There is a wide data set, a simple example is
df<-data.frame("id"=c(1:6),
"ax"=c(1,2,2,3,4,4),
"bx"=c(7,8,8,9,10,10),
"cx"=c(11,12,12,13,14,14))
I'm looking for a way to assign the values in "ax" to column "bx" and "cx". Here, imagine we have thousands of columns we intend to replace with "ax", so I want this to be done in an automated approach using R. The expected output look like
df<-data.frame("id"=c(1:6),
"ax"=c(1,2,2,3,4,4),
"bx"=c(1,2,2,3,4,4),
"cx"=c(1,2,2,3,4,4))
I've thought of, and tried using mutate_at and ends_with, but this has not work for me. For example, I tried
df %>%
mutate_at(vars(ends_with("x")), labels = "ax")
and this prints an error. Not sure what's wrong or what's to be added to get this working, so I would like to request your help on this. Thank you very much!
A simple way using base R would be :
change_cols <- grep('x$', names(df))
df[change_cols] <- df$ax
df
# id ax bx cx
#1 1 1 1 1
#2 2 2 2 2
#3 3 2 2 2
#4 4 3 3 3
#5 5 4 4 4
#6 6 4 4 4
I would suggest this tidyverse approach using across() to select the range of variables you want:
library(tidyverse)
#Data
df<-data.frame("id"=c(1:6),
"ax"=c(1,2,2,3,4,4),
"bx"=c(7,8,8,9,10,10),
"cx"=c(11,12,12,13,14,14))
#Mutate
df %>% mutate(across(c(bx:cx), ~ ax))
Output:
id ax bx cx
1 1 1 1 1
2 2 2 2 2
3 3 2 2 2
4 4 3 3 3
5 5 4 4 4
6 6 4 4 4
Another option with mutate_at()
df %>%
mutate_at(vars(matches("x$")), ~ax)
# id ax bx cx
# 1 1 1 1 1
# 2 2 2 2 2
# 3 3 2 2 2
# 4 4 3 3 3
# 5 5 4 4 4
# 6 6 4 4 4
I have some trouble while trying to get a subset of an initial dataframe as in the end it corrupts my dataframe! Here is an example of what is happening:
Let consider a dataframe :
>test=data.frame("v1"=c(1,2,3,4,-5,-3),"v2"=c(1,2,3,4,5,6))
> print(test)
v1 v2
1 1 1
2 2 2
3 3 3
4 4 4
5 -5 5
6 -3 6
Then I want to take the subset where var1 value is strictly below, let's say -2:
> subtest=test[test$v1<-2,]
> print(subtest)
v1 v2
2 2 2
> print(test)
v1 v2
1 2 1
2 2 2
3 2 3
4 2 4
5 2 5
6 2 6
Not only did the subset operation didn't work but it actually corrupt my initial database by replacing all the v1 values by 2 !
I t is not a bug. It is just a typo!
The operator <- is used to assign values. In the command subtest=test[test$v1<-2,], you have assigned all values of variable v1 to 2.
To do what you want, you should do that instead:
subtest=test[test$v1< -2,]
It is not a bug. You just have to put brackets, because otherwise it will be assignment mark <-. If you want strictly below -2:
subtest=test[test$v1<(-2),]
I'm trying to analyze some date using R but I'm not very familiar with R (yet) and therefore I'm totally stuck.
What I try to do is manipulate my input data so I can use it to calculate Cohen's Kappa.
Now the problem is, that for rater_1, I have several ratings for some of the items and I need to select one. If rater_1 has given the same rate on an item as rater_2, then this rating should be chosen, if not any rating of the list can be used.
I tried
unique(merge(rater_1, rater_2, all.x=TRUE))
which brings me close, but if the ratings between the two raters diverge, only one is kept.
So, my question is, how do I get from
item rating_1
1 3
2 5
3 4
item rating_2
1 2
1 3
2 4
2 1
2 2
3 4
3 2
to
item rating_1 rating_2
1 3 3
2 5 4
3 4 4
?
There are some fancy ways to do this, but I thought it might be helpful to combine a few basic techniques to accomplish this task. Usually, in your question, you should include some easy way to generate your data, like this:
# Create some sample data
set.seed(1)
id<-rep(1:50)
rater_1<-sample(1:5,50,replace=TRUE)
df1<-data.frame(id,rater_1)
id<-rep(1:50,each=2)
rater_2<-sample(1:5,100,replace=TRUE)
df2<-data.frame(id,rater_2)
Now, here is one simple technique for doing this.
# Merge together the data frames.
all.merged<-merge(df1,df2)
# id rater_1 rater_2
# 1 1 2 3
# 2 1 2 5
# 3 2 2 3
# 4 2 2 2
# 5 3 3 1
# 6 3 3 1
# Find the ones that are equal.
same.rating<-all.merged[all.merged$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 have a data frame in R which is similar to the follows. Actually my real ’df’ dataframe is much bigger than this one here but I really do not want to confuse anybody so that is why I try to simplify things as much as possible.
