I am working in R. I have typed in the command :
table(shoppingdata$Identifier, shoppingdata$Coupon)
I have the following data:
FALSE TRUE
197386 0 5
197388 0 2
197390 2 0
197392 0 3
197394 1 0
197397 0 1
197398 1 1
197400 0 4
197402 1 5
197406 0 5
First of all, I cannot name the vectors FALSE and TRUE by something else, e.g couponused.
Most importantly, I want to create a third column which is the sum of FALSE+TRUE( Coupon used+coupon not used= number of visits). The actual columns contain hundreds of entries.
The solution is not obvious at all.
You have stumbled into the abyss of R data types, through no fault of your own.
Assuming that shoppingdata is a data frame,
table(shoppingdata$Identifier, shoppingdata$Coupon)
creates an object of type "table". One would think that using, e.g.
as.data.frame(table(shoppingdata$Identifier, shoppingdata$Coupon))
would turn this into a data frame with the same format as in the printout, but, as the example below shows, it does not!
# example
data <- data.frame(ID=rep(1:5,each=10),coupon=(sample(c(T,F),50,replace=T)))
# creates "contingency table", not a data frame.
t <- table(data)
t
# coupon
# ID FALSE TRUE
# 1 5 5
# 2 3 7
# 3 4 6
# 4 6 4
# 5 3 7
as.data.frame(t) # not useful!!
# ID coupon Freq
# 1 1 FALSE 5
# 2 2 FALSE 3
# 3 3 FALSE 4
# 4 4 FALSE 6
# 5 5 FALSE 3
# 6 1 TRUE 5
# 7 2 TRUE 7
# 8 3 TRUE 6
# 9 4 TRUE 4
# 10 5 TRUE 7
# this works...
coupons <- data.frame(ID=rownames(t),not.used=t[,1],used=t[,2])
# add two columns to make a third
coupons$total <- coupons$used + coupons$not.used
# or, less typing
coupons$ total <- with(coupons,not.used+used)
FWIW, I think yours is a perfectly reasonable question. The reason more people don't use R is that it has an extremely steep learning curve, and the documentation is not very good. On the other hand, once you've climbed that learning curve, R is astonishingly powerful.
Related
I have a table like this
table(mtcars$gear, mtcars$cyl)
I want to rank the rows by the ones with more observations in the 4 cylinder. E.g.
4 6 8
4 8 4 0
5 2 1 2
3 1 2 12
I have been playing with order/sort/rank without much success. How could I order tables output?
We can convert table to data.frame and then order by the column.
sort_col <- "4"
tab <- as.data.frame.matrix(table(mtcars$gear, mtcars$cyl))
tab[order(-tab[sort_col]), ]
# OR tab[order(tab[sort_col], decreasing = TRUE), ]
# 4 6 8
#4 8 4 0
#5 2 1 2
#3 1 2 12
If we don't want to convert it into data frame and want to maintain the table structure we can do
tab <- table(mtcars$gear, mtcars$cyl)
tab[order(-tab[,dimnames(tab)[[2]] == sort_col]),]
# 4 6 8
# 4 8 4 0
# 5 2 1 2
# 3 1 2 12
Could try this. Use sort for the relevant column, specifying decreasing=TRUE; take the names of the sorted rows and subset using those.
table(mtcars$gear, mtcars$cyl)[names(sort(table(mtcars$gear, mtcars$cyl)[,1], dec=T)), ]
4 6 8
4 8 4 0
5 2 1 2
3 1 2 12
In the same scope as Milan, but using the order() function, instead of looking for names() in a sort()-ed list.
The [,1] is to look at the first column when ordering.
table(mtcars$gear, mtcars$cyl)[order(table(mtcars$gear, mtcars$cyl)[,1], decreasing=T),]
I have a few large data sets with many variables. There is a "key" variable that is the ID for the research participant. In these data sets, there are some IDs that are duplicated. I have written code to extract all data for duplicated IDs, but I would like a way to check if the remainder of the variables for those IDs are equal or not. Below is a simplistic example:
ID X Y Z
1 2 3 4
1 2 3 5
2 5 5 4
2 5 5 4
3 1 2 3
3 2 2 3
3 1 2 3
In this example, I would like to be able to identify that the rows for ID 1 and ID 3 are NOT all equal. Is there any way to do this in R?
You can use duplicated for this:
d <- read.table(text='ID X Y Z
1 2 3 4
1 2 3 5
2 5 5 4
2 5 5 4
3 1 2 3
3 2 2 3
3 1 2 3
4 1 1 1', header=TRUE)
tapply(duplicated(d), d[, 1], function(x) all(x[-1]))
## 1 2 3 4
## FALSE TRUE FALSE TRUE
Duplicated returns a vector indicating, for each row of a dataframe, whether it has been encountered earlier in the dataframe. We use tapply over this logical vector, splitting it in to groups based on ID and applying a function to each of these groups. The function we apply is all(x[-1]), i.e. we ask whether all rows for the group, other than the initial row, are duplicated?
