-06 15420
## 6 2012-10-07 11015

Histogram of the total number of steps taken each day

barplot(steps.date$steps, names.arg=steps.date$date, ylim=c(0, 25000), 
        xlab="date", ylab="sum(steps)",)

Mean of total number of steps taken per day

mean(steps.date$steps)

## [1] 10766.19

Median of total number of steps taken per day

median(steps.date$steps)

## [1] 10765

What is the average daily activity pattern?

Time series plot of the 5-minute interval and average number of steps taken averaged across all days

steps.interval <- aggregate(steps ~ interval, activity, mean)
plot(steps.interval, type='l')

The 5-minute interval contains the maximum number of steps

steps.interval$interval[which.max(steps.interval$steps)]

## [1] 835

Imputing missing values

The total number of missing values in the dataset is

sum(is.na(activity$steps))

## [1] 2304

The strategy for filling in all of the missing values in the dataset is to use mean of the day.

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

activity.clean <- merge(activity, steps.date, by="date", suffixes=c("", ".mean"))
nas <- is.na(activity.clean$steps)
activity.clean$steps[nas] <- activity.clean$steps.mean[nas]
activity.clean <- activity.clean[, c(1:3)]
head(activity.clean)

##         date steps interval
## 1 2012-10-02     0     1740
## 2 2012-10-02     0     1715
## 3 2012-10-02     0     1725
## 4 2012-10-02     0     1710
## 5 2012-10-02     0     1735
## 6 2012-10-02     0     1855

Histogram of the total number of steps taken each day

steps.date <- aggregate(steps ~ date, activity.clean, sum)
barplot(steps.date$steps, names.arg=steps.date$date, ylim=c(0, 25000), 
        xlab="date", ylab="sum(steps)",)

Mean of total number of steps taken per day

mean(steps.date$steps)

## [1] 10766.19

Median of total number of steps taken per day

median(steps.date$steps)

## [1] 10765

Data don’t appear to be significant different because imputation uses mean for that particular day but steps are NA for that entire day.

Are there differences in activity patterns between weekdays and weekends?

Add new factor variable dayType with 2 levels – “weekday” and “weekend”

dayType <- function(dates) {
  f <- function(date) {
    if (weekdays(date) %in% c("Saturday", "Sunday")) {
      "weekend"
    }
    else {
      "weekday"
    }
  }
  sapply(dates, f)
}

activity$dayType <- as.factor(dayType(activity$date))
str(activity)

## 'data.frame':    17568 obs. of  4 variables:
##  $ steps   : int  NA NA NA NA NA NA NA NA NA NA ...
##  $ date    : Date, format: "2012-10-01" "2012-10-01" ...
##  $ interval: int  0 5 10 15 20 25 30 35 40 45 ...
##  $ dayType : Factor w/ 2 levels "weekday","weekend": 1 1 1 1 1 1 1 1 1 1 ...

Panel plot containing a time series plot of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all weekdays and weekends

library(lattice)

steps.interval <- aggregate(steps ~ interval + dayType, activity, mean)
xyplot(steps ~ interval | dayType, data=steps.interval, layout=c(2,1), type='l')