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[英]Unquoting fails to find variable in mutate and map2 when renaming column of data in nested tibble R
[英]R nested tibble map2 comparisons
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]>
data.x和data.y中的元素从列名的角度来看是相同的,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(.$val)))
结果应该是:
# A tibble: 3 x 4
ID data.x data.y col1
<chr> <list> <list> <dbl>
1 a <tibble [2 x 2]> <tibble [2 x 2]> 7.
2 b <tibble [2 x 2]> <tibble [2 x 2]> 8.
3 c <tibble [2 x 2]> <tibble [2 x 2]> 8.
数据:
> dput(tbl)
structure(list(ID = c("a", "b", "c"), data.x = list(structure(list(
text = c("Y", "Y"), val = c(1, 1)), .Names = c("text", "val"
), row.names = c(NA, -2L), class = c("tbl_df", "tbl", "data.frame"
)), structure(list(text = c("N", "N"), val = c(2, 2)), .Names = c("text",
"val"), row.names = c(NA, -2L), class = c("tbl_df", "tbl", "data.frame"
)), structure(list(text = c("Y", "Y"), val = c(3, 3)), .Names = c("text",
"val"), row.names = c(NA, -2L), class = c("tbl_df", "tbl", "data.frame"
))), data.y = list(structure(list(text = c("Y", "Y"), val = c(6,
7)), .Names = c("text", "val"), row.names = c(NA, -2L), class = c("tbl_df",
"tbl", "data.frame")), structure(list(text = c("Y", "Y"), val = c(8,
6)), .Names = c("text", "val"), row.names = c(NA, -2L), class = c("tbl_df",
"tbl", "data.frame")), structure(list(text = c("N", "N"), val = c(7,
8)), .Names = c("text", "val"), row.names = c(NA, -2L), class = c("tbl_df",
"tbl", "data.frame")))), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -3L), .Names = c("ID", "data.x", "data.y"
))
根据预期的输出,我们从'data.x'和'data.y' lists
中提取data.frame
的'val'列,将它连接在一起( c
)并获得max
tbl %>%
mutate(col1 = map2_dbl(data.x, data.y, ~ max(c(.x$val, .y$val))))
# A tibble: 3 x 4
# ID data.x data.y col1
# <chr> <list> <list> <dbl>
#1 a <tibble [2 x 2]> <tibble [2 x 2]> 7.00
#2 b <tibble [2 x 2]> <tibble [2 x 2]> 8.00
#3 c <tibble [2 x 2]> <tibble [2 x 2]> 8.00
对于“数据”的多个列,可以使用pmap
tbl %>%
mutate(col1 = pmap_dbl(.[-1], ~ max(c(..1$val, ..2$val))))
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