Question

BI正在尝试将矩阵dataa的每一列除以向量mu_i中的一个元素。

> dput(dataa)
structure(list(`175` = c(2.04, 1.91, 2, 1.92, 1.85, 1.96, 1.88, 
1.9), `194` = c(1.66, 1.71, 1.42, 1.76, 1.66, 1.61, 1.55, 1.66
), `213` = c(1.53, 1.54, 1.38, 1.31, 1.35, 1.27, 1.26, 1.38), 
    `231` = c(1.15, 1.22, 1.17, 1.16, 1.21, 1.28, 1.17, 0), `250` = c(1.26, 
    0.83, 1.08, 1.02, 1.09, 1.06, 0, 0)), .Names = c("175", "194", 
"213", "231", "250"), row.names = c(NA, -8L), class = "data.frame")
> dput(mu_i)
c(1.9325, 1.62875, 1.3775, 1.19428571428571, 1.05666666666667
)

基本上，我想要的是dataa[,i] / mu_i[i]并从1到5迭代i。（即{i} ^ $ $ dataa列除以$ i ^ th $元素mu_i）我知道我可以执行for循环，但我想知道如何使用apply函数完成此操作？

更新：我很感激你的所有答案。我只想问一个问题：我试图使用像这样的mapply函数来做到这一点 mapply(function(d,m,i)d[,i] - m[i],dataa,mu_i,seq_along(mu_i) )。我不明白为什么会失败。出于某些原因，我需要将dataa和mu_i作为参数传递给apply函数中的函数。我怎么能这样做？

Answer 1

这是sweep的用途

sweep(dataa, 2, mu_i, "/")

Answer 2

您可以在不使用apply系列函数的情况下执行此操作。尝试transpose dataa，然后除以mu_i

  t(t(dataa)/mu_i) #returns a matrix output
  #          175       194       213       231       250
  #[1,] 1.0556274 1.0191865 1.1107078 0.9629187 1.1924290
  #[2,] 0.9883571 1.0498849 1.1179673 1.0215311 0.7854890
  #[3,] 1.0349288 0.8718342 1.0018149 0.9796651 1.0220820
  #[4,] 0.9935317 1.0805833 0.9509982 0.9712919 0.9652997
  #[5,] 0.9573092 1.0191865 0.9800363 1.0131579 1.0315457
  #[6,] 1.0142303 0.9884881 0.9219601 1.0717703 1.0031546
  #[7,] 0.9728331 0.9516500 0.9147005 0.9796651 0.0000000
  #[8,] 0.9831824 1.0191865 1.0018149 0.0000000 0.0000000

或者

  dataa/mu_i[col(dataa)] #returns a data.frame output
  #       175       194       213       231       250
  #1 1.0556274 1.0191865 1.1107078 0.9629187 1.1924290
  #2 0.9883571 1.0498849 1.1179673 1.0215311 0.7854890
  #3 1.0349288 0.8718342 1.0018149 0.9796651 1.0220820
  #4 0.9935317 1.0805833 0.9509982 0.9712919 0.9652997
  #5 0.9573092 1.0191865 0.9800363 1.0131579 1.0315457
  #6 1.0142303 0.9884881 0.9219601 1.0717703 1.0031546
  #7 0.9728331 0.9516500 0.9147005 0.9796651 0.0000000
  #8 0.9831824 1.0191865 1.0018149 0.0000000 0.0000000

或者如果您需要基于apply的解决方案

  mapply("/", dataa, mu_i) #returns a matrix
  #         175       194       213       231       250
  #[1,] 1.0556274 1.0191865 1.1107078 0.9629187 1.1924290
  #[2,] 0.9883571 1.0498849 1.1179673 1.0215311 0.7854890
  #[3,] 1.0349288 0.8718342 1.0018149 0.9796651 1.0220820
  #[4,] 0.9935317 1.0805833 0.9509982 0.9712919 0.9652997
  #[5,] 0.9573092 1.0191865 0.9800363 1.0131579 1.0315457
  #[6,] 1.0142303 0.9884881 0.9219601 1.0717703 1.0031546
  #[7,] 0.9728331 0.9516500 0.9147005 0.9796651 0.0000000
  #[8,] 0.9831824 1.0191865 1.0018149 0.0000000 0.0000000

如果您需要基于data.frame解决方案的apply输出

  databa <- dataa
  databa[] <- Map("/", dataa, mu_i) #returns `data.frame`

Answer 3

您可以使用sapply

sapply(1:5, function(i) dataa[,i]/mu_i[i])
          [,1]      [,2]      [,3]      [,4]      [,5]
[1,] 1.0556274 1.0191865 1.1107078 0.9629187 1.1924290
[2,] 0.9883571 1.0498849 1.1179673 1.0215311 0.7854890
[3,] 1.0349288 0.8718342 1.0018149 0.9796651 1.0220820
[4,] 0.9935317 1.0805833 0.9509982 0.9712919 0.9652997
[5,] 0.9573092 1.0191865 0.9800363 1.0131579 1.0315457
[6,] 1.0142303 0.9884881 0.9219601 1.0717703 1.0031546
[7,] 0.9728331 0.9516500 0.9147005 0.9796651 0.0000000
[8,] 0.9831824 1.0191865 1.0018149 0.0000000 0.0000000

R：哪个应用函数用于通过矩阵和向量应用？

3 个答案: