blocks|key|509578|text|plyr解决方案(tmp是您的数据框架)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|509579|library("plyr")
ddply(tmp,+.(id),+function(x)+x[c(1,+nrow(x)),+])
#++++id+d+gr++mm+area
#+1++15+1++2+3.4++++1
#+2++15+1++1+5.5++++2
#+3++21+1++1+4.0++++2
#+4++21+1++2+3.8++++2
#+5++22+1++1+4.0++++2
#+6++22+1++2+4.6++++2
#+7++23+1++1+2.7++++2
#+8++23+1++2+3.0++++2
#+9++24+1++1+3.0++++2
#+10+24+1++2+2.0++++3|code-block|syntax|javascript|509580|或使用dplyr+(另请参阅here)：|509581|library("dplyr")
tmp+%25>%25
group_by(id)+%25>%25
slice(c(1,+n()))+%25>%25
ungroup()
#+#+A+tibble:+10+×+5
#+++++++id+++++d++++gr++++mm++area
#++++<int>+<int>+<int>+<dbl>+<int>
#+1+++++15+++++1+++++2+++3.4+++++1
#+2+++++15+++++1+++++1+++5.5+++++2
#+3+++++21+++++1+++++1+++4.0+++++2
#+4+++++21+++++1+++++2+++3.8+++++2
#+5+++++22+++++1+++++1+++4.0+++++2
#+6+++++22+++++1+++++2+++4.6+++++2
#+7+++++23+++++1+++++1+++2.7+++++2
#+8+++++23+++++1+++++2+++3.0+++++2
#+9+++++24+++++1+++++1+++3.0+++++2
#+10++++24+++++1+++++2+++2.0+++++3|509582|entityMap|0|LINK|mutability|MUTABLE|url|https://cran.r-project.org/package=plyr|1|https://cran.r-project.org/package=dplyr|2|https://stackoverflow.com/questions/31528981/^0|9|3|0|4|0|0|0|3|5|1|E|4|2|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@]|D|@$9|18|A|19|1|1A]|$9|1B|A|1C|1|1D]]|E|$]]|$1|M|3|N|5|H|7|1E|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]|Y|$5|R|S|T|E|$U|Z]]]]

A <a href="https://cran.r-project.org/package=plyr" rel="noreferrer">plyr</a> solution (<code>tmp</code> is your data frame):

<pre><code>library("plyr")
ddply(tmp, .(id), function(x) x[c(1, nrow(x)), ])
# id d gr mm area
# 1 15 1 2 3.4 1
# 2 15 1 1 5.5 2
# 3 21 1 1 4.0 2
# 4 21 1 2 3.8 2
# 5 22 1 1 4.0 2
# 6 22 1 2 4.6 2
# 7 23 1 1 2.7 2
# 8 23 1 2 3.0 2
# 9 24 1 1 3.0 2
# 10 24 1 2 2.0 3
</code></pre>

Or with <a href="https://cran.r-project.org/package=dplyr" rel="noreferrer">dplyr</a> (see also <a href="https://stackoverflow.com/questions/31528981/">here</a>):

<pre><code>library("dplyr")
tmp %&gt;%
group_by(id) %&gt;%
slice(c(1, n())) %&gt;%
ungroup()
# # A tibble: 10 × 5
# id d gr mm area
# &lt;int&gt; &lt;int&gt; &lt;int&gt; &lt;dbl&gt; &lt;int&gt;
# 1 15 1 2 3.4 1
# 2 15 1 1 5.5 2
# 3 21 1 1 4.0 2
# 4 21 1 2 3.8 2
# 5 22 1 1 4.0 2
# 6 22 1 2 4.6 2
# 7 23 1 1 2.7 2
# 8 23 1 2 3.0 2
# 9 24 1 1 3.0 2
# 10 24 1 2 2.0 3
</code></pre>

