在apply

Question

我有这样的事情：

#data.table
#      a b
#aland 1 2
#bland 3 4

freq_all = read.table(file='data.table', header=T,stringsAsFactors = FALSE)
country_names = rownames(freq_all)
blood_types = colnames(freq_all)

func <- function(country,type) {paste(country, type)}

newfr <- freq_all
for (country in country_names){
    for (type in blood_types){
        newfr[country, type] <- func(country, type)
    }
}

我想知道我是否可以使用apply()函数或类似的东西。

Answer 1

我们可以使用outer

freq1 <- freq_all
freq1[] <- outer(rownames(freq_all), colnames(freq_all), FUN= paste)
freq1
#           a       b
#aland aland a aland b
#bland bland a bland b

identical(freq1, newfr)
#[1] TRUE

数据

freq_all <- structure(list(a = c(1L, 3L), b = c(2L, 4L)), .Names = c("a", 
  "b"), class = "data.frame", row.names = c("aland", "bland"))