In this document i describe the steps i did and show the code i used to analyse the given data. I used R version 3.2.2., my OS is Windows 10. As a first task the data has to be imported and processed:
To import the data the path has to be set. I have changed the class of the date-variable from factor to date. Additionally i have also changed the interval-variable from integer to factor as it is not really a continous variable, but an ordinal number representing the time. I haven´t changed it explictly into a timestamp as this is the exploratory phase and i find it very easy to handle it that way.s
#setwd("C:/Coursera/Reprod/")
library(lattice)
tab3l_raw <- read.csv("activity.csv")
tab3l_raw$date <- as.Date(tab3l_raw$date)
tab3l_raw$interval <- as.factor(tab3l_raw$interval)After loading and preprocessing we want to answer the following question:
For this question we firstly discard all rows which have a missing value in any column. To do that the complete.cases-function is used. The data is stored in a new variable 'tab3l' as we might need the (almost) raw data afterwards. At first we show the total number of steps per day.
tab3l <- tab3l_raw[complete.cases(tab3l_raw),]
total_per_day <- tapply(tab3l$steps, tab3l$date,sum,na.rm=T)
total_per_day## 2012-10-02 2012-10-03 2012-10-04 2012-10-05 2012-10-06 2012-10-07
## 126 11352 12116 13294 15420 11015
## 2012-10-09 2012-10-10 2012-10-11 2012-10-12 2012-10-13 2012-10-14
## 12811 9900 10304 17382 12426 15098
## 2012-10-15 2012-10-16 2012-10-17 2012-10-18 2012-10-19 2012-10-20
## 10139 15084 13452 10056 11829 10395
## 2012-10-21 2012-10-22 2012-10-23 2012-10-24 2012-10-25 2012-10-26
## 8821 13460 8918 8355 2492 6778
## 2012-10-27 2012-10-28 2012-10-29 2012-10-30 2012-10-31 2012-11-02
## 10119 11458 5018 9819 15414 10600
## 2012-11-03 2012-11-05 2012-11-06 2012-11-07 2012-11-08 2012-11-11
## 10571 10439 8334 12883 3219 12608
## 2012-11-12 2012-11-13 2012-11-15 2012-11-16 2012-11-17 2012-11-18
## 10765 7336 41 5441 14339 15110
## 2012-11-19 2012-11-20 2012-11-21 2012-11-22 2012-11-23 2012-11-24
## 8841 4472 12787 20427 21194 14478
## 2012-11-25 2012-11-26 2012-11-27 2012-11-28 2012-11-29
## 11834 11162 13646 10183 7047
To give some insights in the total number of steps per day and to see its distrubition, a histogram was made:
hist(total_per_day,10, main="Histogram: Total number of steps per day", xlab = "Number of steps")
We see that the frequency for 10k-12k steps is the highest, but there are also a few days, where very few or many steps were reported. To get a better look on the actual numbers we want to calculate the mean and the median of the total number of steps per day:
mean(total_per_day,na.rm=T)## [1] 10766.19
median(total_per_day,na.rm=T)## [1] 10765
Now we want to investigate what the average daily pattern looks like.
###What is the average daily activity pattern?
Therefore we take the mean value for every interval. Afterwards we plot the timeseries. We use the interval as the "time of the day". Keep in mind 835 means 08:35am. As having a factor as xaxis wasn´t working in a easy fashion, i used a trick: i used an index for the inital plot and added the x-axis by using scales.
average_per_intervall <- tapply(tab3l$steps, tab3l$interval,mean,na.rm=T)
xaxi <- seq(1,length(average_per_intervall),by= 24)
xyplot(average_per_intervall ~ rep(1:length(average_per_intervall),1),layout=c(1,1),ylab ="steps in average", xlab="time of the day", type="l",scales=list(x=list(labels=tab3l$interval[xaxi],at=xaxi)))
Additionally we look for the interval with the highest mean. That means the time of the day the average number of steps is the highest.
max_interval <- as.numeric(as.character(tab3l$interval))[which(average_per_intervall==max(average_per_intervall))]
max_interval## [1] 835
That means around 08:35am, in average the highest number of steps is done. Now we want to analyse what happens if we include substitutes for missing values:
###Imputing missing values
At first we compute the number of rows which have a missing value for at least one variable.
