Question

我列出了下面结构的数十个对象。我想删除每个文件的前三行，只保留前三列。

    x<-structure(c("", "Service", "", "HR", "", "", "Function", "", "Code", 
                "X1", "", "", "", "", "", "Doe, John", "Roe, Jane", 
                "Doe, Jane", "", "Full name", "", "", "", ""), .Dim = c(6L, 4L))

    y<-structure(c("", "Service", "", "IT", "", "", "Function", "", "Code", 
                "X2", "", "", "", "", "", "Doe, Johnny", "Roe, Janette", 
                "Doe, Janette", "", "Full name", "", "", "", ""), .Dim = c(6L, 4L))

    z<-structure(c("", "Service", "", "RD", "", "", "Function", "", "Code", 
                "X3", "", "", "", "", "", "Doe, Johnny", "Roe, Janette", 
                "Roe, Johnny", "", "Full name", "", "", "", ""), .Dim = c(6L, 4L))

l<-list(x,y,z)

如何使用length(l)概括列表中列出的所有对象的内容？

length(l)

x<-bind_rows(
tbl_df(l[[1]][-c(1,2,3),c(1,2,3)]),
tbl_df(l[[2]][-c(1,2,3),c(1,2,3)]),
tbl_df(l[[3]][-c(1,2,3),c(1,2,3)])
)

Answer 1

bind_rows可以接受数据框列表;因此，您可以循环遍历列表，对矩阵进行子集化并将它们转换为数据帧，然后bind_rows应该对结果列表起作用：

library(dplyr)
bind_rows(lapply(l, function(mat) tbl_df(mat[-(1:3), 1:3])))

# A tibble: 9 x 3
#     V1    V2           V3
#  <chr> <chr>        <chr>
#1    HR    X1    Doe, John
#2                Roe, Jane
#3                Doe, Jane
#4    IT    X2  Doe, Johnny
#5             Roe, Janette
#6             Doe, Janette
#7    RD    X3  Doe, Johnny
#8             Roe, Janette
#9              Roe, Johnny

另一种选择是使用do.call(rbind, ...)导致矩阵（如果你不关心矩阵或数据框）：

do.call(rbind, lapply(l, `[`, -(1:3), 1:3))

#      [,1] [,2] [,3]          
# [1,] "HR" "X1" "Doe, John"   
# [2,] ""   ""   "Roe, Jane"   
# [3,] ""   ""   "Doe, Jane"   
# [4,] "IT" "X2" "Doe, Johnny" 
# [5,] ""   ""   "Roe, Janette"
# [6,] ""   ""   "Doe, Janette"
# [7,] "RD" "X3" "Doe, Johnny" 
# [8,] ""   ""   "Roe, Janette"
# [9,] ""   ""   "Roe, Johnny"

R - 使用bind_rows作为子集化对象的列表

1 个答案: