# Joining factors with different levels, coercing to character vector.

1 vote
777 views

## Problem :

I am very new to R. Now a days if I try to use the join function in a dplyr package then I often face following warning:

Warning message:
In left_join_impl(x, y, by$x, by$y) :
joining factors with different levels, coercing to character vector
I am unable to find any relevant information about above warning. I need expert help from R expert to fix above warning.

1 vote

## Solution :

I have good exposure to R so I am trying to help you. The above warning is clearly informing you that one of the columns which are being used in your join is clearly a factor and also that factor had clearly the different levels in your different datasets. So if you do not want to lose any of your information then your factors should be converted to the character values. Please follow below example:

library(dplyr)
x<-data.frame(a=letters[1:7])
y<-data.frame(a=letters[4:10])

class(x$a) # [1] "factor" # NOTE these are very different levels(x$a)
# [1] "a" "b" "c" "d" "e" "f" "g"
levels(y$a) # [1] "d" "e" "f" "g" "h" "i" "j" m <- left_join(x,y) # Joining by: "a" # Warning message: # joining factors with different levels, coercing to character vector class(m$a)
# [1] "character"

You must make sure that both of your factors have a same level before merging them

combined <- sort(union(levels(x$a), levels(y$a)))
n <- left_join(mutate(x, a=factor(a, levels=combined)),
mutate(y, a=factor(a, levels=combined)))
# Joining by: "a"
class(n$a) #[1] "factor" Above solution will surely fix your problem related to R. ## Related questions 0 votes 1 answer 35 views 35 views Problem: Why I have got this issue>data and reference should be factors with the same levels? 0 votes 1 answer 3 views 3 views Problem: I am unable to find a solution for this > The first argument to input must be a character vector. 1 vote 1 answer 3 views 3 views Problem: I&rsquo;m facing a problem while running a mixed effect model. Sample data: data4<-structure(list(code = structure(1:10, .Label = c("10888", "10889", "10890", "10891", "10892", "10893", "10894", "10896", "10897", "10898", "10899", " ... , data4) Error message: Error in checkNlevels(reTrms$flist, n = n, control) : number of levels of each grouping factor must be < number of observations
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