我有一个数据帧,如下所示:
df <- tibble::rownames_to_column(USArrests, "State") %>%
tidyr::pivot_longer(cols = -State)
head(df)
# A tibble: 6 x 3
State name value
<chr> <chr> <dbl>
1 Alabama Murder 13.2
2 Alabama Assault 236
3 Alabama UrbanPop 58
4 Alabama Rape 21.2
5 Alaska Murder 10
6 Alaska Assault 263
在一个单独的列表对象中l
我有列,我需要从数据帧中删除这些列。元素名称是列名称,值对应于我要删除的行:
l <- list(State = c("Alabama", "Pennsylvania", "Texas"),
name = c("Murder", "Assault"))
硬编码它会这样做:
dplyr::filter(df, !State %in% c("Alabama", "Pennsylvania", "Texas"), !name %in% c("Murder", "Assault"))
State name value
<chr> <chr> <dbl>
1 Alaska UrbanPop 48
2 Alaska Rape 44.5
3 Arizona UrbanPop 80
4 Arizona Rape 31
5 Arkansas UrbanPop 50
6 Arkansas Rape 19.5
7 California UrbanPop 91
8 California Rape 40.6
9 Colorado UrbanPop 78
10 Colorado Rape 38.7
# ... with 84 more rows
但是,l
经常更改,所以我不能/不想硬编码。我尝试了以下操作,但只计算了最后一个表达式:
library(purrr)
filter_expr <- imap_chr(l, ~ paste0("! ",
.y,
" %in% c("",
paste(.x, collapse = "",""),
"")")) %>% parse(text = .)
filter(df, eval(filter_expr))
State name value
<chr> <chr> <dbl>
1 Alabama UrbanPop 58
2 Alabama Rape 21.2
3 Alaska UrbanPop 48
4 Alaska Rape 44.5
5 Arizona UrbanPop 80
6 Arizona Rape 31
7 Arkansas UrbanPop 50
8 Arkansas Rape 19.5
9 California UrbanPop 91
10 California Rape 40.6
# ... with 90 more rows
当过滤条件存储在像l
这样对整洁更惯用的结构中时,有没有办法过滤df
?
我认为这个答案是这样的,但是,表达式不是动态的。
我们可以在filter
循环中使用across
遍历"l"的names
,通过使用列名(cur_column()
)中的键对"l"进行子集化来创建逻辑表达式,并否定(!
)。 请注意,cur_column()
目前仅适用于across
,而不适用于if_all/if_any
(dplyr
-1.0.6
R 4.1.0
)
library(dplyr)
df %>%
filter(across(all_of(names(l)), ~ !. %in% l[[cur_column()]]))
-输出
# A tibble: 94 x 3
# State name value
# <chr> <chr> <dbl>
# 1 Alaska UrbanPop 48
# 2 Alaska Rape 44.5
# 3 Arizona UrbanPop 80
# 4 Arizona Rape 31
# 5 Arkansas UrbanPop 50
# 6 Arkansas Rape 19.5
# 7 California UrbanPop 91
# 8 California Rape 40.6
# 9 Colorado UrbanPop 78
#10 Colorado Rape 38.7
# … with 84 more rows
如果我们可以设置一个属性,我们就可以利用if_all
library(magrittr)
df %>%
mutate(across(all_of(names(l)), ~ set_attr(., 'cn', cur_column()))) %>%
filter(if_all(all_of(names(l)), ~ ! . %in% l[[attr(., 'cn')]]))
<小时 />或带imap/reduce
library(purrr)
df %>%
filter(imap(l, ~ !cur_data()[[.y]] %in% .x) %>%
reduce(`&`))
或者另一种选择是anti_join
for(nm in names(l)) df <- anti_join(df, tibble(!! nm := l[[nm]]))
这里另一个可能的选择是使用purrr
创建一个逻辑向量,该向量允许&
与|
条件,并且可以在没有cur_column
的情况下访问当前列名(.y
),这只能在across
内部使用:
df %>%
filter(imap(l, ~ !df[[.y]] %in% .x) %>% reduce(`&`)) # can use magrittr::and
输出
State name value
<chr> <chr> <dbl>
1 Alaska UrbanPop 48
2 Alaska Rape 44.5
3 Arizona UrbanPop 80
4 Arizona Rape 31
5 Arkansas UrbanPop 50
6 Arkansas Rape 19.5
7 California UrbanPop 91
8 California Rape 40.6
9 Colorado UrbanPop 78
10 Colorado Rape 38.7
# ... with 84 more rows
与or变体为:
df %>%
filter(imap(l, ~ !df[[.y]] %in% .x) %>% reduce(`|`)) # can use magrittr::or