Reproducible Research: Assignment 1
by R. Bagnall 22 June 2015
Synopsys
This reproducible research assignment for the Data Science Specialization makes use of data from a personal activity monitoring device. This device collects data at 5 minute intervals through out the day. The data consists of two months worth of data from an anonymous individual collected during the months of October and November 2012 and include the number of step taken in 5 minute intervals each day.
1. Loading and processing the data
The data for this assignment can be downloaded from the course website: Activity monitoring data [52K]
The variables included in this dataset are:
- steps: Number of steps taking in a 5-minute interval (missing values are coded as
NA)
- date: The date on which the measurement was taken in YYYY-MM-DD format
- interval: Identifier for the 5-minute interval in which measurement was taken
The dataset is stored in a comma-separated-value (.csv) file and there are a total of 17,568 observation in this dataset.
Read in the dataset and call it df; show the first few columns.
df <- read.csv("activity.csv", header=TRUE)
head(df)## 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 2012-10-01 252. What is the mean total number of steps taken per day?
Ignoring the missing values, split df by date and calculate the total number of steps taken per day. Make a histogram of the total number of steps taken each day. Here I use breaks=15 to show smaller divisions of the data.
# calculate total number of steps taken per day
StepsPerDay <- as.data.frame(sapply(split(df, df[2], drop=TRUE), function(x) {sum(x$steps)}))
# make histrogram of total steps taken per day
hist(as.numeric(StepsPerDay[,1]), main="Histogram of Steps Per Day", xlab="Total Number of Steps", ylab="Frequency", breaks=15, border="red", xlim=c(0,25000))
mean(StepsPerDay[,1], na.rm=TRUE)## [1] 10766.19The mean number of steps taken per day is 10766.
median(StepsPerDay[,1], na.rm=TRUE)## [1] 10765The median number of steps taken per day is 10765.
3. What is the average daily activity pattern?
Make a time series plot of the five minute intervals and average steps taken, averaged across all days.
# make a boolean vector of na values in steps column of df
na <- is.na(df$steps)
# make a new data frame without na values
dfcomplete <- df[!na, ]
# split dfcomplete on time interval
# calculate the average number of steps taken across all days
AveStepsPerDay <- sapply(split(dfcomplete, dfcomplete[3], drop=TRUE), function(x) {ave(x$steps)})
# turn results into a data frame
dfAveSteps <-as.data.frame(AveStepsPerDay[1,])
# add time intervals shown as row names as a column
dfAveSteps$time <- row.names(dfAveSteps)
# plot the average number of steps taken per day
plot(dfAveSteps$time, AveStepsPerDay[1, ], type="l", col="darkgreen", xlim=c(0,2500), main="Average Number of Steps Per Time Interval", xlab="5 Minute Time Interval", ylab="Average Steps Per Time Interval")
Which five munite interval, averaged across all days, contains the maximum number of steps?
subset(dfAveSteps, AveStepsPerDay[1, ] == max(AveStepsPerDay[1, ]))## AveStepsPerDay[1, ] time
## 835 206.1698 835The 835 interval has the most number of steps (206 steps)
4. Imputing Missing Values
What is the total number of rows with missing values?
