如何迭代 R 中因子的水平？

Question

I would like to create a function that helps me to identify possible mistakes in the levels of a factor by accessing the first letter, so first I am focused on the identification part.

Data Frame '''

alleles<-(c('A*24:02', 'A*11:01', 'blank',  'A*31:01'))
as.factor(alleles)
freq<-c(0.3782, 0.4209, 0.0362, 0.0761)

df<-data.frame(alleles, freq)

'''

My attempt _ '''

for(i in df$alleles){
  if (i != 'A'){
    can<-c()
    append(can, i)
    df$alleles<-df$alleles[-c(can)]
  }
}

''' Error message Error in -c(can) : invalid argument to unary operator

Observations If I do '''print(can)''' the output is "NULL" meaning that it is not working the use of "append".

Answer 1

You can also try:

#Data
alleles<-(c('A*24:02', 'A*11:01', 'blank',  'A*31:01'))
freq<-c(0.3782, 0.4209, 0.0362, 0.0761)
df<-data.frame(alleles, freq)
can<-c()
#Check
for(i in 1:length(df$alleles))
{
  if (substr(df$alleles[i],1,1) != 'A'){
    can <- c(can, as.character(df$alleles[i]))
  }
}
#Apply
df<-df[-which(df$alleles %in% can),]

Output:

df
  alleles   freq
1 A*24:02 0.3782
2 A*11:01 0.4209
4 A*31:01 0.0761

Answer 2

为什么不直接使用正则表达式？

df[grepl("^A", df$alleles),]

Answer 3

我们可以使用grep

df[grep("^A", df$alleles),]