Note that I added a group with a single record to ensure that the solution works in these cases as well.
Alternatively, you can reduce the dataframe to unique records with unique, and then split by ID and check whether each split has only a single row:
sapply(split(unique(d), unique(d)[, 1]), nrow) == 1
## 1 2 3 4
## FALSE TRUE FALSE TRUE
(If it's a big dataframe it's worth calculating unique(d) in advance rather than calling it twice.)
If I have a vector numbers <- c(1,1,2,4,2,2,2,2,5,4,4,4), and I use 'table(numbers)', I get
names 1 2 4 5
counts 2 5 4 1
What if I want it to include 3 also or generally, all numbers from 1:max(numbers) even if they are not represented in numbers. Thus, how would I generate an output as such:
names 1 2 3 4 5
counts 2 5 0 4 1
If you want R to add up numbers that aren't there, you should create a factor and explicitly set the levels. table will return a count for each level.
table(factor(numbers, levels=1:max(numbers)))
# 1 2 3 4 5
# 2 5 0 4 1
For this particular example (positive integers), tabulate would also work:
numbers <- c(1,1,2,4,2,2,2,2,5,4,4,4)
tabulate(numbers)
# [1] 2 5 0 4 1
I'm trying to analyze some date using R but I'm not very familiar with R (yet) and therefore I'm totally stuck.
What I try to do is manipulate my input data so I can use it to calculate Cohen's Kappa.
Now the problem is, that for rater_1, I have several ratings for some of the items and I need to select one. If rater_1 has given the same rate on an item as rater_2, then this rating should be chosen, if not any rating of the list can be used.
I tried
unique(merge(rater_1, rater_2, all.x=TRUE))
which brings me close, but if the ratings between the two raters diverge, only one is kept.
So, my question is, how do I get from
item rating_1
1 3
2 5
3 4
item rating_2
1 2
1 3
2 4
2 1
2 2
3 4
3 2
to
item rating_1 rating_2
1 3 3
2 5 4
3 4 4
?
There are some fancy ways to do this, but I thought it might be helpful to combine a few basic techniques to accomplish this task. Usually, in your question, you should include some easy way to generate your data, like this:
# Create some sample data
set.seed(1)
id<-rep(1:50)
rater_1<-sample(1:5,50,replace=TRUE)
df1<-data.frame(id,rater_1)
id<-rep(1:50,each=2)
rater_2<-sample(1:5,100,replace=TRUE)
df2<-data.frame(id,rater_2)
Now, here is one simple technique for doing this.
# Merge together the data frames.
all.merged<-merge(df1,df2)
# id rater_1 rater_2
# 1 1 2 3
# 2 1 2 5
# 3 2 2 3
# 4 2 2 2
# 5 3 3 1
# 6 3 3 1
# Find the ones that are equal.
same.rating<-all.merged[all.merged$rater_2==all.merged$rater_1,]
# Consider id 44, sometimes they match twice.
# So remove duplicates.
same.rating<-same.rating[!duplicated(same.rating),]
# Find the ones that never matched.
not.same.rating<-all.merged[!(all.merged$id %in% same.rating$id),]
# Pick one. I chose to pick the maximum.
picked.rating<-aggregate(rater_2~id+rater_1,not.same.rating,max)
# Stick the two together.
result<-rbind(same.rating,picked.rating)
result<-result[order(result$id),] # Sort
# id rater_1 rater_2
# 27 1 2 5
# 4 2 2 2
# 33 3 3 1
# 44 4 5 3
# 281 5 2 4
# 11 6 5 5
A fancy way to do this would be like this:
same.or.random<-function(x) {
matched<-which.min(x$rater_1==x$rater_2)
if(length(matched)>0) x[matched,]
else x[sample(1:nrow(x),1),]
}
do.call(rbind,by(merge(df1,df2),id,same.or.random))
I am using R to generate examples of how to deal with missing data for the statistics class I am teaching. One method requires generating a "missing values binary variable", with 0 for cases containing missing values, and 1 with no missing values. For example
n X Y Z
1 4 300 2
2 8 400 4
3 10 500 7
4 18 NA 10
5 20 50 NA
6 NA 1000 5
I would like to generate a variable M, such that
n m
1 1
2 1
3 1
4 0
5 0
6 0
It seems this should be simple, given R's ability to handle missing values. The closest I have found is m <-ifelse(is.na(missguns),0,1), but all this does is generate a new entire data matrix with 0 or 1 indicating missingness. However, I just want one variable indicating if a row contains missing values.
complete.cases does exactly what you want.
complete.cases(x)
## [1] TRUE TRUE TRUE FALSE FALSE FALSE
You can coerce to numeric or integer:
as.integer(complete.cases(x))
## [1] 1 1 1 0 0 0