blocks|key|726069|text|快速而简短的data.table解决方案：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|726070|tmp[,+.SD[c(1,.N)],+by=id]|code-block|syntax|javascript|726071|其中.SD表示(D)ata的每个(S)子集，.N是每个组中的行数，tmp是data.table；例如，由fread()默认提供或通过使用setDT()转换data.frame来提供。|726072|请注意，如果一个组只包含一行，该行将在输出中出现两次，因为该行既是该组的第一行，也是最后一行。为了避免这种情况下的重复，感谢@Thell：|726073|tmp[,+.SD[unique(c(1,.N))],+by=id]|726074|或者，下面的代码为.N==1特殊情况提供了明确的逻辑：|726075|tmp[,+if+(.N==1)+.SD+else+.SD[c(1,.N)],+by=id]|726076|您不需要在if的第一部分中使用.SD[1]，因为在这种情况下，.N为1，所以.SD必须只有一行。|726077|如果愿意，您可以将j包装在{}中，并在{}中包含整个代码页面。只要{}中的最后一个表达式返回一个要堆叠的类似list的对象(比如普通的list、data.table或data.frame)。|726078|tmp[,+{+...;+if+(.N==1)+.SD+else+.SD[c(1,.N)]+}+,+by=id]|726079|entityMap|0|LINK|mutability|MUTABLE|url|http://r-datatable.com/^0|6|A|6|A|0|0|0|2|3|M|2|X|3|11|A|1G|7|1W|7|25|A|0|0|0|9|5|0|0|5|2|F|6|V|2|Y|1|12|3|0|9|1|D|2|J|2|X|2|1I|4|1V|4|20|A|2B|A|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@$9|1B|A|1C|1|1D]]|E|$]]|$1|F|3|G|5|H|7|1E|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1F|8|@$9|1G|A|1H|B|C]|$9|1I|A|1J|B|C]|$9|1K|A|1L|B|C]|$9|1M|A|1N|B|C]|$9|1O|A|1P|B|C]|$9|1Q|A|1R|B|C]|$9|1S|A|1T|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1U|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|1V|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|1W|8|@$9|1X|A|1Y|B|C]]|D|@]|E|$]]|$1|S|3|T|5|H|7|1Z|8|@]|D|@]|E|$I|J]]|$1|U|3|V|5|6|7|20|8|@$9|21|A|22|B|C]|$9|23|A|24|B|C]|$9|25|A|26|B|C]|$9|27|A|28|B|C]|$9|29|A|2A|B|C]]|D|@]|E|$]]|$1|W|3|X|5|6|7|2B|8|@$9|2C|A|2D|B|C]|$9|2E|A|2F|B|C]|$9|2G|A|2H|B|C]|$9|2I|A|2J|B|C]|$9|2K|A|2L|B|C]|$9|2M|A|2N|B|C]|$9|2O|A|2P|B|C]|$9|2Q|A|2R|B|C]]|D|@]|E|$]]|$1|Y|3|Z|5|H|7|2S|8|@]|D|@]|E|$I|J]]|$1|10|3|-4|5|6|7|2T|8|@]|D|@]|E|$]]]|11|$12|$5|13|14|15|E|$16|17]]]]

A fast and short <a href="http://r-datatable.com" rel="noreferrer"><code>data.table</code></a> solution :

<pre><code>tmp[, .SD[c(1,.N)], by=id]
</code></pre>

where <code>.SD</code> represents each (S)ubset of (D)ata, <code>.N</code> is the number of rows in each group and <code>tmp</code> is a <code>data.table</code>; e.g. as provided by <code>fread()</code> by default or by converting a <code>data.frame</code> using <code>setDT()</code>.

Note that if a group only contains one row, that row will appear twice in the output because that row is both the first and last row of that group. To avoid the repetition in that case, thanks to @Thell:

<pre><code>tmp[, .SD[unique(c(1,.N))], by=id]
</code></pre>

Alternatively, the following makes the logic explicit for the <code>.N==1</code> special case :

<pre><code>tmp[, if (.N==1) .SD else .SD[c(1,.N)], by=id]
</code></pre>

You don't need <code>.SD[1]</code> in the first part of the <code>if</code> because in that case <code>.N</code> is <code>1</code> so <code>.SD</code> must be just one row anyway. 

You can wrap <code>j</code> in <code>{}</code> and have a whole page of code inside <code>{}</code> if you like. Just as long as the last expression inside <code>{}</code> returns a <code>list</code>- like object to be stacked (such as a plain <code>list</code>, <code>data.table</code> or <code>data.frame</code>).

