imputing data with median by date in R [duplicate] - r

Question

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How to replace NA with mean by group / subset?
(5 answers)
Closed 3 years ago.
I need to replace the missing values in the field "steps" by the median of "steps" calculated over that particular day (group by "date") with NA values removed. I have already referred this thread but my NA values aren't replaced. Can somebody help me find out where am I going wrong? I would prefer using base package/data table/plyr. The dataset looks approx. like this:-
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
---
17564: NA 2012-11-30 2335
17565: NA 2012-11-30 2340
17566: NA 2012-11-30 2345
17567: NA 2012-11-30 2350
17568: NA 2012-11-30 2355
The structure and summary of the dataset(activity) are as shown below
#str(activity)
Classes ‘data.table’ and '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" "2012-10-01" ...
$ interval: int 0 5 10 15 20 25 30 35 40 45 ...
#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
Things I have tried:
Datatable method:
activityrepNA<-activity[,steps := ifelse(is.na(steps), median(steps, na.rm=TRUE), steps), by=date]
summary(activityrepNA)
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
Using ave
activity$steps[is.na(activity$steps)] <- with(activity, ave(steps,date, FUN = function(x) median(x, na.rm = TRUE)))[is.na(activity$steps)]
> summary(activity)
steps date interval
Min. : 0.00 Min. :2012-10-01 Min. : 0.0
1st Qu.: 0.00 1st Qu.:2012-10-16 1st Qu.: 588.8
Median : 0.00 Median :2012-10-31 Median :1177.5
Mean : 37.38 Mean :2012-10-31 Mean :1177.5
3rd Qu.: 12.00 3rd Qu.:2012-11-15 3rd Qu.:1766.2
Max. :806.00 Max. :2012-11-30 Max. :2355.0
NA's :2304
Attempt at ddply
cleandatapls<-ddply(activity,
+ .(as.character(date)),
+ transform,
+ steps=ifelse(is.na(steps), median(steps, na.rm=TRUE), steps))
> summary(cleandatapls)
as.character(date) steps date interval
Length:17568 Min. : 0.00 Min. :2012-10-01 Min. : 0.0
Class :character 1st Qu.: 0.00 1st Qu.:2012-10-16 1st Qu.: 588.8
Mode :character Median : 0.00 Median :2012-10-31 Median :1177.5
Mean : 37.38 Mean :2012-10-31 Mean :1177.5
3rd Qu.: 12.00 3rd Qu.:2012-11-15 3rd Qu.:1766.2
Max. :806.00 Max. :2012-11-30 Max. :2355.0
NA's :2304
Aggregate for calculating median
whynoclean<-aggregate(activity,by=list(activity$date),FUN=median,na.rm=TRUE)
> summary(whynoclean)
Group.1 steps date interval
Min. :2012-10-01 Min. :0 Min. :2012-10-01 Min. :1178
1st Qu.:2012-10-16 1st Qu.:0 1st Qu.:2012-10-16 1st Qu.:1178
Median :2012-10-31 Median :0 Median :2012-10-31 Median :1178
Mean :2012-10-31 Mean :0 Mean :2012-10-31 Mean :1178
3rd Qu.:2012-11-15 3rd Qu.:0 3rd Qu.:2012-11-15 3rd Qu.:1178
Max. :2012-11-30 Max. :0 Max. :2012-11-30 Max. :1178
NA's :8
EDIT output for the code using mutate
activity %>% group_by(date) %>% mutate(steps = replace(steps, is.na(steps), median(steps, na.rm = T)))
Source: local data table [17,568 x 3]
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
7 NA 2012-10-01 30
8 NA 2012-10-01 35
9 NA 2012-10-01 40
10 NA 2012-10-01 45
.. ... ... ...
UPDATE:
Steven Beaupre helped me realize that my approach for the imputing was flawed as there were specific dates having only NA values that were causing the problem as the median of NA's is NA. Used another suggested approach.

Answer 1

Try:
library(dplyr)
df %>%
group_by(date) %>%
mutate(steps = ifelse(is.na(steps), median(steps, na.rm = T), steps))
If for a given date, all steps are NAs, you could replace them with 0:
df %>%
group_by(date) %>%
mutate(steps = ifelse(all(is.na(steps)), 0,
ifelse(is.na(steps), median(steps, na.rm = T), steps)))