Reproducible Research: Peer Assessment 1

1) Loading and preprocessing the data

if (!file.exists("../ProjectData/activity.csv")) {
    download.file(url = "https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip", 
        destfile = "../activity.zip", method = "auto")
    unzip("../activity.zip", exdir = "../ProjectData")
}
amd <- read.csv("../ProjectData/activity.csv", header = T, sep = ",")
str(amd)

## 'data.frame':    17568 obs. of  3 variables:
##  $ steps   : int  NA NA NA NA NA NA NA NA NA NA ...
##  $ date    : Factor w/ 61 levels "2012-10-01","2012-10-02",..: 1 1 1 1 1 1 1 1 1 1 ...
##  $ interval: int  0 5 10 15 20 25 30 35 40 45 ...

Process/transform the date field in Activity monitoring data into a date format suitable for your analysis

amd$date <- as.Date(amd$date)
str(amd)

## 'data.frame':    17568 obs. of  3 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 ...

2) What is mean total number of steps taken per day?

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

amdsteps <- tapply(amd$steps, amd$date, sum)

library(reshape2)
amdmelt <- melt(amdsteps)
names(amdmelt) <- c("Date", "SumofSteps")
head(amdmelt)

##         Date SumofSteps
## 1 2012-10-01         NA
## 2 2012-10-02        126
## 3 2012-10-03      11352
## 4 2012-10-04      12116
## 5 2012-10-05      13294
## 6 2012-10-06      15420

hist(amdmelt$SumofSteps, main = "Histogram of Total Number of Steps per Day", 
    xlab = "Total Number of Steps per Day", ylab = "Frequency", col = "blue", 
    breaks = 30)

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

mean(amdmelt$SumofSteps, na.rm = T)

## [1] 10766

median(amdmelt$SumofSteps, na.rm = T)

## [1] 10765

3) What is the average daily activity pattern?

i) 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)

amdavg <- tapply(amd$steps, amd$interval, mean, na.rm = T)
amdmelt_avg <- melt(amdavg)
names(amdmelt_avg) <- c("interval", "avg")
nrow(amdmelt_avg)

## [1] 288

plot(avg ~ interval, data = amdmelt_avg, type = "l", main = "Average daily activity pattern")

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

amdmelt_avg[amdmelt_avg$avg == max(amdmelt_avg$avg), ]

##     interval   avg
## 104      835 206.2

4) Imputing missing values

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

colSums(is.na(amd))

##    steps     date interval 
##     2304        0        0

ii) 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.

mean(amd$steps, na.rm = T)

## [1] 37.38

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

amdimpute <- amd
amdimpute$steps[is.na(amdimpute$steps)] <- mean(amdimpute$steps, na.rm = T)
colSums(is.na(amdimpute))

##    steps     date interval 
##        0        0        0

iv) 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?

amdimputesteps <- tapply(amdimpute$steps, amdimpute$date, sum)

library(reshape2)
amdmeltimpute <- melt(amdimputesteps)
names(amdmeltimpute) <- c("Date", "SumofSteps")
head(amdmeltimpute)

##         Date SumofSteps
## 1 2012-10-01      10766
## 2 2012-10-02        126
## 3 2012-10-03      11352
## 4 2012-10-04      12116
## 5 2012-10-05      13294
## 6 2012-10-06      15420

hist(amdmeltimpute$SumofSteps, main = "Histogram of Total Number of Steps per Day on  Impute Value", 
    xlab = "Total Number of Steps per Day", ylab = "Frequency", col = "blue", 
    breaks = 30)

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

mean(amdmeltimpute$SumofSteps, na.rm = T)

## [1] 10766

median(amdmeltimpute$SumofSteps, na.rm = T)

## [1] 10766

(Mean = Median) after imputing missing value with mean value of steps. Now it became less skewed.

5) Are there differences in activity patterns between weekdays and weekends?

amdimpute$weekdays <- weekdays(amdimpute$date)

amdimpute$weeks[(amdimpute$weekdays == "Saturday" | amdimpute$weekdays == "Sunday")] <- "weekend"
amdimpute$weeks[!(amdimpute$weekdays == "Saturday" | amdimpute$weekdays == "Sunday")] <- "weekdays"

library(plyr)
week_comp <- ddply(amdimpute, c("interval", "weeks"), function(x) apply(x[1], 
    2, mean))
head(week_comp)

##   interval    weeks steps
## 1        0 weekdays 7.007
## 2        0  weekend 4.673
## 3        5 weekdays 5.384
## 4        5  weekend 4.673
## 5       10 weekdays 5.140
## 6       10  weekend 4.673

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