So here’s the data frame.
id <-c(1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3)
a <-c(3,1,3,3,1,3,3,3,3,1,3,2,1,2,1,3,3,2,1,1,1,3,1,3,3,3,2,1,1,3)
b <-c(3,2,1,1,1,1,1,1,1,1,1,2,1,3,2,1,1,1,2,1,3,1,2,2,1,3,3,2,3,2)
c <-c(1,3,2,3,2,1,2,3,3,2,2,3,1,2,3,3,3,1,1,2,3,3,1,2,2,3,2,2,3,2)
d <-c(3,3,3,1,3,2,2,1,2,3,2,2,2,1,3,1,2,2,3,2,3,2,3,2,1,1,1,1,1,2)
e <-c(2,3,1,2,1,2,3,3,1,1,2,1,1,3,3,2,1,1,3,3,2,2,3,3,3,2,3,2,1,3)
df <-data.frame(id,a,b,c,d,e)
df
Basically what I would like to do is to get the occurrences of numbers for each column (a,b,c,d,e) and for each id group (1,2,3) (for this latter grouping see my column ’id’).
So, for column ’a’ and for id number ’1’ (for the latter see column ’id’) the code would be something like this:
as.numeric(table(df[1:10,2]))
##The results are:
[1] 3 7
Just to briefly explain my results: in column ’a’ (and regarding only those records which have number ’1’ in column ’id’) we can say that number '1' occured 3 times and number '3' occured 7 times.
Again, just to show you another example. For column ’a’ and for id number ’2’ (for the latter grouping see again column ’id’):
as.numeric(table(df[11:20,2]))
##After running the codes the results are:
[1] 4 3 3
Let me explain a little again: in column ’a’ and regarding only those observations which have number ’2’ in column ’id’) we can say that number '1' occured 4 times, number '2' occured 3 times and number '3' occured 3 times.
So this is what I would like to do. Calculating the occurrences of numbers for each custom-defined subsets (and then collecting these values into a data frame). I know it is not a difficult task but the PROBLEM is that I’m gonna have to change the input ’df’ dataframe on a regular basis and hence both the overall number of rows and columns might change over time…
What I have done so far is that I have separated the ’df’ dataframe by columns, like this:
for (z in (2:ncol(df))) assign(paste("df",z,sep="."),df[,z])
So df.2 will refer to df$a, df.3 will equal df$b, df.4 will equal df$c etc. But I’m really stuck now and I don’t know how to move forward…
Is there a proper, ”automatic” way to solve this problem?
How about -
> library(reshape)
> dftab <- table(melt(df,'id'))
> dftab
, , value = 1
variable
id a b c d e
1 3 8 2 2 4
2 4 6 3 2 4
3 4 2 1 5 1
, , value = 2
variable
id a b c d e
1 0 1 4 3 3
2 3 3 3 6 2
3 1 4 5 3 4
, , value = 3
variable
id a b c d e
1 7 1 4 5 3
2 3 1 4 2 4
3 5 4 4 2 5
So to get the number of '3's in column 'a' and group '1'
you could just do
> dftab[3,'a',1]
[1] 4
A combination of tapply and apply can create the data you want:
tapply(df$id,df$id,function(x) apply(df[id==x,-1],2,table))
However, when a grouping doesn't have all the elements in it, as in 1a, the result will be a list for that id group rather than a nice table (matrix).
$`1`
$`1`$a
1 3
3 7
$`1`$b
1 2 3
8 1 1
$`1`$c
1 2 3
2 4 4
$`1`$d
1 2 3
2 3 5
$`1`$e
1 2 3
4 3 3
$`2`
a b c d e
1 4 6 3 2 4
2 3 3 3 6 2
3 3 1 4 2 4
$`3`
a b c d e
1 4 2 1 5 1
2 1 4 5 3 4
3 5 4 4 2 5
I'm sure someone will have a more elegant solution than this, but you can cobble it together with a simple function and dlply from the plyr package.
ColTables <- function(df) {
counts <- list()
for(a in names(df)[names(df) != "id"]) {
counts[[a]] <- table(df[a])
}
return(counts)
}
results <- dlply(df, "id", ColTables)
This gets you back a list - the first "layer" of the list will be the id variable; the second the table results for each column for that id variable. For example:
> results[['2']]['a']
$a
1 2 3
4 3 3
For id variable = 2, column = a, per your above example.
A way to do it is using the aggregate function, but you have to add a column to your dataframe
> df$freq <- 0
> aggregate(freq~a+id,df,length)
a id freq
1 1 1 3
2 3 1 7
3 1 2 4
4 2 2 3
5 3 2 3
6 1 3 4
7 2 3 1
8 3 3 5
Of course you can write a function to do it, so it's easier to do it frequently, and you don't have to add a column to your actual data frame
> frequency <- function(df,groups) {
+ relevant <- df[,groups]
+ relevant$freq <- 0
+ aggregate(freq~.,relevant,length)
+ }
> frequency(df,c("b","id"))
b id freq
1 1 1 8
2 2 1 1
3 3 1 1
4 1 2 6
5 2 2 3
6 3 2 1
7 1 3 2
8 2 3 4
9 3 3 4
You didn't say how you'd like the data. The by function might give you the output you like.
by(df, df$id, function(x) lapply(x[,-1], table))