<pre><code>tmp[, { ...; if (.N==1) .SD else .SD[c(1,.N)] } , by=id]
</code></pre>

blocks|key|509740|text|另一种利用dplyr的方法可以是：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|509741|tmp+%25>%25
+group_by(id)+%25>%25
+filter(1:n()+%25in%25+range(1:n()))

++++++id+++++d++++gr++++mm++area
+++<int>+<int>+<int>+<dbl>+<int>
+1++++15+++++1+++++2+++3.4+++++1
+2++++15+++++1+++++1+++5.5+++++2
+3++++21+++++1+++++1+++4+++++++2
+4++++21+++++1+++++2+++3.8+++++2
+5++++22+++++1+++++1+++4+++++++2
+6++++22+++++1+++++2+++4.6+++++2
+7++++23+++++1+++++1+++2.7+++++2
+8++++23+++++1+++++2+++3+++++++2
+9++++24+++++1+++++1+++3+++++++2
10++++24+++++1+++++2+++2+++++++3|code-block|syntax|javascript|509742|或者与使用row_number()的想法相同|509743|tmp+%25>%25
+group_by(id)+%25>%25
+filter(row_number()+%25in%25+range(row_number()))|509744|或者使用slice()执行操作|509745|tmp+%25>%25
+group_by(id)+%25>%25
+slice(c(which.min(1:n()),+which.max(1:n())))|509746|entityMap^0|5|5|0|0|5|C|0|0|4|7|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|T|$]]

Another approach utilizing <code>dplyr</code> could be:

<pre><code>tmp %&gt;%
 group_by(id) %&gt;%
 filter(1:n() %in% range(1:n()))

 id d gr mm area
 &lt;int&gt; &lt;int&gt; &lt;int&gt; &lt;dbl&gt; &lt;int&gt;
 1 15 1 2 3.4 1
 2 15 1 1 5.5 2
 3 21 1 1 4 2
 4 21 1 2 3.8 2
 5 22 1 1 4 2
 6 22 1 2 4.6 2
 7 23 1 1 2.7 2
 8 23 1 2 3 2
 9 24 1 1 3 2
10 24 1 2 2 3
</code></pre>

Or the same idea with using <code>row_number()</code>:

<pre><code>tmp %&gt;%
 group_by(id) %&gt;%
 filter(row_number() %in% range(row_number()))
</code></pre>

Or performing the operation with <code>slice()</code>:

<pre><code>tmp %&gt;%
 group_by(id) %&gt;%
 slice(c(which.min(1:n()), which.max(1:n())))
</code></pre>

blocks|key|509835|text|使用slice_head()和slice_tail()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|509836|library(tidyverse)

tmp+<-+structure(list(id+=+c(15L,+15L,+15L,+15L,+21L,+21L,+22L,+22L,+
22L,+23L,+23L,+23L,+24L,+24L,+24L,+24L),+d+=+c(1L,+1L,+1L,+1L,+
1L,+1L,+1L,+1L,+1L,+1L,+1L,+1L,+1L,+1L,+1L,+1L),+gr+=+c(2L,+1L,+
1L,+1L,+1L,+2L,+1L,+1L,+2L,+1L,+1L,+2L,+1L,+1L,+1L,+2L),+mm+=+c(3.4,+
4.9,+4.4,+5.5,+4,+3.8,+4,+4.9,+4.6,+2.7,+4,+3,+3,+2,+4,+2),+area+=+c(1L,+
2L,+1L,+2L,+2L,+2L,+2L,+2L,+2L,+2L,+2L,+2L,+2L,+3L,+2L,+3L)),+class+=+"data.frame",+row.names+=+c(NA,+
-16L))

tmp+%25>%25
++group_by(id)+%25>%25
++slice_head()|code-block|syntax|javascript|509837|#+A+tibble:+5+x+5
#+Groups:+++id+[5]
+++++id+++++d++++gr++++mm++area
++<int>+<int>+<int>+<dbl>+<int>
1++++15+++++1+++++2+++3.4+++++1
2++++21+++++1+++++1+++4+++++++2
3++++22+++++1+++++1+++4+++++++2
4++++23+++++1+++++1+++2.7+++++2
5++++24+++++1+++++1+++3+++++++2|509838|tmp+%25>%25
++group_by(id)+%25>%25
++slice_tail()|509839|#+A+tibble:+5+x+5
#+Groups:+++id+[5]
+++++id+++++d++++gr++++mm++area
++<int>+<int>+<int>+<dbl>+<int>
1++++15+++++1+++++1+++5.5+++++2
2++++21+++++1+++++2+++3.8+++++2
3++++22+++++1+++++2+++4.6+++++2
4++++23+++++1+++++2+++3+++++++2
5++++24+++++1+++++2+++2+++++++3|509840|请注意：|509841|默认情况下，slice_head()和slice_tail()返回1行，但是您也可以分别使用slice+a+number+of+row或一定比例的行来指定参数n和prop。有关更多详细信息，请参阅?slice。|509842|entityMap^0|2|C|F|C|0|0|0|0|0|0|6|C|J|C|27|1|29|4|2Q|6|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|M|3|N|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|15|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|V|$]]

