The data was downloaded from this link provided in the assignment. The data in the file is described in the README.md document. Briefly, there are three columns in the CSV named 'steps', 'date', and 'interval'. The read.csv function, along with specified colClasses, can be used to read the data into R. The package data.table is also used to turn the data frame into a data.table for easier manipulation.
activity <- read.csv(unz('activity.zip', 'activity.csv'),
colClasses = c('numeric', 'Date', 'numeric'))
library(data.table)
activity <-data.table(activity)For this part of the assignment, the missing values in the dataset are ignored
(i.e. use na.rm = TRUE as necessary and don't impute missing values).
First step is to calculate the total number of steps taken per day. The sum
function will be applied to the data table with the option by=date to get
the sum of steps per day. This will be saved to a new variable called 'total'.
The first few lines of the total data table is shown below.
total <- activity[,list(steps = sum(steps, na.rm = TRUE)), by = date]
head(total)## date steps
## 1: 2012-10-01 0
## 2: 2012-10-02 126
## 3: 2012-10-03 11352
## 4: 2012-10-04 12116
## 5: 2012-10-05 13294
## 6: 2012-10-06 15420
A histogram of the total number of steps taken each day is then made. There was no specified break points.
hist(total$steps, xlab = 'Total steps per day', main = 'Total Steps Per Day')
The histogram indicates that most of the days had a total number of steps between 10,000 and 15,000.
The mean and median of the total number of steps taken per day are then calculated and reported.
mean_steps <- mean(total$steps)
median_steps <- median(total$steps)Echoing the above
mean_steps## [1] 9354.23
median_steps## [1] 10395
To create a time series plot (i.e. type = "l") of the 5-minute interval (x-axis) and the average number of steps taken averaged across all days (y-axis), a data frame that summarizes the mean steps per 5-minute interval is created. The data frame (here saved as ```by_interval``) is then plotted.
by_interval <- activity[,list(steps = mean(steps, na.rm = TRUE)), by = interval]
with(by_interval, plot(interval,
steps,
type = 'l',
ylab = 'Average Nunber of Steps',
xlab = '5-minute Interval'))
To find out Which 5-minute interval, on average across all the days in the
dataset, contains the maximum number of steps, the function max can be
used. As shown below, the interval 835 has the maximum number of steps on
average across all the days in the dataset.
by_interval[steps == max(by_interval$steps)]## interval steps
## 1: 835 206.1698
To calculate and report the total number of missing values in the dataset (i.e.
the total number of rows with NAs), the summary or the is.na
functions can be used.
summary(activity)## steps date interval
## Min. : 0.00 Min. :2012-10-01 Min. : 0.0
## 1st Qu.: 0.00 1st Qu.:2012-10-16 1st Qu.: 588.8
## Median : 0.00 Median :2012-10-31 Median :1177.5
## Mean : 37.38 Mean :2012-10-31 Mean :1177.5
## 3rd Qu.: 12.00 3rd Qu.:2012-11-15 3rd Qu.:1766.2
## Max. :806.00 Max. :2012-11-30 Max. :2355.0
## NA's :2304
nrow(activity[is.na(activity$steps)])## [1] 2304
A total of 2304 rows have NA. A new dataset (here called imputed) that is
equal to the original dataset but with the missing data filled in with the mean
for that day is created. To do this, a data frame containing all the means for
each interval was made (here called invterval_means). The imputed
dataset is then created by looking for the rows with NAs in the activity
dataset and filling that in with the corresponding mean for that interval using
the interval_means data frame.
interval_means <- activity[, list(steps=mean(steps, na.rm=TRUE)), by=interval]
imputed <- transform(activity,
steps = ifelse(is.na(steps),
interval_means[interval_means$interval == interval,
steps],
steps))
summary(imputed)## steps date interval
## Min. : 0.00 Min. :2012-10-01 Min. : 0.0
## 1st Qu.: 0.00 1st Qu.:2012-10-16 1st Qu.: 588.8
## Median : 0.00 Median :2012-10-31 Median :1177.5
## Mean : 37.38 Mean :2012-10-31 Mean :1177.5
## 3rd Qu.: 27.00 3rd Qu.:2012-11-15 3rd Qu.:1766.2
## Max. :806.00 Max. :2012-11-30 Max. :2355.0
Compare the summary of the imputed data with the summary of the original data.
There are no longer any NAs in the steps column and the third quartile has
changed. Interestingly, the mean number of steps for the entire dataset remains
the same.
The total steps taken for each day can be computed in the same way that the
total steps taken for each day was computed for the dataset with no imputation.
A histogram can also be created.
imputed_total <- imputed[,list(steps = sum(steps)), by = date]
hist(imputed_total$steps,
main = 'Total Steps per Day (Missing Values Imputed)',
xlab = 'Total steps per day')
The histogram of the imputed data set shows that most days still have a total number of steps between 10,000 and 15,000. There are now less days with total between 0 and 5,000 steps compared to the dataset where no imputation was done.
The meand and median of the imputed dataset is computed as follows.
imputed_mean <- mean(imputed_total$steps)
imputed_median <- median(imputed_total$steps)Echoing the above, we get the new mean and median for the imputed dataset of total steps per day. This is different from the estimates obtained in the first part of the assignment where no imputation was done. The mean and median of the imputed dataset are higher than the dataset with no imputation. Also, the mean and median of the imputed dataset are the same.
imputed_mean## [1] 10766.19
imputed_median## [1] 10766.19
To determine whether there are differences in activity patterns between weekdays
and weekends, a new data frame (here called weekends) was created. In this
data frame, a new column (called days) was added which stated whether the
date for a row was on a weekend or a weekday. The days column was then
used as a factor to summarize the data and create plots with the interval in the
x-axis and the number of steps in the y-axis.
weekends <- transform(activity,
days = ifelse(weekdays(date) == 'Saturday' |
weekdays(date) == 'Sunday',
'weekend',
'weekday'))
weekends$days <- factor(weekends$days)
by_interval_w <- weekends[,list(steps = mean(steps, na.rm = TRUE)),
by = c('interval', 'days')]
par(mfrow = c(2, 1))
with(by_interval_w[days == 'weekend'],
plot(interval,
steps,
type = 'l',
ylim = c(0, 200),
main = 'Weekend'))
with(by_interval_w[days == 'weekday'],
plot(interval,
steps,
type = 'l',
ylim = c(0, 200),
main = 'Weekday'))
The plots above show that on the weekend, there seems to be more intervals >1000 that have higher number of steps. Does this mean that people walk/ run/ jog more in the weekends? Perhaps.