by Mauricio G. Melara Camargo

Loading and preprocessing the data

data = read.csv("~/work/Coursera/Reproducible_Research/RepData_PeerAssessment1/activity.csv", 
    na.strings = "NA")

Let's take a quick load on the data format:

str(data)

## '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 ...

The “date”“ field is loaded as a factor, so lets convert it to a Date format for later analysis:

data[2] = as.Date(data[[2]])

What is mean total number of steps taken per day?

library(reshape2)
meltData <- melt(data, id = c("date"), measure.vars = c("steps"))
stepsPerDay <- dcast(meltData, date ~ ., sum)

## removing NA values
stepsPerDay <- stepsPerDay[!is.na(stepsPerDay[2]), ]
head(stepsPerDay)

##         date     .
## 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
## 7 2012-10-07 11015

hist(stepsPerDay[[2]], breaks = 20, col = "red", xlab = "Steps per Day", main = "Steps per Day Distribution")

Total steps per day mean:

mean(stepsPerDay[[2]])

## [1] 10766

Total steps per day median:

median(stepsPerDay[[2]])

## [1] 10765

What is the average daily activity pattern?

meltData <- melt(data, id = c("interval"), measure.vars = c("steps"))
meltData <- meltData[!is.na(meltData[3]), ]
avgStepsPerInterval <- dcast(meltData, interval ~ ., mean)
plot(avgStepsPerInterval[[1]], avgStepsPerInterval[[2]], type = "l", ylab = "Average number of steps", 
    xlab = "Interval", main = "Average number of steps per interval", col = "red")

Let us check which time period has the highest average number of steps:

maxIndex = which.max(avgStepsPerInterval[[2]])
avgStepsPerInterval[maxIndex, 1]

## [1] 835

That is, at 8:35 AM there is the maximum mean number of steps.

Imputing missing values

To calculate the number of input lines with NA values

table(is.na(data[1]))

## 
## FALSE  TRUE 
## 15264  2304

That is, 2304 lines have NA values, which corresponds approximately to 15% of the dataset.

Now, let's replace the NA values by the mean of that 5 minutes period, as calculated on previous session.

library(plyr)
## add the column related to the averages steps per 5 minutes period
dataMerged <- merge(avgStepsPerInterval, data)
## sort by date
dataMerged <- arrange(dataMerged, date)
## add new collumn, based on steps value
dataMerged$filledSteps <- ifelse(is.na(dataMerged$steps), dataMerged$., dataMerged$steps)
# remove adittional colluns, to get the same format as the original table
dataMerged <- dataMerged[c(5, 4, 1)]
head(dataMerged)

##   filledSteps       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

We will now evaluate what was the impact of imputing data into the dataset.

meltData <- melt(dataMerged, id = c("date"), measure.vars = c("filledSteps"))
stepsPerDay_posFill <- dcast(meltData, date ~ ., sum)

hist(stepsPerDay_posFill[[2]], breaks = 20, col = "red", xlab = "Steps per Day", 
    main = "Steps per Day Distribution after imputing missing values")

Total steps per day mean, after inputing missing values:

mean(stepsPerDay_posFill[[2]])

## [1] 10766

Total steps per day median, after inputing missing values:

median(stepsPerDay_posFill[[2]])

## [1] 10766

By comparing these values to the mean and median of steps per day, without inputing missing values, we can see that:

• Mean value remains the same;
• Median value had a slight increase (1 step - from 10765 to 10766) and now it has the same value as the mean.
• The general effect of inputing data using the described strategy can be seen on the histogram. The distribution has a higher peak on the center, which is explained by more entries with the mean value. This will also impact on the variance and standard deviation of this distribution, which will be both reduced.

Are there differences in activity patterns between weekdays and weekends?

## this command will make week days to be written in English language,
## instead of my default Portuguese
Sys.setlocale("LC_TIME", "C")

## [1] "C"

## creates a new factor to distinguish weekends from weekdays
dataMerged$dayLabel <- ifelse(weekdays(dataMerged$date) == "Sunday" | weekdays(dataMerged$date) == 
    "Saturday", "weekend", "weekday")

dataMerged = transform(dataMerged, dayLabel = factor(dayLabel))

meltData <- melt(dataMerged, id = c("dayLabel", "interval"), measure.vars = c("filledSteps"))
dayTypeAvg <- dcast(meltData, dayLabel + interval ~ variable, mean)

library(lattice)
xyplot(filledSteps ~ interval | dayLabel, data = dayTypeAvg, layout = c(1, 2), 
    ylab = "Number of Steps", type = "l", main = "Comparison between Weekend and Weekday Average Steps per Interval")

We can see that there is a clear difference between weekend and weekdays activity pattern.