Peer_graded assingment, Week 2, Reproducible Research
Aranda N. Alfredo
Reading, cleaning and undesrtanding data set files
# Undertanding datasets
activity <- read.csv("activity.csv", sep = ',', header = TRUE)
View(activity)
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 25dim(activity)## [1] 17568 3class(activity)## [1] "data.frame"str(activity)## '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 ...activity$steps <- as.numeric(activity$steps)
activity$interval <- as.numeric(activity$interval)
activityNaRm <- activity[complete.cases(activity), ]What is mean total number of steps taken per day?
meanSteps <-aggregate(x = activityNaRm[c("steps")],
FUN = mean,
by = list(activityNaRm$date))
names(meanSteps) <- c("dates", "steps")1.Calculate the total number of steps taken per day
sumSteps <-aggregate(x = activityNaRm[c("steps")],
FUN = sum,
by = list(activityNaRm$date))
names(sumSteps) <- c("dates", "steps")2.Make a histogram of the total number of steps taken each day
hist(sumSteps$steps,
breaks=53,
main = "Total number of steps taken each day",
xlab = "Steps",
col = "lightgreen",
border = "black"
)
3.Calculate and report the mean and median of the total number of steps taken per day
dim <- length(sumSteps$steps)
totalMean <- sum(sumSteps$steps)/dim
totalMedian <- median(sum(sumSteps$steps)/dim)The total mean of the number of steps taken per day is 1.076618910^{4} and the total median of the number of steps taken per day is 1.076618910^{4}
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)
library(ggplot2)## Warning: package 'ggplot2' was built under R version 3.4.3averages <- aggregate(x=list(steps=activity$steps), by=list(interval=activity$interval),FUN=mean, na.rm=TRUE)
ggplot(data=averages, aes(x=interval, y=steps)) +
geom_line() +
ggtitle("Time Series: average number of steps") +
xlab("5-minute interval") +
ylab("average number of steps taken")
2.Which 5-minute interval, on average across all the days in the dataset, contains the maximum number of steps?
maxSteps <- max(meanSteps$steps)
maxS <- meanSteps[meanSteps$steps == maxSteps,]
maxD <- maxS$datesThe day it constains the maximum number of steps is 2012-11-23
Imputing missing values
1.Calculate and report the total number of missing values in the dataset (i.e. the total number of rows with NAs)
totalNA <- length(activity$steps) - length(activityNaRm$steps)The total numbers of missing values (NA) in the datasets is 2304
2.Devise a strategy for filling in all of the missing values in the dataset. The strategy does not need to be sophisticated. For example, you could use the mean/median for that day, or the mean for that 5-minute interval, etc.
I realized a strategy as follow: If the first or last day have NA then replace it with the global average without NA. Next, if the position i (do not consider first and last positions) has NA then this one takes the averege between i+1 and i-1. Finally, if 1+i or i-1 have NA then replace it with the global average without NA too.
activity2 <- activity
d <- length(activity$steps)
for (i in 1:d) {
if (is.na(activity2[i,1])) {
activity2[i,1] <- averages[averages$interval == activity2[i,3],2]
activity2[i,4] <- "check"
}
}3.Create a new dataset that is equal to the original dataset but with the missing data filled in.
activity2 <- activity24.Make a histogram of the total number of steps taken each day and Calculate and report the mean and median total number of steps taken per day.
# Preparing data for histogram
actsteps <- tapply(activity$steps, activity$date, FUN=sum, na.rm=TRUE)
stepsMean <- mean(actsteps, na.rm=TRUE)
actsteps2 <- tapply(activity2$steps, activity2$date, FUN=sum)
stepsMean2 <- mean(actsteps2)
dif <-abs(stepsMean - stepsMean2)
library(gridExtra)## Warning: package 'gridExtra' was built under R version 3.4.3library(ggplot2)
# Plots
p1<-qplot(actsteps,
binwidth=1000,
ylim=c(0,15),
main="NA REMOVED",
xlab="total number of steps taken each day")
p2<-qplot(actsteps2,
binwidth=1000,
ylim=c(0,15),
main="COMPLETED CASE",
xlab="total number of steps taken each day")
grid.arrange(p1, p2, ncol=2)
Are there differences in activity patterns between weekdays and weekends?
1.Create a new factor variable in the dataset with two levels - “weekday” and “weekend” indicating whether a given date is a weekday or weekend day.
# Taking as factor date
activity2$date <- as.Date(activity2$date)
activity2$WD <- weekdays(activity2$date)
activity2$WDG <- "week"
# Days names in Chile
for (i in 1:length(activity2$steps)) {
if (activity2[i,5] == "sábado" | activity2[i,5] == "domingo") {
activity2[i,6] <- "weekend"
}
}
activity2[,6] <- as.factor(activity2[,6])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)
# Compute means by intervals
activity2w <-subset(activity2,activity2[,6]=="week")
activity2we <-subset(activity2,activity2[,6]=="weekend")
averagesW <- aggregate(steps ~ interval, activity2w, FUN=mean)
averagesWe <- aggregate(steps ~ interval, activity2we, FUN=mean)
# Preparing plots
plot1 <- ggplot(data=averagesW, aes(x=interval, y=steps)) +
geom_line() +
ylim(0, 250) +
ggtitle("Weekdays") +
xlab("5-minute interval") +
ylab("average number of steps taken")
plot2 <- ggplot(data=averagesWe, aes(x=interval, y=steps)) +
geom_line() +
ylim(0, 250) +
ggtitle("Weekend Days") +
xlab("5-minute interval") +
ylab("average number of steps taken")
library(gridExtra)
grid.arrange(plot1, plot2, ncol=2)
The maximum steps is in Weekdays.