# Reproducible Research - Peer Assessment 1
by Gregorio Ambrosio Cestero
Loading and preprocessing the data
- Load the data
activity <- read.csv(unz("activity.zip", "activity.csv"))
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- Process/transform the data into a format suitable for the analysis
No process needed
head (activity)## steps date interval
## 1 NA 2012-10-01 0
## 2 NA 2012-10-01 5
## 3 NA 2012-10-01 10
## 4 NA 2012-10-01 15
## 5 NA 2012-10-01 20
## 6 NA 2012-10-01 25What is mean total number of steps taken per day?
- Calculate the total number of steps taken per day
stepsPerDay <- aggregate(steps ~ date,data = activity, sum, na.rm = TRUE)- Make a histogram of the total number of steps taken each day
hist(stepsPerDay$steps, main = "Histogram of total steps per day", xlab = "Steps", col = "royalblue2")
3 .Calculate and report the mean and median of the total number of steps taken per day
meanOfStepsPerDay <- mean(stepsPerDay$steps)
cat("Mean steps per day = ",meanOfStepsPerDay)## Mean steps per day = 10766.19medianOfStepsPerDay <- median(stepsPerDay$steps)
cat("Median steps per day = ",medianOfStepsPerDay)## Median steps per day = 10765summary(stepsPerDay)## date steps
## 2012-10-02: 1 Min. : 41
## 2012-10-03: 1 1st Qu.: 8841
## 2012-10-04: 1 Median :10765
## 2012-10-05: 1 Mean :10766
## 2012-10-06: 1 3rd Qu.:13294
## 2012-10-07: 1 Max. :21194
## (Other) :47The steps per day mean is 10766.19 steps.
The steps per day median is 10765.00 steps.
What is the average daily activity pattern?
- Make a time series plot (i.e. type = “l”) of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all days (y-axis)
stepsPerInterval <- aggregate(steps ~ interval, data = activity, mean, na.rm = TRUE)
library(lattice)
xyplot(stepsPerInterval$steps ~ stepsPerInterval$interval,
type = 'l',
main = "Average number of steps over across days",
xlab = "5-minute interval",
ylab = "Average number of steps")
- Which 5-minute interval, on average across all the days in the dataset, contains the maximum number of steps?
maxNumOfStepsInterval <- stepsPerInterval$interval[which.max(stepsPerInterval$steps)]
print(maxNumOfStepsInterval)## [1] 835The interval 835 contains the maximun number of steps
Imputing missing values
- Calculate and report the total number of missing values in the dataset (i.e. the total number of rows with NAs)
rowsWithNAs <- sum(!complete.cases(activity))
print(rowsWithNAs)## [1] 2304Total number of missing values in the dataset: 2304
- Devise a strategy for filling in all of the missing values in the dataset.
All NA’s are replaced with the mean of 5-minute interval
- Create a new dataset that is equal to the original dataset but with the missing data filled in.
activityFilled <- merge(activity, stepsPerInterval, by="interval", suffixes=c("",".mean"))
nas <- is.na(activityFilled$steps)
activityFilled$steps[nas] <- activityFilled$steps.mean[nas]
activityFilled <- activityFilled[,c(2,3,1)]
activityFilled <- activityFilled[with(activityFilled, order(date)),]- 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?
stepsPerDayFilled <- aggregate(steps ~ date,data = activityFilled, sum, na.rm = TRUE)
hist(stepsPerDayFilled$steps, main = "Histogram of total steps per day without NA's", xlab="Steps", col="royalblue2")
meanOfStepsPerDayFilled <- mean(stepsPerDayFilled$steps)
cat("Mean steps per day = ",meanOfStepsPerDayFilled)## Mean steps per day = 10766.19medianOfStepsPerDayFilled <- median(stepsPerDayFilled$steps)
cat("Median steps per day = ",medianOfStepsPerDayFilled)## Median steps per day = 10766.19summary(stepsPerDayFilled)## date steps
## 2012-10-01: 1 Min. : 41
## 2012-10-02: 1 1st Qu.: 9819
## 2012-10-03: 1 Median :10766
## 2012-10-04: 1 Mean :10766
## 2012-10-05: 1 3rd Qu.:12811
## 2012-10-06: 1 Max. :21194
## (Other) :55The steps per day mean is 10766.19 steps.
The steps per day median is 10766.19 steps.
The mean value is the same as the previous value because of the strategy based on the mean for 5-minute intervals. The median is very slightly different.
Are there differences in activity patterns between weekdays and weekends?
- Create a new factor variable in the dataset with two levels - “weekday” and “weekend” indicating whether a given date is a weekday or weekend day.
Sys.setlocale("LC_TIME", "English")## [1] "English_United States.1252"daytype <- function(date) {
ifelse(weekdays(as.Date(date)) %in% c("Saturday","Sunday"),"weekend", "weekday")
}
activityFilled$daytype <- as.factor(sapply(activityFilled$date,daytype))
head (activityFilled)## steps date interval daytype
## 1 1.7169811 2012-10-01 0 weekday
## 63 0.3396226 2012-10-01 5 weekday
## 128 0.1320755 2012-10-01 10 weekday
## 205 0.1509434 2012-10-01 15 weekday
## 264 0.0754717 2012-10-01 20 weekday
## 327 2.0943396 2012-10-01 25 weekday- 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).
stepsPerIntervalPerType <- aggregate(steps~interval+daytype,activityFilled,mean)
library(lattice)
xyplot(steps ~ interval | factor(daytype),
data = stepsPerIntervalPerType,
aspect = 1/2,
type = "l",
main = "Average number of steps over across all weekday days or weekend days",
ylab = "Average number of steps",
xlab = "5-minute interval")