blocks|key|2551962|text|实际上你可以使用data.frame+%2B+t来实现，也就是，|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2551963|dflong+<-+data.frame(t(dfwide))|code-block|syntax|javascript|2551964|entityMap^0|8|A|L|1|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Actually you can use <code>data.frame</code> + <code>t</code> to make it, i.e.,

<pre><code>dflong &lt;- data.frame(t(dfwide))
</code></pre>

blocks|key|2414971|text|以下是pivot_longer的一种选择|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2414972|library(dplyr)
library(tidyr)
df1+%25>%25
+++pivot_longer(cols+=+everything(),+names_to+=+'candidate',+values_to+=+'votes')
#+A+tibble:+5+x+2
#++candidate+++votes
#++<chr>+++++++<dbl>
#1+candidate_1++++49
#2+candidate_2++++33
#3+candidate_3++++19
#4+candidate_4++++93
#5+candidate_5++++42|code-block|syntax|javascript|2414973|利用更新后的数据，|2414974|df2+%25>%25
+++pivot_longer(cols+=+-c(location,+total_voters),
+++++++++++++++++names_to+=+'candidate',+values_to+=+'votes')
#+A+tibble:+8+x+4
#++location+total_voters+candidate+++votes
#++<chr>+++++++++++<int>+<chr>+++++++<int>
#1+Dayton++++++++++++236+candidate_1++++49
#2+Dayton++++++++++++236+candidate_2++++33
#3+Dayton++++++++++++236+candidate_3++++19
#4+Dayton++++++++++++236+candidate_5++++42
#5+Toledo++++++++++++332+candidate_1++++61
#6+Toledo++++++++++++332+candidate_2++++78
#7+Toledo++++++++++++332+candidate_3++++71
#8+Toledo++++++++++++332+candidate_5++++52|2414975|2414976|或者在base+R中，这可以通过stack来完成|2414977|stack(df1)[2:1]|2414978|或通过转换为table|2414979|as.data.frame.table(as.matrix(df1))[,-1]|2414980|或者像@markus建议的那样|2414981|reshape2::melt(df1)|2414982|数据|2414983|df1+<-+data.frame(candidate_1+=+49,+candidate_2+=+33,+
+++++++++candidate_3+=+19,+candidate_4+=+93,+candidate_5+=+42)

df2+<-+structure(list(location+=+c("Dayton",+"Toledo"),+total_voters+=+c(236L,+
332L),+candidate_1+=+c(49L,+61L),+candidate_2+=+c(33L,+78L),+
++++candidate_3+=+c(19L,+71L),+candidate_5+=+c(42L,+52L)),+row.names+=+c(NA,+
-2L),+class+=+"data.frame")|2414984|entityMap^0|3|C|0|0|0|0|0|3|6|G|5|0|0|6|5|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1B|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1E|8|@$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]]|D|@]|E|$]]|$1|R|3|S|5|H|7|1J|8|@]|D|@]|E|$I|J]]|$1|T|3|U|5|6|7|1K|8|@$9|1L|A|1M|B|C]]|D|@]|E|$]]|$1|V|3|W|5|H|7|1N|8|@]|D|@]|E|$I|J]]|$1|X|3|Y|5|6|7|1O|8|@]|D|@]|E|$]]|$1|Z|3|10|5|H|7|1P|8|@]|D|@]|E|$I|J]]|$1|11|3|12|5|6|7|1Q|8|@]|D|@]|E|$]]|$1|13|3|14|5|H|7|1R|8|@]|D|@]|E|$I|J]]|$1|15|3|-4|5|6|7|1S|8|@]|D|@]|E|$]]]|16|$]]

Here is one option with <code>pivot_longer</code>

<pre><code>library(dplyr)
library(tidyr)
df1 %&gt;%
 pivot_longer(cols = everything(), names_to = 'candidate', values_to = 'votes')
# A tibble: 5 x 2
# candidate votes
# &lt;chr&gt; &lt;dbl&gt;
#1 candidate_1 49
#2 candidate_2 33
#3 candidate_3 19
#4 candidate_4 93
#5 candidate_5 42
</code></pre>