table(na)## na
## FALSE TRUE
## 15264 2304The number of rows with missing values is 2304.
Imputation of missing values will be performed by: * 1. Calculate the mean number of steps for each 5 minute interval (excluding NA) * 2. Replce NA step values with the average step value for that interval
### Make a data frame of average steps per time interval ###
# Make a vector of mean steps per interval
stepmeans <- as.data.frame(sapply(split(df, df[3]), function(x) {as.integer(mean(x$steps, na.rm=TRUE))}), col.names="steps")
# change column name
names(stepmeans) <- "meansteps"
# add in time intervals
stepmeans$interval <- row.names(stepmeans)
### Impute missing values ###
# split df on date to produce a list of data frames
splitted <- split(df, df[2])
# make an empty list
final=list()
# loop over length of the list
for (i in 1:length(splitted)){
# for each list, split by day
newsplit <- as.data.frame(splitted[i])
# rename header
names(newsplit) <- c("steps", "date", "interval")
# if there are no NA values, return the list
if(!anyNA(newsplit[,1])) {
final[[i]] <- newsplit
next
}
# else, merge list with data frame of mean steps
intermediate <- merge(newsplit, stepmeans)
# retain only required columns and rename headers
intermediate <- as.data.frame(intermediate[c(4,3,1)])
names(intermediate) <-c("steps", "date", "interval")
# return the imputed data back to the list
final[[i]] <- intermediate
}
# bind all lists into one data frame
imputed_df <-do.call("rbind", final)Make a histogram of the total number of steps taken each day.
# split data frame on day and calculate the total number of steps taken each day
imputed_steps <- sapply(split(imputed_df, imputed_df[2], drop=TRUE), function(x) {sum(x$steps)})
# coerce to data frame
imputed_steps <- as.data.frame(imputed_steps)
# plot as histogram
hist(as.numeric(imputed_steps[,1]), main="Histogram of Steps with Imputation", xlab="Number of Steps with Imputation", ylab="Frequency", breaks=15, col="blue", border="white", xlim=c(0,25000))
The mean number of steps taken per day before imputation is 10766 and after imputation is 10749. These values differ by 17 steps. Imputation has decreased the mean number of steps taken per day.
The median number of steps taken per day before imputation is 10765 and after imputation is 10641. These values differ by 124 steps. Imputation has decreased the average number of steps taken per day.
5. Are there differences in activity patterns between weekdays and weekends?
create a new factor variable in the data set with two levels - “weekdays” and “weekends”
# turn date column into class(date)
imputed_df$date <- as.Date(as.character(imputed_df$date), format="%Y-%m-%d")
# add wekday/weekend as a factor
imputed_df$weekend <- as.factor(ifelse(weekdays(imputed_df$date) %in% c("Saturday","Sunday"), "Weekend", "Weekday"))
str(imputed_df)## 'data.frame': 17568 obs. of 4 variables:
## $ steps : int 1 0 0 0 0 2 0 0 0 1 ...
## $ date : Date, format: "2012-10-01" "2012-10-01" ...
## $ interval: int 0 5 10 15 20 25 30 35 40 45 ...
## $ weekend : Factor w/ 2 levels "Weekday","Weekend": 1 1 1 1 1 1 1 1 1 1 ...Make a panel plot containing a time series of the 5 minute intervals and average number of steps taken, averaged across all weekdays or weekend days.
# split dataframe on weekday
imputed_week <- split(imputed_df, imputed_df$weekend)
#calculate the average number of steps taken across 5 minute intervals for weekdays
weekday <- as.data.frame(imputed_week[1])
weekdaysteps <- as.data.frame(sapply(split(weekday, weekday[3]), function(x) {mean(x$Weekday.steps)}))
weekdaysteps$interval <- row.names(weekdaysteps)
names(weekdaysteps) <- c("Steps", "Interval")
weekdaysteps$day <- rep("weekday", nrow(weekdaysteps))
#calculate the average number of steps taken across 5 minute intervals for weekend days
weekend <- as.data.frame(imputed_week[2])
weekendsteps <- as.data.frame(sapply(split(weekend, weekend[3]), function(x) {mean(x$Weekend.steps)}))
weekendsteps$interval <- row.names(weekendsteps)
names(weekendsteps) <- c("Steps", "Interval")
weekendsteps$day <- rep("weekend", nrow(weekendsteps))
# make plot
finalplotdf <-rbind(weekendsteps, weekdaysteps)
finalplotdf$Interval <- as.numeric(finalplotdf$Interval)
#finalplotdf$day <- as.factor(finalplotdf$day)
library(lattice)
xyplot(Steps ~ Interval | factor(day), data=finalplotdf, type="l", layout=(c(1,2)))