blocks|key|1280250|text|df+<-+data.frame(a+=+1:2,+b+=+letters[1:2])+
df[rep(seq_len(nrow(df)),+each+=+2),+]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1280251|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>df &lt;- data.frame(a = 1:2, b = letters[1:2]) 
df[rep(seq_len(nrow(df)), each = 2), ]
</code></pre>

blocks|key|1280272|text|请尝试使用以下示例|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1280273|N=2
rep(1:4,+each+=+N)+|code-block|syntax|javascript|1280274|作为索引|1280275|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

try using for example

<pre><code>N=2
rep(1:4, each = N) 
</code></pre>

as an index

blocks|key|2916210|text|如果您可以重复整个过程，或者先将其子集，然后再重复，那么this+similar+question可能会有所帮助。再一次：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2916211|library(mefa)
rep(mtcars,10)+|code-block|syntax|javascript|2916212|或者简单地说|2916213|mefa:::rep.data.frame(mtcars)|2916214|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/13275260/how-can-i-repeat-a-data-frame/13316594#13316594^0|S|L|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

If you can repeat the whole thing, or subset it first then repeat that, then <a href="https://stackoverflow.com/questions/13275260/how-can-i-repeat-a-data-frame/13316594#13316594">this similar question</a> may be helpful. Once again:

<pre><code>library(mefa)
rep(mtcars,10) 
</code></pre>

or simply

<pre><code>mefa:::rep.data.frame(mtcars)
</code></pre>

blocks|key|3004240|text|rep.row函数有时似乎会列出列，这会导致错误的内存劫持。我写了以下内容，看起来效果很好：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3004241|library(plyr)
rep.row+<-+function(r,+n){
++colwise(function(x)+rep(x,+n))(r)
}|code-block|syntax|javascript|3004242|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|$]]

The rep.row function seems to sometimes make lists for columns, which leads to bad memory hijinks. I have written the following which seems to work well:

<pre><code>library(plyr)
rep.row &lt;- function(r, n){
 colwise(function(x) rep(x, n))(r)
}
</code></pre>

blocks|key|3004283|text|除了@dardisco提到的mefa::rep.data.frame()之外，它还非常灵活。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3004284|您可以将每行重复N次|3004285|rep(df,+each=N)|code-block|syntax|javascript|3004286|或者将整个数据帧重复N次(想想:就像你回收一个矢量化的参数一样)|BOLD|3004287|rep(df,+times=N)|3004288|为mefa竖起大拇指！我从来没有听说过它，直到现在，我必须编写手动代码来做这件事。|3004289|entityMap^0|E|M|0|0|0|2|A|0|0|1|4|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|Z|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|10|8|@$9|11|A|12|B|O]]|D|@]|E|$]]|$1|P|3|Q|5|J|7|13|8|@]|D|@]|E|$K|L]]|$1|R|3|S|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|T|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|U|$]]

Adding to what @dardisco mentioned about <code>mefa::rep.data.frame()</code>, it's very flexible.

You can either repeat each row N times:

<pre><code>rep(df, each=N)
</code></pre>

or repeat the entire dataframe N times (think: like when you recycle a vectorized argument)

<pre><code>rep(df, times=N)
</code></pre>

Two thumbs up for <code>mefa</code>! I had never heard of it until now and I had to write manual code to do this.

blocks|key|3004325|text|另一种方法是首先获取行索引，附加额外的df副本，然后按索引排序：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3004326|df$index+=+1:nrow(df)
df+=+rbind(df,df)
df+=+df[order(df$index),][,-ncol(df)]|code-block|syntax|javascript|3004327|尽管其他解决方案可能更短，但此方法在某些情况下可能更有利。|3004328|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Another way to do this would to first get row indices, append extra copies of the df, and then order by the indices:

<pre><code>df$index = 1:nrow(df)
df = rbind(df,df)
df = df[order(df$index),][,-ncol(df)]
</code></pre>

Although the other solutions may be shorter, this method may be more advantageous in certain situations.

blocks|key|1692979|text|为了参考和补充引用mefa的答案，如果你不想包含整个包，那么看看mefa::rep.data.frame()的实现可能是值得的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1692980|>+data+<-+data.frame(a=letters[1:3],+b=letters[4:6])
>+data
++a+b
1+a+d
2+b+e
3+c+f
>+as.data.frame(lapply(data,+rep,+2))
++a+b
1+a+d
2+b+e
3+c+f
4+a+d
5+b+e
6+c+f|code-block|syntax|javascript|1692981|entityMap^0|W|M|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

