blocks|key|2189254|text|我想这是一种潮流：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2189255|dat+%25>%25
++bind_cols(as_tibble(`colnames<-`(t(apply(dat[-1],+1,+rank)),+paste0("rank_x",+1:3))))
#>+++id+x1+x2+x3+rank_x1+rank_x2+rank_x3
#>+1++a++1++4++2+++++1.0+++++3.0+++++2.0
#>+2++b++3++2++2+++++3.0+++++1.5+++++1.5
#>+3++c++5++6++5+++++1.5+++++3.0+++++1.5
#>+4++d++7++0++9+++++2.0+++++1.0+++++3.0|code-block|syntax|javascript|2189256|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I guess this is sort of tidyverse:
<pre class="lang-r prettyprint-override"><code>dat %&gt;%
 bind_cols(as_tibble(`colnames&lt;-`(t(apply(dat[-1], 1, rank)), paste0(&quot;rank_x&quot;, 1:3))))
#&gt; id x1 x2 x3 rank_x1 rank_x2 rank_x3
#&gt; 1 a 1 4 2 1.0 3.0 2.0
#&gt; 2 b 3 2 2 3.0 1.5 1.5
#&gt; 3 c 5 6 5 1.5 3.0 1.5
#&gt; 4 d 7 0 9 2.0 1.0 3.0
</code></pre>

blocks|key|1057851|text|我们可以使用unnest_wider|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1057852|library(dplyr)
library(tidyr)
library(stringr)
dat+%25>%25
+++rowwise()+%25>%25
+++mutate(my_ranks+=+list(rank(c_across(starts_with("x")))))+%25>%25
+++unnest_wider(c(my_ranks))+%25>%25
+++rename_at(vars(starts_with("...")),+~+str_replace(.,+fixed("..."),+"rank_x"))
#+A+tibble:+4+x+7
#++id+++++++x1++++x2++++x3++rank_x1+rank_x2+rank_x3
#++<chr>+<dbl>+<dbl>+<dbl>++++<dbl>++++<dbl>++++<dbl>
#1+a+++++++++1+++++4+++++2++++++1++++++++3++++++++2++
#2+b+++++++++3+++++2+++++2++++++3++++++++1.5++++++1.5
#3+c+++++++++5+++++6+++++5++++++1.5++++++3++++++++1.5
#4+d+++++++++7+++++0+++++9++++++2++++++++1++++++++3++|code-block|syntax|javascript|1057853|1057854|另一个选择是pmap/as_tibble_row|1057855|library(tibble)
library(purrr)
dat+%25>%25+
+++++mutate(my_ranks+=+pmap(select(.,+starts_with('x')),+~+
+++++++++++as_tibble_row(rank(c(...)),
++++++++++++.name_repair+=+~+str_c('rank',+seq_along(.)))))+%25>%25
+++++unnest(c(my_ranks))
#+A+tibble:+4+x+7
#++id+++++++x1++++x2++++x3+rank1+rank2+rank3
#++<chr>+<dbl>+<dbl>+<dbl>+<dbl>+<dbl>+<dbl>
#1+a+++++++++1+++++4+++++2+++1+++++3+++++2++
#2+b+++++++++3+++++2+++++2+++3+++++1.5+++1.5
#3+c+++++++++5+++++6+++++5+++1.5+++3+++++1.5
#4+d+++++++++7+++++0+++++9+++2+++++1+++++3++|1057856|1057857|使用来自rowRanks的matrixStats可以完成更简单的工作。|1057858|library(matrixStats)
nm1+<-+names(dat)[-1]
dat[paste0('rank',+nm1)]+<-+rowRanks(as.matrix(dat[nm1]),+ties.method+=+'average')|1057859|entityMap^0|6|C|0|0|0|6|I|0|0|0|4|8|D|B|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|10|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|N|3|O|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|P|3|-4|5|6|7|15|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|S|3|T|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|V|$]]

We can use <code>unnest_wider</code>
<pre><code>library(dplyr)
library(tidyr)
library(stringr)
dat %&gt;%
 rowwise() %&gt;%
 mutate(my_ranks = list(rank(c_across(starts_with(&quot;x&quot;))))) %&gt;%
 unnest_wider(c(my_ranks)) %&gt;%
 rename_at(vars(starts_with(&quot;...&quot;)), ~ str_replace(., fixed(&quot;...&quot;), &quot;rank_x&quot;))
# A tibble: 4 x 7
# id x1 x2 x3 rank_x1 rank_x2 rank_x3
# &lt;chr&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt;
#1 a 1 4 2 1 3 2 
#2 b 3 2 2 3 1.5 1.5
#3 c 5 6 5 1.5 3 1.5
#4 d 7 0 9 2 1 3 
</code></pre>
<hr />
Another option is <code>pmap/as_tibble_row</code>
<pre><code>library(tibble)
library(purrr)
dat %&gt;% 
 mutate(my_ranks = pmap(select(., starts_with('x')), ~ 
 as_tibble_row(rank(c(...)),
 .name_repair = ~ str_c('rank', seq_along(.))))) %&gt;%
 unnest(c(my_ranks))
# A tibble: 4 x 7
# id x1 x2 x3 rank1 rank2 rank3
# &lt;chr&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt;
#1 a 1 4 2 1 3 2 
#2 b 3 2 2 3 1.5 1.5
#3 c 5 6 5 1.5 3 1.5
#4 d 7 0 9 2 1 3 
</code></pre>
<hr />
It can be done more straightforward with <code>rowRanks</code> from <code>matrixStats</code>
<pre><code>library(matrixStats)
nm1 &lt;- names(dat)[-1]
dat[paste0('rank', nm1)] &lt;- rowRanks(as.matrix(dat[nm1]), ties.method = 'average')
</code></pre>

I have the following data frame:
<pre><code>dat &lt;- data.frame(id = c(&quot;a&quot;, &quot;b&quot;, &quot;c&quot;, &quot;d&quot;),
 x1 = c(1, 3, 5, 7),
 x2 = c(4, 2, 6, 0),
 x3 = c(2, 2, 5, 9))
</code></pre>
I now want to calculate the ranking per row across my three x columns and want to store that result into my <code>dat</code> data frame.
So the result could be stored in two ways:
a) ideally, there will be 4 new columns with the respective ranks or 
b) there will be a new nested column that I probably need to unnest somehow.
I tried the following which at least gives me a list column.
<pre><code>dat %&gt;%
 rowwise() %&gt;%
 mutate(my_ranks = list(rank(c_across(starts_with(&quot;x&quot;)))))
</code></pre>
But when I try to unnest, it will give me the ranks but it does so by creating new rows (i.e. each original case now appears four times). Although I guess I could somehow reshape this result with <code>pivot_wider</code>, it feels wrong to follow that route.
Any better/easier idea? Thanks.

tidyverse calculate ranking per row across several columns

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有以下数据框架：dat <- data.frame(id = c("a", "b", "c", "d"), x1 = c(1, 3, 5, 7), x2 = c(4, 2, 6, 0), x3 = c(2, 2, 5...

问tidyverse计算几个列的每行排序
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