blocks|key|590972|text|编辑:更新到一个更好的现代R答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|590973|您可以使用replicate()，然后将结果重新组合在一起。行名将自动更改为从1:nrow运行。|offset|length|style|CODE|590974|d+<-+data.frame(a+=+c(1,2,3),b+=+c(1,2,3))
n+<-+3
do.call("rbind",+replicate(n,+d,+simplify+=+FALSE))|code-block|syntax|javascript|590975|一种更传统的方法是使用索引，但是这里的行名更改并不是很整洁(但更有信息)：|590976|+d[rep(seq_len(nrow(d)),+n),+]|590977|下面是对上面的改进，前两个使用purrr函数式编程，惯用purrr：|590978|purrr::map_dfr(seq_len(3),+~d)|590979|和不那么惯用的咕噜声(相同的结果，虽然更尴尬)：|590980|purrr::map_dfr(seq_len(3),+function(x)+d)|590981|最后，通过索引而不是列表使用dplyr应用|590982|d+%25>%25+slice(rep(row_number(),+3))|590983|entityMap^0|0|5|B|0|0|0|0|F|5|0|0|0|0|E|5|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@$D|16|E|17|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|18|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|19|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|1A|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|1B|8|@$D|1C|E|1D|F|G]]|9|@]|A|$]]|$1|S|3|T|5|J|7|1E|8|@]|9|@]|A|$K|L]]|$1|U|3|V|5|6|7|1F|8|@]|9|@]|A|$]]|$1|W|3|X|5|J|7|1G|8|@]|9|@]|A|$K|L]]|$1|Y|3|Z|5|6|7|1H|8|@$D|1I|E|1J|F|G]]|9|@]|A|$]]|$1|10|3|11|5|J|7|1K|8|@]|9|@]|A|$K|L]]|$1|12|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]]|13|$]]

EDIT: updated to a better modern R answer. 

You can use <code>replicate()</code>, then <code>rbind</code> the result back together. The rownames are automatically altered to run from 1:nrows. 

<pre><code>d &lt;- data.frame(a = c(1,2,3),b = c(1,2,3))
n &lt;- 3
do.call("rbind", replicate(n, d, simplify = FALSE))
</code></pre>

A more traditional way is to use indexing, but here the rowname altering is not quite so neat (but more informative): 

<pre><code> d[rep(seq_len(nrow(d)), n), ]
</code></pre>

Here are improvements on the above, the first two using <code>purrr</code> functional programming, idiomatic purrr: 

<pre><code>purrr::map_dfr(seq_len(3), ~d)
</code></pre>

and less idiomatic purrr (identical result, though more awkward): 

<pre><code>purrr::map_dfr(seq_len(3), function(x) d)
</code></pre>

and finally via indexing rather than list apply using <code>dplyr</code>:

<pre><code>d %&gt;% slice(rep(row_number(), 3))
</code></pre>

blocks|key|590997|text|d+<-+data.frame(a+=+c(1,2,3),b+=+c(1,2,3))
r+<-+Reduce(rbind,+list(d)[rep(1L,+times=3L)])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|590998|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>d &lt;- data.frame(a = c(1,2,3),b = c(1,2,3))
r &lt;- Reduce(rbind, list(d)[rep(1L, times=3L)])
</code></pre>

blocks|key|287593|text|对于data.frame对象，该解决方案比@mdsummer和@wojciech的快几倍。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|287594|d[rep(seq_len(nrow(d)),+n),+]|code-block|syntax|javascript|287595|对于data.table对象，@mdsummer的应用比转换为data.frame后的应用要快一些。对于大的n，这可能会翻转。|287596|📷|atomic|287597|。|287598|完整代码：|287599|packages+<-+c("data.table",+"ggplot2",+"RUnit",+"microbenchmark")
lapply(packages,+require,+character.only=T)

Repeat1+<-+function(d,+n)+{
++return(do.call("rbind",+replicate(n,+d,+simplify+=+FALSE)))
}

Repeat2+<-+function(d,+n)+{
++return(Reduce(rbind,+list(d)[rep(1L,+times=n)]))
}

Repeat3+<-+function(d,+n)+{
++if+("data.table"+%25in%25+class(d))+return(d[rep(seq_len(nrow(d)),+n)])
++return(d[rep(seq_len(nrow(d)),+n),+])
}

Repeat3.dt.convert+<-+function(d,+n)+{
++if+("data.table"+%25in%25+class(d))+d+<-+as.data.frame(d)
++return(d[rep(seq_len(nrow(d)),+n),+])
}

