为什么替换函数对数据帧有效,但对r中的小滴无效?

时间:2019-09-12 10:10:34

标签: r dataframe tibble

我查看了讨论What is the difference between as.tibble(), as_data_frame(), and tbl_df()?,以弄清楚replace_na函数(如下所示)为什么适用于数据帧而不适用于小标题。您能帮我理解为什么它对小球不起作用吗?如何修改功能,使其同时适用于data.frametibble

数据

library(dplyr)

#dput(df1)
df1 <- structure(list(id = c(1, 2, 3, 4), gender = c("M", "F", NA, "F"
), grade = c("A", NA, NA, NA), age = c(2, NA, 2, NA)), row.names = c(NA, 
-4L), class = c("tbl_df", "tbl", "data.frame"))

#dput(df2)
df2 <- structure(list(id = c(1, 2, 3, 4), gender = c("M", "F", "M", 
"F"), grade = c("A", "A", "B", "NG"), age = c(22, 23, 21, 19)), row.names = c(NA, 
-4L), class = c("tbl_df", "tbl", "data.frame"))

替换功能

replace_na <- function(df_to, df_from) {
  replace(df_to, is.na(df_to), df_from[is.na(df_to)])
}

用法

replace_na(df1,df2)

  

错误:必须在[中使用向量,而不是类矩阵的对象。

     

致电rlang::last_error()查看回溯

     

从以下位置调用:abort(error_dim_column_index(j))

但是;将arglist强制为数据帧会产生所需的输出,如下所示。

replace_na(as.data.frame(df1), as.data.frame(df2))
#   id gender grade age
# 1  1      M     A   2
# 2  2      F     A  23
# 3  3      M     B   2
# 4  4      F    NG  19

谢谢。

1 个答案:

答案 0 :(得分:3)

is.na()返回数据帧的逻辑矩阵:

is.na(df1) 
#>         id gender grade   age
#> [1,] FALSE  FALSE FALSE FALSE
#> [2,] FALSE  FALSE  TRUE  TRUE
#> [3,] FALSE   TRUE  TRUE FALSE
#> [4,] FALSE  FALSE  TRUE  TRUE

基础data.frame类支持矩阵子集; tbl_df更为严格,但没有。

as.data.frame(df2)[is.na(df1)]
#> [1] "M"  "A"  "B"  "NG" "23" "19"
df2[is.na(df1)]
#> Must use a vector in `[`, not an object of class matrix.

要使replace_na()函数与tbl_df一起使用,您需要对每一列分别进行操作。例如,使用递归:

replace_na <- function(x, y) {
  if (is.data.frame(x)) {
    x[] <- Map(replace_na, x, y)
    return(x)
  }

  replace(x, is.na(x), y[is.na(x)])
}

replace_na(df1, df2)
#> # A tibble: 4 x 4
#>      id gender grade   age
#>   <dbl> <chr>  <chr> <dbl>
#> 1     1 M      A         2
#> 2     2 F      A        23
#> 3     3 M      B         2
#> 4     4 F      NG       19

此方法通常也更快:

replace_na_vec <- function(x, y) {
  replace(x, is.na(x), y[is.na(x)])
}

df1_10k <- do.call("rbind", replicate(10000, df1, simplify = FALSE))
df2_10k <- do.call("rbind", replicate(10000, df2, simplify = FALSE))

bench::mark(
  check = FALSE,
  new = replace_na(df1, df2),
  old = replace_na_vec(as.data.frame(df1), as.data.frame(df2)),
  new_10k = replace_na(df1_10k, df2_10k),
  old_10k = replace_na_vec(as.data.frame(df1_10k), as.data.frame(df2_10k))
)
#> # A tibble: 4 x 6
#>   expression      min   median `itr/sec` mem_alloc `gc/sec`
#>   <bch:expr> <bch:tm> <bch:tm>     <dbl> <bch:byt>    <dbl>
#> 1 new         74.01us  97.79us   7295.          0B    12.6 
#> 2 old        269.97us 529.93us   1845.     81.02KB     8.23
#> 3 new_10k      1.82ms   2.75ms    338.      4.27MB    32.3 
#> 4 old_10k     94.29ms 104.05ms      9.68   10.24MB     2.42

reprex package(v0.3.0)于2019-09-12创建