问题

以下是您要翻译的代码部分：

library(dplyr)
d %>%
  filter(if_all(c(c,e), ~ b == .b))

希望这对您有所帮助。如果您有任何其他问题，请随时提出。

a   b   c   d   e
1   x   x   A   x
2   y   y   A   NA
3   z   v   B   NA
4   x   w   T   w
5   s   NA  K   NA

a   b   c   d   e
3   z   v   B   NA
4   x   w   T   w

library(dplyr)
d %>% 
  filter(if_all(c(c,e), ~ b == .b))

答案1

得分: 1

以下是使用dplyr的一个想法，

library(dplyr)
df %>%
  rowwise() %>%
  filter(sum(!is.na(c_across(c('b', 'c', 'e')))) > 1, length(unique(na.omit(c_across(c('b', 'c', 'e'))))) > 1) %>%
  ungroup()

一个 tibble: 2 × 5

  a b     c     d     e    

1 3 z v B NA
2 4 x w T w

<details>
<summary>英文:</summary>
Here is an idea using `dplyr`,
    library(dplyr)
    
    df %&gt;%
         rowwise() %&gt;%
         filter(sum(!is.na(c_across(c(&#39;b&#39;, &#39;c&#39;, &#39;e&#39;)))) &gt; 1, length(unique(na.omit(c_across(c(&#39;b&#39;, &#39;c&#39;, &#39;e&#39;))))) &gt; 1) %&gt;%
         ungroup()
    
    # A tibble: 2 &#215; 5
          a b     c     d     e    
      &lt;int&gt; &lt;chr&gt; &lt;chr&gt; &lt;chr&gt; &lt;chr&gt;
    1     3 z     v     B     NA   
    2     4 x     w     T     w    
</details>
# 答案2
**得分**: 1

apply(
df[, c("b", "c", "e")],
1,
function(row) {
row <- row[!is.na(row)]
any(row != row[1])
}
)

#> [1] FALSE FALSE TRUE TRUE FALSE

---
Where `df` is:

df <- read.table(text =
'a b c d e

1 x x A x
2 y y A NA
3 z v B NA
4 x w T w
5 s NA K NA',
header = TRUE)

<details>
<summary>英文:</summary>

apply(
df[, c("b", "c", "e")],
1,
function(row) {
row <- row[!is.na(row)]
any(row != row[1])
}
)

#> [1] FALSE FALSE TRUE TRUE FALSE

---
Where `df` is:

df <- read.table(text =
'a b c d e

1 x x A x
2 y y A NA
3 z v B NA
4 x w T w
5 s NA K NA',
header = TRUE)

</details>
# 答案3
**得分**: 0
我相信OP想要输出对于所有值都是唯一的行而不包括NAs的情况下为TRUE。我们可以使用`table`逐行进行操作，如果表的所有值都为1（没有重复），则输出TRUE。
请记得`pick`所需的列来供给这个函数。
根据这个新的索引变量进行筛选很简单。

df <- data.frame(
id = c(1:4),
a = c('a', 'a', 'a', 'z'),
b = c('b', 'b', 'c', 'a'),
c = c('c', 'b', NA, 'd'))

library(dplyr)

df |>
mutate(index = apply(pick(a:c), 1, table) |>
lapply((x) all(x ==1)
)
)

a b c index
1 a b c TRUE
2 a b b FALSE
3 a c <NA> TRUE
4 z a d TRUE

一个修改后的、更简单的版本，使用`purrr::pmap`:

df |>
mutate(index = pmap(pick(a:c), (...) all(table(c(...)) == 1)))

id a b c index
1 1 a b c TRUE
2 2 a b b FALSE
3 3 a c <NA> TRUE
4 4 z a d TRUE

<details>
<summary>英文:</summary>
I believe the OP wants to output TRUE for rows in which all values are unique, exluding NAs. We can use `table` rowwise and output TRUE if `all` values of the table are `1`(no duplicates). 
Remember to `pick`the desired columns to feed the function.
Filtering on this new index variable is straightforward.

df <- data.frame(
id = c(1:4),
a = c('a', 'a', 'a', 'z'),
b = c('b', 'b', 'c', 'a'),
c = c('c', 'b', NA, 'd'))

library(dplyr)

df |>
mutate(index = apply(pick(a:c), 1, table) |>
lapply((x) all(x ==1)
)
)

a b c index
1 a b c TRUE
2 a b b FALSE
3 a c <NA> TRUE
4 z a d TRUE

A modified, simpler version, with `purrr::pmap`:

df |>
mutate(index = pmap(pick(a:c), (...) all(table(c(...)) == 1)))

id a b c index
1 1 a b c TRUE
2 2 a b b FALSE
3 3 a c <NA> TRUE
4 4 z a d TRUE

答案4

得分: 0

使用一个辅助函数：

library(tidyverse)
data <- tibble(
  a = c(1, 2, 3, 4, 5),
  b = c("x", "y", "z", "x", "s"),
  c = c("x", "y", "v", "w", NA),
  d = c("A", "A", "B", "T", "K"),
  e = c("x", NA, NA, "w", NA)
)
unique_row <- function(input) {
  result <- input %>%
    na.omit() %>%
    unique()
  
  return(length(result) != 1)
}
data %>%
rowwise() %>%
  filter(unique_row(c(b, c, e)))

(Note: I've removed the HTML encoding for characters like "<" and """ to make the code more readable in Chinese. The code should work as expected without these encodings.)

library(tidyverse)
data <- tibble(
  a = c(1, 2, 3, 4, 5),
  b = c("x", "y", "z", "x", "s"),
  c = c("x", "y", "v", "w", NA),
  d = c("A", "A", "B", "T", "K"),
  e = c("x", NA, NA, "w", NA)
)
unique_row <- function(input) {
  result <- input %>% 
    na.omit() %>%
    unique()
  
  return(length(result) != 1)
}
data %>%
rowwise() %>%
  filter(unique_row(c(b, c, e)))