#+Try+with+data.frames
mtcars1+<-+Repeat1(mtcars,+3)
mtcars2+<-+Repeat2(mtcars,+3)
mtcars3+<-+Repeat3(mtcars,+3)

checkEquals(mtcars1,+mtcars2)
#++Only+difference+is+row.names+having+".k"+suffix+instead+of+"k"+from+1+&+2
checkEquals(mtcars1,+mtcars3)

#+Works+with+data.tables+too
mtcars.dt+<-+data.table(mtcars)
mtcars.dt1+<-+Repeat1(mtcars.dt,+3)
mtcars.dt2+<-+Repeat2(mtcars.dt,+3)
mtcars.dt3+<-+Repeat3(mtcars.dt,+3)

#+No+row.names+mismatch+since+data.tables+don't+have+row.names
checkEquals(mtcars.dt1,+mtcars.dt2)
checkEquals(mtcars.dt1,+mtcars.dt3)

#+Time+test
res+<-+microbenchmark(Repeat1(mtcars,+10),
++++++++++++++++++++++Repeat2(mtcars,+10),
++++++++++++++++++++++Repeat3(mtcars,+10),
++++++++++++++++++++++Repeat1(mtcars.dt,+10),
++++++++++++++++++++++Repeat2(mtcars.dt,+10),
++++++++++++++++++++++Repeat3(mtcars.dt,+10),
++++++++++++++++++++++Repeat3.dt.convert(mtcars.dt,+10))
print(res)
ggsave("repeat_microbenchmark.png",+autoplot(res))|287600|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://developer.qcloudimg.com/http-save/yehe-900000/3bbca3b73d7a43168fe2fe69a01bfc83.png|imageAlt^0|2|A|0|0|2|A|V|A|0|0|1|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|M|3|N|5|O|7|1D|8|@]|D|@$9|1E|A|1F|1|1G]]|E|$]]|$1|P|3|Q|5|6|7|1H|8|@]|D|@]|E|$]]|$1|R|3|S|5|6|7|1I|8|@]|D|@]|E|$]]|$1|T|3|U|5|H|7|1J|8|@]|D|@]|E|$I|J]]|$1|V|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]]|W|$X|$5|Y|Z|10|E|$11|12|13|-4]]]]

For <code>data.frame</code> objects, this solution is several times faster than @mdsummer's and @wojciech-sobala's. 

<pre><code>d[rep(seq_len(nrow(d)), n), ]
</code></pre>

For <code>data.table</code> objects, @mdsummer's is a bit faster than applying the above after converting to <code>data.frame</code>. For large n this might flip.
<img src="https://i.stack.imgur.com/hyrhD.png" alt="microbenchmark">.

Full code:

<pre><code>packages &lt;- c("data.table", "ggplot2", "RUnit", "microbenchmark")
lapply(packages, require, character.only=T)

Repeat1 &lt;- function(d, n) {
 return(do.call("rbind", replicate(n, d, simplify = FALSE)))
}

Repeat2 &lt;- function(d, n) {
 return(Reduce(rbind, list(d)[rep(1L, times=n)]))
}

Repeat3 &lt;- function(d, n) {
 if ("data.table" %in% class(d)) return(d[rep(seq_len(nrow(d)), n)])
 return(d[rep(seq_len(nrow(d)), n), ])
}

Repeat3.dt.convert &lt;- function(d, n) {
 if ("data.table" %in% class(d)) d &lt;- as.data.frame(d)
 return(d[rep(seq_len(nrow(d)), n), ])
}

# Try with data.frames
mtcars1 &lt;- Repeat1(mtcars, 3)
mtcars2 &lt;- Repeat2(mtcars, 3)
mtcars3 &lt;- Repeat3(mtcars, 3)

checkEquals(mtcars1, mtcars2)
# Only difference is row.names having ".k" suffix instead of "k" from 1 &amp; 2
checkEquals(mtcars1, mtcars3)

# Works with data.tables too
mtcars.dt &lt;- data.table(mtcars)
mtcars.dt1 &lt;- Repeat1(mtcars.dt, 3)
mtcars.dt2 &lt;- Repeat2(mtcars.dt, 3)
mtcars.dt3 &lt;- Repeat3(mtcars.dt, 3)

# No row.names mismatch since data.tables don't have row.names
checkEquals(mtcars.dt1, mtcars.dt2)
checkEquals(mtcars.dt1, mtcars.dt3)