<h2>use <code>slice_head()</code> and <code>slice_tail()</code></h2>
<pre><code>library(tidyverse)

tmp &lt;- structure(list(id = c(15L, 15L, 15L, 15L, 21L, 21L, 22L, 22L, 
22L, 23L, 23L, 23L, 24L, 24L, 24L, 24L), d = c(1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), gr = c(2L, 1L, 
1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L), mm = c(3.4, 
4.9, 4.4, 5.5, 4, 3.8, 4, 4.9, 4.6, 2.7, 4, 3, 3, 2, 4, 2), area = c(1L, 
2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 3L)), class = &quot;data.frame&quot;, row.names = c(NA, 
-16L))

tmp %&gt;%
 group_by(id) %&gt;%
 slice_head()
</code></pre>
<pre><code># A tibble: 5 x 5
# Groups: id [5]
 id d gr mm area
 &lt;int&gt; &lt;int&gt; &lt;int&gt; &lt;dbl&gt; &lt;int&gt;
1 15 1 2 3.4 1
2 21 1 1 4 2
3 22 1 1 4 2
4 23 1 1 2.7 2
5 24 1 1 3 2
</code></pre>
<pre><code>tmp %&gt;%
 group_by(id) %&gt;%
 slice_tail()
</code></pre>
<pre><code># A tibble: 5 x 5
# Groups: id [5]
 id d gr mm area
 &lt;int&gt; &lt;int&gt; &lt;int&gt; &lt;dbl&gt; &lt;int&gt;
1 15 1 1 5.5 2
2 21 1 2 3.8 2
3 22 1 2 4.6 2
4 23 1 2 3 2
5 24 1 2 2 3
</code></pre>
<h2>Note that:</h2>
By default, <code>slice_head()</code> and <code>slice_tail()</code> return 1 row, but you can also specify the arguments <code>n</code> and <code>prop</code> with slice a number of rows or a proportion of rows respectively. See <code>?slice</code> for more details.

How can I select the first and last row for each unique <code>id</code> in the following dataframe?
<pre><code>tmp &lt;- structure(list(id = c(15L, 15L, 15L, 15L, 21L, 21L, 22L, 22L, 
22L, 23L, 23L, 23L, 24L, 24L, 24L, 24L), d = c(1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), gr = c(2L, 1L, 
1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L), mm = c(3.4, 
4.9, 4.4, 5.5, 4, 3.8, 4, 4.9, 4.6, 2.7, 4, 3, 3, 2, 4, 2), area = c(1L, 
2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 3L)), .Names = c(&quot;id&quot;, 
&quot;d&quot;, &quot;gr&quot;, &quot;mm&quot;, &quot;area&quot;), class = &quot;data.frame&quot;, row.names = c(NA, 
-16L))
tmp
#&gt; id d gr mm area
#&gt; 1 15 1 2 3.4 1
#&gt; 2 15 1 1 4.9 2
#&gt; 3 15 1 1 4.4 1
#&gt; 4 15 1 1 5.5 2
#&gt; 5 21 1 1 4.0 2
#&gt; 6 21 1 2 3.8 2
#&gt; 7 22 1 1 4.0 2
#&gt; 8 22 1 1 4.9 2
#&gt; 9 22 1 2 4.6 2
#&gt; 10 23 1 1 2.7 2
#&gt; 11 23 1 1 4.0 2
#&gt; 12 23 1 2 3.0 2
#&gt; 13 24 1 1 3.0 2
#&gt; 14 24 1 1 2.0 3
#&gt; 15 24 1 1 4.0 2
#&gt; 16 24 1 2 2.0 3
</code></pre>

Select the first and last row by group in a data frame

如何为以下数据帧中的每个唯一id选择第一行和最后一行？tmp <- structure(list(id = c(15L, 15L, 15L, 15L, 21L, 21L, 22L, 22L, 22L, 23L, 23L, 23L, 24L, 24L, 24L, 24L), d = c(1L, 1L, 1L, 1L, 1...

问按组选择数据框中的第一行和最后一行
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