Question

我有两个数组

chemArr = c("water","oil",...(lots of values omitted)...,"hydrogen")
timeArr = c("0", "8hr", "12hr", "24hr", "36hr", "48hr", "72hr")

我想构建一个数据框

chem  0               8hr               ... 72hr
water f("water", "0") f("water", "8hr") ...
...

其中f是我写的函数。有没有一种很好的方法在R中做到这一点？

在我的特定情况下，使用chem函数并每次返回一列更有效，因为每个函数共享计算。但由于所需的总时间很短，如果它更方便的话，我可以用其他方式做。

Answer 1

说明评论中讨论的解决方案：

one <- letters[1:3]
two <- letters[4:6]

one
[1] "a" "b" "c"
two
[1] "d" "e" "f"

mapply(paste0, one, two)
   a    b    c 
"ad" "be" "cf" 

mapply(paste0, sort(rep(one, length(two))), two)
   a    a    a    b    b    b    c    c    c 
"ad" "ae" "af" "bd" "be" "bf" "cd" "ce" "cf" 

mapply(paste0, one, list(two)) # courtesy of @thelatemail
     a    b    c   
[1,] "ad" "bd" "cd"
[2,] "ae" "be" "ce"
[3,] "af" "bf" "cf"

outer(one, two, paste0) # courtesy of @akrun
     [,1] [,2] [,3]
[1,] "ad" "ae" "af"
[2,] "bd" "be" "bf"
[3,] "cd" "ce" "cf"

Answer 2

我建议使用expand.grid，这会创建一个＆＃34;长形式＆＃34;所有双向组合，然后使用mapply从函数创建值：

chemArr = c("water","oil","hydrogen")
timeArr = c("0", "8hr", "12hr", "24hr", "36hr", "48hr", "72hr")
mygrid <- expand.grid(chemArr, timeArr)
mygrid <- expand.grid(chems = chemArr, times = timeArr)
str(mygrid)
#'data.frame':  21 obs. of  2 variables:
# $ chems: Factor w/ 3 levels "water","oil",..: 1 2 3 1 2 3 1 2 3 1 ...
# $ times: Factor w/ 7 levels "0","8hr","12hr",..: 1 1 1 2 2 2 3 3 3 4 ...
# - attr(*, "out.attrs")=List of 2
#  ..$ dim     : Named int  3 7
#  .. ..- attr(*, "names")= chr  "chems" "times"
#  ..$ dimnames:List of 2
#  .. ..$ chems: chr  "chems=water" "chems=oil" "chems=hydrogen"
#  .. ..$ times: chr  "times=0" "times=8hr" "times=12hr" "times=24hr" ... 

mygrid$f_value <- mapply(f, mygrid$chems, mygrid$times)

二维适用于R？

2 个答案: