Question

我正在尝试使用map2来比较嵌套的tibble列。这是我的数据格式：

> tbl
# A tibble: 3 x 3
  ID    data.x           data.y          
  <chr> <list>           <list>          
1 a     <tibble [2 x 2]> <tibble [2 x 2]>
2 b     <tibble [2 x 2]> <tibble [2 x 2]>
3 c     <tibble [2 x 2]> <tibble [2 x 2]>

从列名的角度来看，data.x中的数据和data.y是相同的，值可能不同。我想从val列获得最大价值。我认为这可行，但只返回data.x的max。我并不完全了解map2的工作原理。

tbl %>%
  mutate(col1 = map2_dbl(data.x, data.y, ~ max(.$val)))

结果应为：

# A tibble: 3 x 4
  ID    data.x           data.y            col1
  <chr> <list>           <list>           <dbl>
1 a     <tibble [2 x 2]> <tibble [2 x 2]>    7.
2 b     <tibble [2 x 2]> <tibble [2 x 2]>    8.
3 c     <tibble [2 x 2]> <tibble [2 x 2]>    8.

数据：

> dput(tbl)
structure(list(ID = c("a", "b", "c"), data.x = list(structure(list(
    text = c("Y", "Y"), val = c(1, 1)), .Names = c("text", "val"
), row.names = c(NA, -2L), class = c("tbl_df", "tbl", "data.frame"
)), structure(list(text = c("N", "N"), val = c(2, 2)), .Names = c("text", 
"val"), row.names = c(NA, -2L), class = c("tbl_df", "tbl", "data.frame"
)), structure(list(text = c("Y", "Y"), val = c(3, 3)), .Names = c("text", 
"val"), row.names = c(NA, -2L), class = c("tbl_df", "tbl", "data.frame"
))), data.y = list(structure(list(text = c("Y", "Y"), val = c(6, 
7)), .Names = c("text", "val"), row.names = c(NA, -2L), class = c("tbl_df", 
"tbl", "data.frame")), structure(list(text = c("Y", "Y"), val = c(8, 
6)), .Names = c("text", "val"), row.names = c(NA, -2L), class = c("tbl_df", 
"tbl", "data.frame")), structure(list(text = c("N", "N"), val = c(7, 
8)), .Names = c("text", "val"), row.names = c(NA, -2L), class = c("tbl_df", 
"tbl", "data.frame")))), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -3L), .Names = c("ID", "data.x", "data.y"
))

Answer 1

根据预期的输出，我们正在提取“＆＃39; val＆＃39;两个＆＃39; data.x＆＃39; data.frame中的列和＆＃39; data.y＆＃39; lists，将它连接在一起（c）并获取max值

tbl %>% 
    mutate(col1 = map2_dbl(data.x, data.y, ~ max(c(.x$val, .y$val))))
# A tibble: 3 x 4     
#    ID    data.x           data.y            col1
#   <chr> <list>           <list>           <dbl>
#1 a     <tibble [2 x 2]> <tibble [2 x 2]>  7.00
#2 b     <tibble [2 x 2]> <tibble [2 x 2]>  8.00
#3 c     <tibble [2 x 2]> <tibble [2 x 2]>  8.00

对于多列，可以使用＆＃39;数据＆＃39;，pmap

tbl %>%
    mutate(col1 = pmap_dbl(.[-1], ~ max(c(..1$val, ..2$val))))

R嵌套的tibble map2比较

1 个答案: