Analysis of Activity Pattern

Reproducible Research :Project 1 (By Sougata Biswas)

JOHN HOPHINS DATA SCIENCE SPECIALIZATION

Loading and preprocessing the data

data <- read.csv("activity.csv", header = T)

# create a subset with NA removed
dataNaOmit <- subset(data, is.na(data$steps) == F)

What is mean total number of steps taken per day?

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

calculating the total number of steps taken each day

library(plyr)
totalPerDay <- ddply(dataNaOmit, .(date), summarise, steps=sum(steps))

creating the plot

hist(totalPerDay$steps, breaks = 20, main="Number of Steps", 
     xlab="Total number of steps taken each day", ylab = "Number of Days", col="blue")

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

mean

mean(totalPerDay$steps)

## [1] 10766

median

median(totalPerDay$steps)

## [1] 10765

What is the average daily activity pattern?

1. 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)

calcluating the average number of steps taken in each 5-minite intervals

averagePerInterval <- ddply(dataNaOmit, .(interval), summarise, steps=mean(steps))

creating the plot

plot(averagePerInterval$interval, averagePerInterval$steps,axes = F, type="l", col="blue", xlab="Time", ylab="Average Number of Steps",
     main="Average Daily Activity Pattern")
axis(1,at=c(0,600,1200,1800,2400), label = c("0:00","6:00","12:00","18:00","24:00"))
axis(2)

2. Which 5-minute interval, on average across all the days in the dataset, contains the maximum number

of steps?

averagePerInterval[which.max(averagePerInterval$steps),]

##     interval steps
## 104      835 206.2

The result is 835. So it is the interval from 8:35 to 8:40

Imputing missing values ———————–

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

sum(is.na(data$steps))

## [1] 2304

2.Devise a strategy for filling in all of the missing values in the dataset. The strategy does not
need to be sophisticated.

I will fill the NA with average value for that 5-min interval

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

imputed <- data

for (i in 1:nrow(imputed)){
    if (is.na(imputed$steps[i])){
        imputed$steps[i] <- averagePerInterval$steps[which(imputed$interval[i] == averagePerInterval$interval)]}
}

imputed <- arrange(imputed, interval)

3. Make a histogram

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?

calculating the total number of steps taken each day

totalPerDayImputed <- ddply(imputed, .(date), summarise, steps=sum(steps))

creating the plot

hist(totalPerDayImputed$steps, breaks = 20, main="Number of Steps", xlab="Total number of steps taken each day", ylab = "Number of Days", col="blue")

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

mean(totalPerDayImputed$steps)

## [1] 10766

median(totalPerDayImputed$steps)

## [1] 10766

test does these values differ from thoes in the first part

abs(mean(totalPerDay$steps)-mean(totalPerDayImputed$steps))

## [1] 0

abs(median(totalPerDay$steps)- median(totalPerDayImputed$steps))/median(totalPerDay$steps)

## [1] 0.0001104

so the mean didn’t change after the imputing, the median slightly changed about 0.1% of the original value.

test how total steps taken per day differ

totalDifference <- sum(imputed$steps) - sum(dataNaOmit$steps)
totalDifference

## [1] 86130

Impute the dataset cause the estimation on total steps per day to increase

Are there differences in activity patterns between weekdays and weekends?

1. 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.

Sys.setlocale("LC_TIME", "English") 

## [1] "English_United States.1252"

imputed$weekdays <- weekdays(as.Date(imputed$date))
imputed$weekdays <- ifelse(imputed$weekdays %in% c("Saturday", "Sunday"),"weekend", "weekday")

2.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).

calcluating the average for each interval

average <- ddply(imputed, .(interval, weekdays), summarise, steps=mean(steps))

creating the plot

library(lattice)
xyplot(steps ~ interval | weekdays, data = average, layout = c(1, 2), type="l", xlab = "Interval", ylab = "Number of steps")