# Time test
res &lt;- microbenchmark(Repeat1(mtcars, 10),
 Repeat2(mtcars, 10),
 Repeat3(mtcars, 10),
 Repeat1(mtcars.dt, 10),
 Repeat2(mtcars.dt, 10),
 Repeat3(mtcars.dt, 10),
 Repeat3.dt.convert(mtcars.dt, 10))
print(res)
ggsave("repeat_microbenchmark.png", autoplot(res))
</code></pre>

blocks|key|287609|text|包dplyr包含函数bind_rows()，该函数直接组合列表中的所有数据帧，因此不需要将do.call()与rbind()一起使用。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|287610|df+<-+data.frame(a+=+c(1,+2,+3),+b+=+c(1,+2,+3))
library(dplyr)
bind_rows(replicate(3,+df,+simplify+=+FALSE))|code-block|syntax|javascript|287611|对于大量重复，bind_rows()也比rbind()快得多|287612|library(microbenchmark)
microbenchmark(rbind+=+do.call("rbind",+replicate(1000,+df,+simplify+=+FALSE)),
+++++++++++++++bind_rows+=+bind_rows(replicate(1000,+df,+simplify+=+FALSE)),
+++++++++++++++times+=+20)
##+Unit:+milliseconds
##+++++++expr+++++++min++++++++lq++++++mean+++median++++++++uq+++++++max+neval+cld
##++++++rbind+31.796100+33.017077+35.436753+34.32861+36.773017+43.556112++++20+++b
##++bind_rows++1.765956++1.818087++1.881697++1.86207++1.898839++2.321621++++20++a+|287613|entityMap^0|1|5|A|B|19|9|1J|7|0|0|7|B|K|7|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|P|$]]

The package <code>dplyr</code> contains the function <code>bind_rows()</code> that directly combines all data frames in a list, such that there is no need to use <code>do.call()</code> together with <code>rbind()</code>:

<pre><code>df &lt;- data.frame(a = c(1, 2, 3), b = c(1, 2, 3))
library(dplyr)
bind_rows(replicate(3, df, simplify = FALSE))
</code></pre>

For a large number of repetions <code>bind_rows()</code> is also much faster than <code>rbind()</code>:

<pre><code>library(microbenchmark)
microbenchmark(rbind = do.call("rbind", replicate(1000, df, simplify = FALSE)),
 bind_rows = bind_rows(replicate(1000, df, simplify = FALSE)),
 times = 20)
## Unit: milliseconds
## expr min lq mean median uq max neval cld
## rbind 31.796100 33.017077 35.436753 34.32861 36.773017 43.556112 20 b
## bind_rows 1.765956 1.818087 1.881697 1.86207 1.898839 2.321621 20 a 
</code></pre>

blocks|key|591073|text|甚至更简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|591074|library(data.table)
my_data+<-+data.frame(a+=+c(1,2,3),b+=+c(1,2,3))
rbindlist(replicate(n+=+3,+expr+=+my_data,+simplify+=+FALSE)|code-block|syntax|javascript|591075|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|$]]

Even simpler:

<pre><code>library(data.table)
my_data &lt;- data.frame(a = c(1,2,3),b = c(1,2,3))
rbindlist(replicate(n = 3, expr = my_data, simplify = FALSE)
</code></pre>

blocks|key|591121|text|使用data.table-package，可以将特殊符号.I与rep一起使用。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|591122|df+<-+data.frame(a+=+c(1,2,3),+b+=+c(1,2,3))
dt+<-+as.data.table(df)

n+<-+3

dt[rep(dt[,+.I],+n)]|code-block|syntax|javascript|591123|这意味着：|591124|a+b+1:+1+1+2:+2+2+3:+3+3+4:+1+1+5:+2+2+6:+3+3+7:+1+1+8:+2+2+9:+3+3|blockquote|591125|entityMap|0|LINK|mutability|MUTABLE|url|/questions/tagged/data.table^0|S|2|V|3|2|A|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@$9|12|A|13|1|14]]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|16|8|@]|D|@]|E|$]]|$1|M|3|N|5|O|7|17|8|@]|D|@]|E|$]]|$1|P|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|Q|$R|$5|S|T|U|E|$V|W]]]]

With the <a href="/questions/tagged/data.table" class="post-tag" title="show questions tagged &#39;data.table&#39;" rel="tag">data.table</a>-package, you could use the special symbol <code>.I</code> together with <code>rep</code>:

