将 DataFrame 除以 FUN(left, right) 中的滞后错误：二元运算符的非数字参数

Question

我有一个包含一些资产价格的数据框。我想通过将每列除以其滞后来获得 returns。

我有以下代码：

df=structure(list(A = structure(c(19060, 19061, 19062, 19065, 19066, 
                                  19067, 19068, 19069, 19072, 19073), class = "Date"), B = list(
                                    8359.87382308, 8361.93190783, 8364.50494403, 8368.10699178, 
                                    8372.3097342, 8376.12364242, 8380.18626964, 8384.41865593, 
                                    8387.38878791, 8390.92762529), C = list(48216.51809143, 48205.4923213, 
                                                                            48141.39976957, 48018.63247926, 48051.45532043, 48091.22412151, 
                                                                            48331.57538464, 48445.10388053, 48469.49369979, 48416.3347324), 
                  D = list(17.34859405, 17.34882281, 17.34760762, 17.35130834, 
                           17.36756919, 17.37318196, 17.39286354, 17.40391027, 17.40949275, 
                           17.41696558), E = list(43.16646812, 43.17551867, 43.16216113, 
                                                  43.17305155, 43.25924615, 43.33736433, 43.40667531, 43.45909204, 
                                                  43.55074062, 43.60648949), F = list(25.83271054, 25.84295702, 
                                                                                      25.83724976, 25.85374492, 25.94544031, 26.02025781, 26.07185775, 
                                                                                      26.1007222, 26.18709194, 26.23574844), G = list(28.510282, 
                                                                                                                                      28.525803, 28.539868, 28.55125, 28.563602, 28.576996, 
                                                                                                                                      28.596242, 28.613565, 28.629501, 28.646878), H = list(
                                                                                                                                        11.88687581, 11.89326214, 11.89119058, 11.88914284, 11.8977977, 
                                                                                                                                        11.90032756, 11.90797575, 11.91538244, 11.91918169, 11.92787376), 
                  I = list(29.45064815, 29.46287037, 29.47530093, 29.4877662, 
                           29.49959491, 29.5124632, 29.526301, 29.5401692, 29.5535557, 
                           29.5686575), J = list(15.61856767, 15.62738944, 15.63486662, 
                                                 15.61386311, 15.64214889, 15.64560406, 15.69008278, 15.73562823, 
                                                 15.7482602, 15.78539406)), row.names = c(NA, -10L), class = c("tbl_df", 
                                                                                                               "tbl", "data.frame"))


df2=mutate(df[,c(2:ncol(df))]/lag(df[,c(2:ncol(df))]))

但是我收到错误：FUN(left, right) 中的错误：二元运算符的非数字参数

我不知道为什么这会给我带来问题。

Answer 1

dput中的列都是list列。我们可以先 unnest 它们并遍历数字列 (across) 并除以 lag.

library(dplyr)
library(tidyr)
df2 <- df %>% 
    unnest(where(is.list)) %>% 
    mutate(across(where(is.numeric), ~ .x/lag(.x)))

-输出

df2
# A tibble: 10 × 10
   A              B      C     D     E     F     G     H     I      J
   <date>     <dbl>  <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>  <dbl>
 1 2022-03-09 NA    NA     NA    NA    NA    NA    NA    NA    NA    
 2 2022-03-10  1.00  1.00   1.00  1.00  1.00  1.00  1.00  1.00  1.00 
 3 2022-03-11  1.00  0.999  1.00  1.00  1.00  1.00  1.00  1.00  1.00 
 4 2022-03-14  1.00  0.997  1.00  1.00  1.00  1.00  1.00  1.00  0.999
 5 2022-03-15  1.00  1.00   1.00  1.00  1.00  1.00  1.00  1.00  1.00 
 6 2022-03-16  1.00  1.00   1.00  1.00  1.00  1.00  1.00  1.00  1.00 
 7 2022-03-17  1.00  1.00   1.00  1.00  1.00  1.00  1.00  1.00  1.00 
 8 2022-03-18  1.00  1.00   1.00  1.00  1.00  1.00  1.00  1.00  1.00 
 9 2022-03-21  1.00  1.00   1.00  1.00  1.00  1.00  1.00  1.00  1.00 
10 2022-03-22  1.00  0.999  1.00  1.00  1.00  1.00  1.00  1.00  1.00

可以使用 lag 一次将数据划分为多列，但我们不需要 mutate 即

df2 <- unnest(df, where(is.list))
df2[-1] <- df2[-1]/lag(df2[-1])

将 DataFrame 除以 FUN(left, right) 中的滞后错误：二元运算符的非数字参数

Divide DataFrame by its lag Error in FUN(left, right) : non-numeric argument to binary operator

r

dataframe

dplyr