PA1_template.md

Reproducible Research: Peer Assessment 1

Loading and preprocessing the data

activityZip <- 'activity.zip';
activityFilename <- 'activity.csv';
if (!file.exists(activityFilename))
        unzip(activityZip);

activity <- read.csv(activityFilename, header = TRUE, na.strings = "NA");

summary(activity)

##      steps                date          interval     
##  Min.   :  0.00   2012-10-01:  288   Min.   :   0.0  
##  1st Qu.:  0.00   2012-10-02:  288   1st Qu.: 588.8  
##  Median :  0.00   2012-10-03:  288   Median :1177.5  
##  Mean   : 37.38   2012-10-04:  288   Mean   :1177.5  
##  3rd Qu.: 12.00   2012-10-05:  288   3rd Qu.:1766.2  
##  Max.   :806.00   2012-10-06:  288   Max.   :2355.0  
##  NA's   :2304     (Other)   :15840

What is mean total number of steps taken per day?

Aggregate activity by date - sum steps for each day

aggregatedByDate <- aggregate(. ~ date, data=activity, FUN = sum);

Histogram of total steps per day

hist(aggregatedByDate$steps, breaks = 30, xlab = "Total steps per day", main = "Histogramm of total steps per day", col=3);

Calculate mean of total steps per day

meanTotalSteps <- mean(aggregatedByDate$steps);
meanTotalSteps

## [1] 10766.19

Calculate median of total steps per day

medianTotalSteps <- median(aggregatedByDate$steps)
medianTotalSteps

## [1] 10765

What is the average daily activity pattern?

Aggregate activity by intervals - average steps for each interval

aggregatedByInterval <- aggregate(steps ~ interval, data=activity, FUN = mean);

Plot of average steps per interval

plot(x = aggregatedByInterval$interval, y = aggregatedByInterval$steps, type = "l", xlab = "Interval", ylab="Average steps" )

Which interval has the maximum number of steps averaged by day

aggregatedByInterval[which.max(aggregatedByInterval$steps), ];

##     interval    steps
## 104      835 206.1698

Imputing missing values

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

sum(!complete.cases(activity))

## [1] 2304

Fill out missing values with average value for interval

activityNoNA <- activity;
for (i in 1:nrow(activityNoNA)) {
    if (is.na(activityNoNA$steps[i])) {
        activityNoNA$steps[i] <- aggregatedByInterval[activityNoNA$interval[i] == aggregatedByInterval$interval, ]$steps;
    }
}

summary(activityNoNA)

##      steps                date          interval     
##  Min.   :  0.00   2012-10-01:  288   Min.   :   0.0  
##  1st Qu.:  0.00   2012-10-02:  288   1st Qu.: 588.8  
##  Median :  0.00   2012-10-03:  288   Median :1177.5  
##  Mean   : 37.38   2012-10-04:  288   Mean   :1177.5  
##  3rd Qu.: 27.00   2012-10-05:  288   3rd Qu.:1766.2  
##  Max.   :806.00   2012-10-06:  288   Max.   :2355.0  
##                   (Other)   :15840

Histogram of the total number of steps taken each day

aggregatedByDateImputed <- aggregate(. ~ date, data=activityNoNA, FUN = sum);
hist(aggregatedByDateImputed$steps, breaks = 30, xlab = "Total steps per day", main = "Histogramm of total steps per day", col=3);

Imputing effect on estimates

Mean imputed total steps:

meanImputedTotalSteps <- mean(aggregatedByDateImputed$steps)
meanImputedTotalSteps

## [1] 10766.19

Median imputed total steps:

medianImputedTotalSteps <- median(aggregatedByDateImputed$steps)
medianImputedTotalSteps

## [1] 10766.19

Difference of mean value:

meanabsdiff = abs((meanTotalSteps - meanImputedTotalSteps)/meanTotalSteps*100);
meanTotalSteps - meanImputedTotalSteps

## [1] 0

mean differs in percentage 0 % - so mean is not changed

Difference of median value:

medianabsdiff = abs((medianTotalSteps - medianImputedTotalSteps)/medianTotalSteps*100);
medianTotalSteps - medianImputedTotalSteps

## [1] -1.188679

median differs in percentage 0.0110421 %

Are there differences in activity patterns between weekdays and weekends?

Introduce new variable dayType weekend or weekday

activityNoNA['dayType']  <- weekdays(as.Date(activityNoNA$date));
activityNoNA[activityNoNA$dayType %in% c('Saturday','Sunday'), ]$dayType <- "weekend";
activityNoNA[activityNoNA$dayType != "weekend", ]$dayType <- "weekday";
activityNoNA$dayType <- as.factor(activityNoNA$dayType);
summary(activityNoNA$dayType)

## weekday weekend 
##   12960    4608

Average steps in interval by day type across all days and make a plots of steps by day type

aggregatedByDateTypeImputed <- aggregate(steps ~ interval + dayType, activityNoNA, mean);

library(ggplot2)

qplot(interval, steps, data = aggregatedByDateTypeImputed, geom=c("line"), xlab = "Interval", ylab = "Number of steps" ) + facet_wrap(~ dayType, ncol = 1)