With the updated data,

<pre><code>df2 %&gt;%
 pivot_longer(cols = -c(location, total_voters),
 names_to = 'candidate', values_to = 'votes')
# A tibble: 8 x 4
# location total_voters candidate votes
# &lt;chr&gt; &lt;int&gt; &lt;chr&gt; &lt;int&gt;
#1 Dayton 236 candidate_1 49
#2 Dayton 236 candidate_2 33
#3 Dayton 236 candidate_3 19
#4 Dayton 236 candidate_5 42
#5 Toledo 332 candidate_1 61
#6 Toledo 332 candidate_2 78
#7 Toledo 332 candidate_3 71
#8 Toledo 332 candidate_5 52
</code></pre>

<hr>

Or in <code>base R</code>, this can be done with <code>stack</code>

<pre><code>stack(df1)[2:1]
</code></pre>

Or by converting to a <code>table</code> 

<pre><code>as.data.frame.table(as.matrix(df1))[,-1]
</code></pre>

Or as @markus suggested

<pre><code>reshape2::melt(df1)
</code></pre>

<h3>data</h3>

<pre><code>df1 &lt;- data.frame(candidate_1 = 49, candidate_2 = 33, 
 candidate_3 = 19, candidate_4 = 93, candidate_5 = 42)

df2 &lt;- structure(list(location = c("Dayton", "Toledo"), total_voters = c(236L, 
332L), candidate_1 = c(49L, 61L), candidate_2 = c(33L, 78L), 
 candidate_3 = c(19L, 71L), candidate_5 = c(42L, 52L)), row.names = c(NA, 
-2L), class = "data.frame")
</code></pre>

blocks|key|2546739|text|如果列名称是candidate_1、candidate_2等，那么您可以只使用reshape2包中的melt函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2546740|a=data.frame(candidate_1=49,
+++++++++++++candidate_2=33,
+++++++++++++candidate_3=19,
+++++++++++++candidate_4=93,
+++++++++++++candidate_5=42)

b=reshape2::melt(a)|code-block|syntax|javascript|2546741|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|$]]

If candidate_1 candidate_2 etc are the columns name you can just use melt function from reshape2 packages.

<pre><code>a=data.frame(candidate_1=49,
 candidate_2=33,
 candidate_3=19,
 candidate_4=93,
 candidate_5=42)

b=reshape2::melt(a)
</code></pre>

<pre><code>df2 &lt;- structure(list(location = c("Dayton", "Toledo"), total_voters = c(236L, 
332L), candidate_1 = c(49L, 61L), candidate_2 = c(33L, 78L), 
 candidate_3 = c(19L, 71L), candidate_5 = c(42L, 52L)), row.names = c(NA, 
-2L), class = "data.frame")
</code></pre>

I have data coming from a SQL query that is shaped as such:

<pre><code>+----------+--------------+-------------+-------------+-------------+-------------+-------------+
| location | total_voters | candidate_1 | candidate_2 | candidate_3 | candidate_4 | candidate_5 |
+----------+--------------+-------------+-------------+-------------+-------------+-------------+
| Dayton | 236 | 49 | 33 | 19 | 93 | 42 |
| Toledo | 332 | 61 | 78 | 71 | 70 | 52 |
+----------+--------------+-------------+-------------+-------------+-------------+-------------+
</code></pre>

The numbers represent the number of votes for each candidate. So what am I trying to do? I'd like to use R (I imagine through either dplyr or tidyr) to pivot this data so it looks as such:

<pre><code>+-------------+-------+----------+--------------+
| candidate | votes | location | total_voters |
+-------------+-------+----------+--------------+
| candidate_1 | 49 | Dayton | 236 |
| candidate_2 | 33 | Dayton | 236 |
| candidate_3 | 19 | Dayton | 236 |
| candidate_4 | 93 | Dayton | 236 |
| candidate_5 | 42 | Dayton | 236 |
| candidate_1 | 61 | Toledo | 332 |
| candidate_2 | 78 | Toledo | 332 |
| candidate_3 | 71 | Toledo | 332 |
| candidate_4 | 70 | Toledo | 332 |
| candidate_5 | 52 | Toledo | 332 |
+-------------+-------+----------+--------------+
</code></pre>

What would be the most effective way to accomplish this in R?

Pivoting columns into rows

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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df2 <- structure(list(location = c("Dayton", "Toledo"), total_voters = c(236L, 332L), candidate_1 = c(49L, 61L), candidate_2 = c(33L, 78L), candidate_3 = c(...

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