<pre><code>df &lt;- data.frame(a = c(1,2,3), b = c(1,2,3))
dt &lt;- as.data.table(df)

n &lt;- 3

dt[rep(dt[, .I], n)]
</code></pre>

which gives:

<blockquote>
<pre><code> a b
1: 1 1
2: 2 2
3: 3 3
4: 1 1
5: 2 2
6: 3 3
7: 1 1
8: 2 2
9: 3 3
</code></pre>
</blockquote>

blocks|key|591132|text|为了时间执行的目的，我建议比较rbind的不同方式。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|591133|​|591134|>+mydata+<-+data.frame(a=1:200,b=201:400,c=301:500)
>+microbenchmark(rbind+=+do.call("rbind",replicate(n=100,mydata,simplify+=+FALSE)),
%2B++++++++++++++++bind_rows+=+bind_rows(replicate(n=100,mydata,simplify+=+FALSE)),
%2B++++++++++++++++rbindlist+=+rbindlist(replicate(n=100,exp=+mydata,simplify+=+FALSE)),
%2B++++++++++++++++times=+2000)
Unit:+microseconds
++++++expr++++min++++++lq++++++mean++median++++++uq++++++max+neval
+++++rbind+5760.7+6723.10+8642.6930+7132.30+7761.05+240720.3++2000
+bind_rows++976.4+1186.90+1430.7741+1308.85+1469.80++15817.9++2000
+rbindlist++263.6++347.85++465.5894++392.90++459.95++10974.2++2000|code-block|syntax|javascript|591135|591136|entityMap^0|F|5|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|T|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|U|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|O|$]]

For time execution purposes, i would like to suggest a comparison of different way of rbind:
<div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-html lang-html prettyprint-override"><code>&gt; mydata &lt;- data.frame(a=1:200,b=201:400,c=301:500)
&gt; microbenchmark(rbind = do.call("rbind",replicate(n=100,mydata,simplify = FALSE)),
+ bind_rows = bind_rows(replicate(n=100,mydata,simplify = FALSE)),
+ rbindlist = rbindlist(replicate(n=100,exp= mydata,simplify = FALSE)),
+ times= 2000)
Unit: microseconds
 expr min lq mean median uq max neval
 rbind 5760.7 6723.10 8642.6930 7132.30 7761.05 240720.3 2000
 bind_rows 976.4 1186.90 1430.7741 1308.85 1469.80 15817.9 2000
 rbindlist 263.6 347.85 465.5894 392.90 459.95 10974.2 2000</code></pre>
</div>
</div>

blocks|key|591149|text|您可以使用tidyr::uncount|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|591150|data.frame(a+=+c(1,2,3),b+=+c(1,2,3))+%25>%25+
++tidyr::uncount(3)|code-block|syntax|javascript|591151|++a+b
1+1+1
2+1+1
3+1+1
4+2+2
5+2+2
6+2+2
7+3+3
8+3+3
9+3+3|591152|entityMap^0|5|E|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|H|7|S|8|@]|D|@]|E|$I|J]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

You can use <code>tidyr::uncount</code>:
<pre class="lang-r prettyprint-override"><code>data.frame(a = c(1,2,3),b = c(1,2,3)) %&gt;% 
 tidyr::uncount(3)
</code></pre>
<pre class="lang-r prettyprint-override"><code> a b
1 1 1
2 1 1
3 1 1
4 2 2
5 2 2
6 2 2
7 3 3
8 3 3
9 3 3
</code></pre>

I have the following data frame:
<pre><code>data.frame(a = c(1,2,3),b = c(1,2,3))
 a b
1 1 1
2 2 2
3 3 3
</code></pre>
I want to repeat the rows n times. For example, here the rows are repeated 3 times:
<pre><code> a b
1 1 1
2 2 2
3 3 3
4 1 1
5 2 2
6 3 3
7 1 1
8 2 2
9 3 3
</code></pre>
Is there an easy function to do this in R? Thanks!

Repeat rows of a data.frame N times

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：data.frame(a = c(1,2,3),b = c(1,2,3))  a b1 1 12 2 23 3 3我想重复这几行。例如，在这里，行被重复3次：  a b1 1 12 2 23 3 34 1 15 2 26 3 37 1 18 2 29 3 3在R中有一个简单的函数可以这样做吗？谢谢!

问重复data.frame N次的行
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问重复data.frame N次的行EN