Question

我有下面的tibble，我想从中创建第4列，它是来自A，B和B的联合载体。 C.我知道dplyr :: unite（）可以创建一个新的字符向量，但是我希望用向量创建一个列表列。

现在rowwise工作，但不保持输入tibble。有关将A_Vector列保存到C_Vector的建议吗？

以下是代码：

library(tidyverse)

My_Data <- tibble(A_Vector = rnorm(10),
                  B_Vector = rnorm(10),
                  C_Vector = rnorm(10)) %>% 
           rowwise() %>% 
           do(Port_Weights = matrix(c(.$A_Vector,.$B_Vector,.$C_Vector),3,1))

结果：

Source: local data frame [10 x 1]
Groups: <by row>

# A tibble: 10 x 1
    Port_Weights
 *        <list>
 1 <dbl [3 x 1]>
 2 <dbl [3 x 1]>
 3 <dbl [3 x 1]>
 4 <dbl [3 x 1]>
 5 <dbl [3 x 1]>
 6 <dbl [3 x 1]>
 7 <dbl [3 x 1]>
 8 <dbl [3 x 1]>
 9 <dbl [3 x 1]>
10 <dbl [3 x 1]>

这不起作用：

My_Data <- tibble(A_Vector = rnorm(10),
                  B_Vector = rnorm(10),
                  C_Vector = rnorm(10)) %>% 
  mutate(Port_Weights = rowwise() %>% do(matrix(c(.$A_Vector,.$B_Vector,.$C_Vector),3,1)))

长版本，显然没有意义：

My_Data <- tibble(A_Vector = rnorm(10),
                  B_Vector = rnorm(10),
                  C_Vector = rnorm(10))

Data_Unite <-   My_Data %>% 
  rowwise() %>% 
  do(Port_Weights = matrix(c(.$A_Vector,.$B_Vector,.$C_Vector),3,1))

My_Data <- as.tibble(cbind(My_Data,Data_Unite))

但确实提供了追捧的结果：

# A tibble: 10 x 4
      A_Vector   B_Vector   C_Vector  Port_Weights
 *       <dbl>      <dbl>      <dbl>        <list>
 1 -1.23504457 -0.3750408 -0.4214122 <dbl [3 x 1]>
 2 -0.90678699  0.5261914  1.1191229 <dbl [3 x 1]>
 3 -0.62944085  0.5995529  0.2096462 <dbl [3 x 1]>
 4  2.06171633  1.5399094  2.2972950 <dbl [3 x 1]>
 5  0.08761555  0.1424207 -1.4758585 <dbl [3 x 1]>
 6 -1.07334432 -1.9112787  0.4820864 <dbl [3 x 1]>
 7 -0.18655423 -1.3698855  0.6672621 <dbl [3 x 1]>
 8 -0.97961789 -0.8194373 -0.4158516 <dbl [3 x 1]>
 9  0.68112936 -1.9864507  1.0193449 <dbl [3 x 1]>
10  0.61455438  0.5885380 -1.0925312 <dbl [3 x 1]>

Answer 1

数据：

library(tidyverse)

my_tibble <- tibble(A_Vector = rnorm(10),
                    B_Vector = rnorm(10),
                    C_Vector = rnorm(10))

要向数据框添加列，请使用mutate代替do，并使用Map并行循环显示三个向量，并构造每行的矩阵：

my_tibble %>% 
   mutate(Port_Weights = Map(function(...) matrix(c(...), 3, 1), A_Vector, B_Vector, C_Vector))

# A tibble: 10 x 4
#      A_Vector   B_Vector    C_Vector  Port_Weights
#         <dbl>      <dbl>       <dbl>        <list>
# 1  0.62674726 -0.5432169 -1.66763618 <dbl [3 x 1]>
# 2 -0.47346722 -0.4436020 -1.04892634 <dbl [3 x 1]>
# 3  0.19059238 -1.6733052  2.79275828 <dbl [3 x 1]>
# 4 -0.23501873 -1.1664704 -0.19324676 <dbl [3 x 1]>
# 5  0.66552642 -1.3328070 -1.53575954 <dbl [3 x 1]>
# 6 -0.41251920 -0.2056882  1.66537220 <dbl [3 x 1]>
# 7  0.48396052  0.3968486  0.16110407 <dbl [3 x 1]>
# 8  0.43035213 -0.6433268  1.61640228 <dbl [3 x 1]>
# 9  0.06747126 -1.0146385 -0.47824193 <dbl [3 x 1]>
#10  0.79916411 -1.2349901 -0.05151402 <dbl [3 x 1]>

如果元素不必是矩阵：

my_tibble %>% mutate(Port_Weights = Map(c, A_Vector, B_Vector, C_Vector))

相当于（与data.table::transpose）：

my_tibble %>% mutate(Port_Weights = data.table::transpose(as.list(.)))

Answer 2

由于您使用的是tidyverse，因此您还可以考虑pmap包中的purrr功能，该功能是tidyverse的一部分。

set.seed(123)

library(tidyverse)

My_Data <- tibble(A_Vector = rnorm(10),
                  B_Vector = rnorm(10),
                  C_Vector = rnorm(10))

My_Data2 <- My_Data %>%
  mutate(Port_Weights = pmap(.l = list(A_Vector, B_Vector, C_Vector),
                             .f = function(x, y, z) matrix(c(x, y, z), 3, 1)))

My_Data2
# A tibble: 10 x 4
      A_Vector   B_Vector   C_Vector  Port_Weights
         <dbl>      <dbl>      <dbl>        <list>
 1 -0.56047565  1.2240818 -1.0678237 <dbl [3 x 1]>
 2 -0.23017749  0.3598138 -0.2179749 <dbl [3 x 1]>
 3  1.55870831  0.4007715 -1.0260044 <dbl [3 x 1]>
 4  0.07050839  0.1106827 -0.7288912 <dbl [3 x 1]>
 5  0.12928774 -0.5558411 -0.6250393 <dbl [3 x 1]>
 6  1.71506499  1.7869131 -1.6866933 <dbl [3 x 1]>
 7  0.46091621  0.4978505  0.8377870 <dbl [3 x 1]>
 8 -1.26506123 -1.9666172  0.1533731 <dbl [3 x 1]>
 9 -0.68685285  0.7013559 -1.1381369 <dbl [3 x 1]>
10 -0.44566197 -0.4727914  1.2538149 <dbl [3 x 1]>

来自其他向量列的新列表列，带有dplyr和rowwise

2 个答案: