R 划分两个矩阵列表,其中一个列表中的第i个矩阵元素除以第二个列表中的第i个矩阵元素
我有两个矩阵列表。以下是它们的结构示例:R 划分两个矩阵列表,其中一个列表中的第i个矩阵元素除以第二个列表中的第i个矩阵元素,r,arrays,list,matrix,R,Arrays,List,Matrix,我有两个矩阵列表。以下是它们的结构示例: list1<- list(structure(c(1, 2, 7, 1, 3, 0, 0, 0, 1, 4, 1, 3, 2, 3, 4, 6, 0, 0, 0, 3, 3), .Dim = c(7L, 3L), .Dimnames = list(c("lepA", "lepB", "lepC", "lepD", "lepE", "lepF", "lepG"), NULL)), structure(c(1, 3, 7, 1, 3, 2, 3,
list1<- list(structure(c(1, 2, 7, 1, 3, 0, 0, 0, 1, 4, 1, 3, 2, 3, 4,
6, 0, 0, 0, 3, 3), .Dim = c(7L, 3L), .Dimnames = list(c("lepA",
"lepB", "lepC", "lepD", "lepE", "lepF", "lepG"), NULL)), structure(c(1,
3, 7, 1, 3, 2, 3, 4, 6, 4, 1, 3, 3, 3), .Dim = c(7L, 2L), .Dimnames = list(
c("lepA", "lepB", "lepC", "lepD", "lepE", "lepF", "lepG"),
NULL)), structure(c(5, 8, 7, 1, 3, 3, 3), .Dim = c(7L, 1L
), .Dimnames = list(c("lepA", "lepB", "lepC", "lepD", "lepE",
"lepF", "lepG"), NULL)))
list2<-list(structure(c(6, 1, 51, 13, 15, 0, 0, 0, 6, 50, 13, 15, 6,
5, 5, 9, 0, 0, 0, 7, 5), .Dim = c(7L, 3L), .Dimnames = list(c("lepA",
"lepB", "lepC", "lepD", "lepE", "lepF", "lepG"), NULL)), structure(c(6,
7, 51, 13, 15, 6, 5, 5, 9, 50, 13, 15, 7, 5), .Dim = c(7L, 2L
), .Dimnames = list(c("lepA", "lepB", "lepC", "lepD", "lepE",
"lepF", "lepG"), NULL)), structure(c(11, 10, 51, 13, 15, 7, 5
), .Dim = c(7L, 1L), .Dimnames = list(c("lepA", "lepB", "lepC",
"lepD", "lepE", "lepF", "lepG"), NULL)))
使用
映射
:
Map(`/`, list1, list2)
#[[1]]
# [,1] [,2] [,3]
#lepA 0.16666667 NaN 0.8000000
#lepB 2.00000000 0.16666667 0.6666667
#lepC 0.13725490 0.08000000 NaN
#lepD 0.07692308 0.07692308 NaN
#lepE 0.20000000 0.20000000 NaN
#lepF NaN 0.33333333 0.4285714
#lepG NaN 0.60000000 0.6000000
#[[2]]
# [,1] [,2]
#lepA 0.16666667 0.80000000
#lepB 0.42857143 0.66666667
#lepC 0.13725490 0.08000000
#lepD 0.07692308 0.07692308
#lepE 0.20000000 0.20000000
#lepF 0.33333333 0.42857143
#lepG 0.60000000 0.60000000
#[[3]]
# [,1]
#lepA 0.45454545
#lepB 0.80000000
#lepC 0.13725490
#lepD 0.07692308
#lepE 0.20000000
#lepF 0.42857143
#lepG 0.60000000
或purr中的map2
purrr::map2(list1, list2, `/`)
我们可以使用seq\u和lappy
lapply(seq_along(list1), function(i) list1[[i]]/list2[[i]])
mapply(“/”,list1,list2)
作为编码风格建议(请求):请不要在名为list
、list1
和list2
的变量旁边使用函数list
。这在语法上当然是合法的,但在我看来这很难。你在处理列表时使用什么语法?
purrr::map2(list1, list2, `/`)
lapply(seq_along(list1), function(i) list1[[i]]/list2[[i]])