Reproducible Research: Peer Assessment 1

Loading and preprocessing the data

suppressMessages(library(dplyr))
suppressMessages(library(lubridate))
stringsAsFactors=FALSE
activity <- read.table(unz("repdata_data_activity.zip", "activity.csv"),
                       header = TRUE, sep = ",")

What is mean total number of steps taken per day?

stepsPerDay <- activity %>% select(date, steps) %>% group_by(date) %>% 
  summarize(mean(steps, na.rm = TRUE))
names(stepsPerDay) <- c("date", "sum")

1. Make a histogram of the total number of steps taken each day

hist(stepsPerDay$sum, 
     freq = TRUE,
     col = 'red', 
     xlab = 'Total Number of Steps per day',
     breaks = 10,
     main = 'Number of Steps per Day')

dev.copy(png, file = 'figures/plot1.png', height=480, width=480)

png 3

dev.off()

png 2

stepsPerDayMean <- mean(stepsPerDay$sum, na.rm = TRUE)
stepsPerDayMedian <- median(stepsPerDay$sum, na.rm = TRUE)

2. Calculate and report the mean and median total number of steps taken per day

Variable	mean	median
Steps Per Day	37.3826	37.3785

What is the average daily activity pattern?

1. Make a time series plot of the 5-minute interval and the ave number steps

intervalPerDayMean <- activity %>% group_by(interval) %>% 
  summarize(mean(steps, na.rm = TRUE))
names(intervalPerDayMean) <- c("interval", "MeanSteps")
intervalLevels <- levels(factor(activity$interval))
par(mar = c(4.5,4.5,1,1))
plot(intervalPerDayMean$MeanSteps ~ intervalLevels, type = "l",
     ylab = "number of steps", xlab = "interval levels", col = "red")

dev.copy(png, file = 'figures/plot2.png', height=480, width=960)

png 3

dev.off()

png 2

2. Which 5-minute interval on average contains the maximum number of steps

max5MinMeanValue <- max(intervalPerDayMean$MeanSteps)
max5MinMeanRow <- which(intervalPerDayMean$MeanSteps == max5MinMeanValue)
max5MinMeanInterval <- intervalPerDayMean[row(intervalPerDayMean) == 
                                            max5MinMeanRow,][1,1]

The (835) 5-minute interval contains 206 maximum number of steps in average across all days:

Imputing missing values

1. Calculate and report the total number of missing values in the dataset

iSteps <- which(is.na(activity[,1]) == TRUE)
jInterval <- activity[iSteps,3]
activity[iSteps, 1] <- intervalPerDayMean[,2]

Total number of missing values rows in the dataset is: 2304

2. Devise a strategy for filling in all of the missing values in the dataset.

• Strategy is to fill in the missing values with the average of all steps values in a particular interval

missingvals = c(NA)

meanVector <- activity %>% group_by(interval) %>% summarize(avg_steps = 
              mean(steps, na.rm = TRUE, na.inf = na.omit))

3. Create a new dataset that is equal to the original dataset but missing values filled in

newActivity <- tbl_df(activity)

newActivity <- cbind(newActivity, meanVector)

newActivityVector <- data.frame(newActivity$avg_steps)
newActivityVector <- cbind(newActivityVector, newActivity$date)
newActivityVector <- cbind(newActivityVector, newActivity$interval)

names(newActivityVector) <- c("steps", "date", "interval")

4. 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.

hist(newActivityVector$steps, 
     freq = TRUE,
     col = 'red', 
     xlab = 'Total Number of Steps per day',
     breaks = 10,
     main = 'Number of Steps per Day')

dev.copy(png, file = 'figures/plot3.png', height=480, width=480)

png 3

dev.off()

png 2

meanf <- newActivityVector %>% group_by(date) %>% 
  summarize(avg_steps = mean(steps)) 
medianf <- newActivityVector %>% group_by(date) %>% summarize(med_steps = median(steps))

Variable	mean	median
Steps Per Day	37.3826	34.1132

Are there differences in activity patterns between weekdays and weekends?

Yes. It depends how reliable may a technique for filling be in regards to the NA data. Although, in this case mean values are matched, median value is biased to the left.

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.

weekdays1 <- c('Mon', 'Tue', 'Wed', 'Thu', 'Fri')
activity <- activity %>% mutate(daytype = 
                                  factor((weekdays(as.POSIXlt(activity$date),
                                                   abbr = TRUE) %in% 
                                            weekdays1)+1L, levels = 1:2,
                                         labels = c("weekend", "weekday")))

2. Make a panel plot containing a time series plot of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all weekday days or weekend days (y-axis).

par(mfrow = c(2, 1))
par(mar = c(2.5,2.5,1,1))

intervalLevels <- levels(factor(activity$daytype, activity$interval))

## Warning in `levels<-`(`*tmp*`, value = if (nl == nL) as.character(labels)
## else paste0(labels, : duplicated levels in factors are deprecated

plot(activity$steps ~ intervalLevels, type = "l",
     ylab = "number of steps", xlab = "interval levels", col = "red")
dev.copy(png, file = 'figures/plot4.png', height=480, width=960)

png 3

dev.off()

png 2