num_of_NA_rows <- sum(!complete.cases(tab3l_raw))
num_of_NA_rows## [1] 2304
That means there are 2304 rows with a missing value. To include this rows we have to substitute the missing values of steps. As we see in further analysis that the only values which are missing are regarding the variable steps we only have to think about a strategy for this variable. The strategy to substitute a missing value for steps is the following: If one value is missing it is replaced by the average number of the reported steps taken on this day. If all values are missing for one day, we replace the values of that day with 0. The new data is stored in tab3l_clean.
tab3l_clean <- tab3l_raw
for (i in seq(1,length(tab3l_clean$steps))){
if(is.na(tab3l_clean$steps[i])){
subset <- tab3l_clean$steps[tab3l_clean$date==tab3l_clean$date[i]]
if(is.nan(mean(subset,na.rm=T))){
tab3l_clean$steps[i]<-0
}else{
tab3l_clean$steps[i]<-mean(subset,na.rm=T)
}
}
}For the "cleaned" data we now check if there are different values for the mean and the median. Afterwards we plot the histogram for this data.
total_per_day_clean <- tapply(tab3l_clean$steps, tab3l_clean$date,sum,na.rm=T)
mean_per_days_clean <- mean(total_per_day_clean,na.rm=T)
mean_per_days_clean## [1] 9354.23
median_per_days_clean <- median(total_per_day_clean,na.rm=T)
median_per_days_clean## [1] 10395
hist(total_per_day_clean,10, main="Histogram: Total number of steps per day", xlab = "Number of steps")
We can see that the histogram looks different. Especially we see that there are a quite a lot days now which have very few steps in total. This is probably because we set the value of steps to 0, if all values are missing for one day. Furthermoe the median and the mean values are lower. This can also be explained by the days with total numer of days equal to 0 now.
###Are there differences in activity patterns between weekdays and weekends?
Now we want to see if there are any differences between weekdays and weekends. Therefore we create a logical vector indicating if we have a weekend-day or not. For reproducibility i use the following expression to get the English names for the weekdays. We can also create a factor for that. Though i don´t really use it.
Sys.setlocale("LC_TIME", "C")## [1] "C"
days <- weekdays(tab3l$date,abbreviate = F)
days_log <- days == "Saturday" | days == "Sunday"
days_fac <- character(length(days_log))
days_fac[days_log] <- "weekend"
days_fac[!days_log] <- "weekday"
days_fac <- as.factor(days_fac)
summary(days_fac)## weekday weekend
## 11232 4032
Now we can run a separate analysis for both categories, we therefore create a big table, in the first rows we have the values (average steps per intervall) for weekend-days and then those of the weekdays. Afterwards we plot them with the lattice-library and in the same way as the other timeseries-plot.
average_per_intervall_WE<- tapply(tab3l$steps[days_log], tab3l$interval[days_log],mean,na.rm=T)
average_per_intervall_WD<- tapply(tab3l$steps[!days_log], tab3l$interval[!days_log],mean,na.rm=T)
average_per_intervall_WE <- cbind(steps=average_per_intervall_WE,wday=rep("weekend",length(average_per_intervall_WE)))
average_per_intervall_WD <- cbind(steps=average_per_intervall_WD,wday=rep("weekday",length(average_per_intervall_WD)))
average_both <- rbind(average_per_intervall_WE,average_per_intervall_WD)
average_both_intervall <- data.frame(cbind(average_both,interval=rep(unique(as.character(tab3l$interval)),2)),row.names = 1:length(average_both[,1]))
average_both_intervall$steps <- as.numeric(as.character(average_both_intervall$steps))
average_both_intervall$interval <- as.character(average_both_intervall$interval)
xaxi <- seq(1,length(average_per_intervall_WE[,1]),by= 24)
xyplot(steps ~ rep(1:length(average_per_intervall_WE[,1]),2)| factor(wday),data=average_both_intervall,layout=c(1,2),xlab="time of the day", type="l",scales=list(x=list(labels=average_both_intervall$interval[xaxi],at=xaxi)))
That was my analysis. I hope you appreciate it. Thanks for rewiewing!