PA1.md

Reproducible Research: Peer Assessment 1

Loading and preprocessing the data

unzip('activity.zip')
activityData <- read.csv(file="activity.csv")

What is mean total number of steps taken per day?

Make a histogram of the total number of steps taken each day

stepsData<- aggregate(steps ~ date, data = activityData, FUN=sum)
barplot(stepsData$steps, names.arg=stepsData$date, ylab='No. of Steps',las=2)

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

The mean number of steps taken per days is:

mean(stepsData$steps)

## [1] 10766.19

The median number of steps taken per days is:

median(stepsData$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)

stepsData.interval <- aggregate(steps ~ interval, data = activityData, FUN=mean)
plot(stepsData.interval,type="l")

Which 5-minute interval, on average across all the days in the dataset,

contains the maximum number of steps?

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

## [1] 835

Imputing missing values

Calculate and report the total number of missing values in the dataset

sum(is.na(avtivityData))

Devise a strategy for filling in all of the missing values in the dataset. The

strategy does not need to be sophisticated.

Mean for 5-minute interval will be used as fillers

imputedActivityData <- activityData
imputedActivityData <- merge(activityData, stepsData.interval, by="interval", suffixes=c("",".temp"))
imputedActivityData$steps[is.na(activityData$steps)] <- imputedActivityData$steps.temp[is.na(activityData$steps)]

Create a new dataset that is equal to the original dataset but with the

missing data filled in.

imputedActivityData <- imputedActivityData[,1:3]

Make a histogram of the total number of steps taken each day

stepsData.date <- aggregate(steps ~ date, data = imputedActivityData, FUN = sum)
barplot(stepsData.date$steps, names.arg=stepsData.date$date, ylab='No. of Steps',las=2)

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

The mean number of steps taken per days is:

mean(stepsData.date$steps)

## [1] 9563.93

The median number of steps taken per days is:

median(stepsData.date$steps)

## [1] 11215.68

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?

Yes, there is a difference. The number of steps is not biased anymore by missing values. The mean number of steps has decreased from before, whereas the median has increased.

Are there differences in activity patterns between weekdays and weekends?

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.

typeOfDay <- function(date) {
    if (weekdays(as.Date(date)) %in% c("Monday", "Tuesday","Wednesday","Thursday","Friday"))
        return("Weekday")
    else
        return("Weekend")
}
imputedActivityData$typeOfDay <- as.factor(sapply(imputedActivityData$date, typeOfDay))

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).

stepsData.Weekday <- aggregate(steps ~ interval, data = imputedActivityData, 
    subset = imputedActivityData$typeOfDay == "Weekday", FUN = mean)

stepsData.Weekend <- aggregate(steps ~ interval, data = imputedActivityData, 
    subset = imputedActivityData$typeOfDay == "Weekend", FUN = mean)

par(mfrow=c(2,1))

plot(stepsData.Weekday, type = "l", main = "Weekday")
plot(stepsData.Weekend, type = "l", main = "Weekend")