Reproducible Research: Peer Assessment 1

Loading and preprocessing the data

Show any code that is needed to

• Load the data (i.e. read.csv())
• Process/transform the data (if necessary) into a format suitable for your analysis

activity <- read.csv("activity.csv", header=TRUE)
names(activity)

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

head(activity)

##   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       25

#remove all NA's
activity_no_na <- activity[!is.na(activity$steps),]
head(activity_no_na)

##     steps       date interval
## 289     0 2012-10-02        0
## 290     0 2012-10-02        5
## 291     0 2012-10-02       10
## 292     0 2012-10-02       15
## 293     0 2012-10-02       20
## 294     0 2012-10-02       25

What is mean total number of steps taken per day?

For this part of the assignment, you can ignore the missing values in the dataset.

• Make a histogram of the total number of steps taken each day
• Calculate and report the mean and median total number of steps taken per day

number_of_steps <- tapply(activity_no_na$steps, activity_no_na$date, sum)
number_of_steps <- number_of_steps[!is.na(number_of_steps)]
hist(number_of_steps, col=rainbow(10))

mean(number_of_steps)

## [1] 10766

median(number_of_steps)

## [1] 10765

What is the average daily activity pattern?

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

mean_steps_interval <- tapply(activity_no_na$steps, activity_no_na$interval, mean)
plot(mean_steps_interval, type="l", xlab="5-minute interval indexes", ylab="average steps taken")

max(mean_steps_interval)

## [1] 206.2

which.max(mean_steps_interval)

## 835 
## 104

As can be seen, the time interval 830-835 (which is the 104 th time interval) corresponds to the maximum number of steps.

Imputing missing values

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

Strategy for imputing missing data

The missing time interval value will be filled by the mean for that 5-minute interval in the entire dataset.

nrow(activity[is.na(activity),])

## [1] 2304

nrow(activity[is.na(activity$steps),])

## [1] 2304

intervals <- unique(activity$interval)
new_activity <- activity
for (int in intervals) {
  num_missing <- nrow(new_activity[new_activity$interval == int & is.na(new_activity$steps),])
  if (num_missing > 0) {
    new_activity[new_activity$interval == int & is.na(new_activity$steps),]$steps <- rep(mean_steps_interval[as.character(int)], num_missing)
  }
}
head(new_activity)

##     steps       date interval
## 1 1.71698 2012-10-01        0
## 2 0.33962 2012-10-01        5
## 3 0.13208 2012-10-01       10
## 4 0.15094 2012-10-01       15
## 5 0.07547 2012-10-01       20
## 6 2.09434 2012-10-01       25

new_number_of_steps <- tapply(new_activity$steps, new_activity$date, sum)
hist(new_number_of_steps, col=rainbow(10))

mean(new_number_of_steps)

## [1] 10766

median(new_number_of_steps)

## [1] 10766

sum(number_of_steps)

## [1] 570608

sum(new_number_of_steps)

## [1] 656738

As can be seen the median got changed after imputation.

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.

• Create a new factor variable in the dataset with two levels - “weekday” and “weekend” indicating whether a given date is a weekday or weekend day.
• 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). The plot should look something like the following, which was creating using simulated data:

new_activity$weekday_or_weekend <- ifelse(weekdays(as.Date(new_activity$date)) %in% c("Saturday", "Sunday"), "weekend", "weekday")  
new_activity_weekday <- new_activity[new_activity$weekday_or_weekend == "weekday",]
mean_steps_weekday <- tapply(new_activity_weekday$steps, new_activity_weekday$interval, mean)
new_activity_weekend <- new_activity[new_activity$weekday_or_weekend == "weekend",]
mean_steps_weekend <- tapply(new_activity_weekend$steps, new_activity_weekend$interval, mean)
mean_steps_weekday <- cbind(intervals=rownames(mean_steps_weekday), mean_steps=mean_steps_weekday, type="weekday")
mean_steps_weekend <- cbind(intervals=rownames(mean_steps_weekend), mean_steps=mean_steps_weekend, type="weekend")
d <- as.data.frame(rbind(mean_steps_weekday, mean_steps_weekend))
d$mean_steps <- as.numeric(as.character(d$mean_steps))
d$intervals <- as.integer(as.character(d$intervals))
library (lattice)
xyplot (mean_steps~intervals|type, data=d, type="l",layout=c(1, 2), as.table=T, xlab="5-minute interval indexes")