Load data file.
D <- read.csv("activity.csv", colClasses = c("integer", "Date", "integer"))
str(D)## '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 ...
head(D)## 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 25
A histogram of the total number of steps taken each day.
D.step <- tapply(D$steps, D$date, sum, na.rm = TRUE)
hist(D.step, xlab = "Total number of steps taken each days", ylab = "Number of days",
breaks = 10)
D.sum <- summary(D.step)
str(D.sum)## Classes 'summaryDefault', 'table' Named num [1:6] 0 6780 10400 9350 12800 21200
## ..- attr(*, "names")= chr [1:6] "Min." "1st Qu." "Median" "Mean" ...
The mean and median total number of steps taken per day are respectively 9350 and 10400.
Time series plot of the average steps taken in each 5-minute interval across all the days.
D.interval <- tapply(D$steps, D$interval, mean, na.rm = TRUE)
str(D.interval)## num [1:288(1d)] 1.717 0.3396 0.1321 0.1509 0.0755 ...
## - attr(*, "dimnames")=List of 1
## ..$ : chr [1:288] "0" "5" "10" "15" ...
plot(D.interval, type = "l", xlab = "5-minute interval", ylab = "average of steps taken")
Which 5-minute interval on avarage contains the maximum?
D.asc <- D.interval[order(-D.interval)]
D.maxintvl <- D.asc[1]The 5-minute interval of 835 contains 206.1698 as the maximum value.
D.na <- sum(is.na(D$steps))The total number of missing values is 2304.
Assign median of 5-minute interval to missing values of data.
D.medianintvl <- tapply(D$steps, D$interval, median, na.rm = TRUE)
D.filter <- D
for (i in names(D.medianintvl)) {
D.filter$steps <- replace(D.filter$steps, which(is.na(D.filter$steps)),
D.medianintvl[[i]])
}The histogram of the total number of steps taken each days.
D.step2 <- tapply(D.filter$steps, D.filter$date, sum)
hist(D.step2, xlab = "Total number of steps taken each days", ylab = "Number of days",
breaks = 10)
The mean and median total number of steps taken per days.
D.mean <- tapply(D$steps, D$date, mean)
D.median <- tapply(D$steps, D$date, median)
D.mean2 <- tapply(D.filter$steps, D.filter$date, mean)
D.median2 <- tapply(D.filter$steps, D.filter$date, median)
hist(D.mean, breaks = 10)
hist(D.mean2, breaks = 10)
When we compare two histogram of median total number of steps taken per days, we found that higher distribution around steps of 0 indicates that most of the missing values are treated as 0 for median value. Imputing missing values changes the distribution to higher density near 0.
Add a new column for weekdays with two levels of "weekday" and "weekend".
D.filter$week <- !(weekdays(D.filter$date, abbreviate = TRUE) %in% c("土",
"日"))
D.filter$week <- factor(D.filter$week, levels = c(TRUE, FALSE), labels = c("weekday",
"weekend"))
head(D.filter)## steps date interval week
## 1 0 2012-10-01 0 weekday
## 2 0 2012-10-01 5 weekday
## 3 0 2012-10-01 10 weekday
## 4 0 2012-10-01 15 weekday
## 5 0 2012-10-01 20 weekday
## 6 0 2012-10-01 25 weekday
The time series plot of 5-minute interval and steps taken all the weekdays and weekends.
library(lattice)
D.aggr <- aggregate(steps ~ interval + week, D.filter, mean)
head(D.aggr)## interval week steps
## 1 0 weekday 2.02222
## 2 5 weekday 0.40000
## 3 10 weekday 0.15556
## 4 15 weekday 0.17778
## 5 20 weekday 0.08889
## 6 25 weekday 1.31111
xyplot(steps ~ interval | week, data = D.aggr, type = "l", xlab = "5-minute interval",
ylab = "average steps taken")