For reference and adding to answers citing mefa, it might worth to take a look on the implementation of <code>mefa::rep.data.frame()</code> in case you don't want to include the whole package:

<pre><code>&gt; data &lt;- data.frame(a=letters[1:3], b=letters[4:6])
&gt; data
 a b
1 a d
2 b e
3 c f
&gt; as.data.frame(lapply(data, rep, 2))
 a b
1 a d
2 b e
3 c f
4 a d
5 b e
6 c f
</code></pre>

blocks|key|2916400|text|我的解决方案类似于mefa:::rep.data.frame，但速度更快，并且关心行名：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2916401|rep.data.frame+<-+function(x,+times)+{
++++rnames+<-+attr(x,+"row.names")
++++x+<-+lapply(x,+rep.int,+times+=+times)
++++class(x)+<-+"data.frame"
++++if+(!is.numeric(rnames))
++++++++attr(x,+"row.names")+<-+make.unique(rep.int(rnames,+times))
++++else
++++++++attr(x,+"row.names")+<-+.set_row_names(length(rnames)+*+times)
++++x
}|code-block|syntax|javascript|2916402|比较解决方案：|2916403|library(Lahman)
library(microbenchmark)
microbenchmark(
++++mefa:::rep.data.frame(Batting,+10),
++++rep.data.frame(Batting,+10),
++++Batting[rep.int(seq_len(nrow(Batting)),+10),+],
++++times+=+10
)
#>+Unit:+milliseconds
#>++++++++++++++++++++++++++++++++++++++++++++expr+++++++min+++++++lq+++++mean+++median++++++++uq+++++++max+neval+cld
#>++++++++++++++mefa:::rep.data.frame(Batting,+10)+127.77786+135.3480+198.0240+148.1749++278.1066++356.3210++++10++a+
#>+++++++++++++++++++++rep.data.frame(Batting,+10)++79.70335++82.8165+134.0974++87.2587++191.1713++307.4567++++10++a+
#>++Batting[rep.int(seq_len(nrow(Batting)),+10),+]+895.73750+922.7059+981.8891+956.3463+1018.2411+1127.3927++++10+++b|2916404|entityMap^0|9|L|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

My solution similar as <code>mefa:::rep.data.frame</code>, but a little faster and cares about row names:

<pre><code>rep.data.frame &lt;- function(x, times) {
 rnames &lt;- attr(x, "row.names")
 x &lt;- lapply(x, rep.int, times = times)
 class(x) &lt;- "data.frame"
 if (!is.numeric(rnames))
 attr(x, "row.names") &lt;- make.unique(rep.int(rnames, times))
 else
 attr(x, "row.names") &lt;- .set_row_names(length(rnames) * times)
 x
}
</code></pre>

Compare solutions:

<pre><code>library(Lahman)
library(microbenchmark)
microbenchmark(
 mefa:::rep.data.frame(Batting, 10),
 rep.data.frame(Batting, 10),
 Batting[rep.int(seq_len(nrow(Batting)), 10), ],
 times = 10
)
#&gt; Unit: milliseconds
#&gt; expr min lq mean median uq max neval cld
#&gt; mefa:::rep.data.frame(Batting, 10) 127.77786 135.3480 198.0240 148.1749 278.1066 356.3210 10 a 
#&gt; rep.data.frame(Batting, 10) 79.70335 82.8165 134.0974 87.2587 191.1713 307.4567 10 a 
#&gt; Batting[rep.int(seq_len(nrow(Batting)), 10), ] 895.73750 922.7059 981.8891 956.3463 1018.2411 1127.3927 10 b
</code></pre>

blocks|key|3004412|text|取自here的干净的dplyr解决方案|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3004413|library(dplyr)
df+<-+tibble(x+=+1:2,+y+=+c("a",+"b"))
df+%25>%25+slice(rep(1:n(),+each+=+2))|code-block|syntax|javascript|3004414|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/38237350/repeating-rows-of-data-frame-in-dplyr#comment63896747_38237805^0|A|5|2|4|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

A clean <code>dplyr</code> solution, taken from <a href="https://stackoverflow.com/questions/38237350/repeating-rows-of-data-frame-in-dplyr#comment63896747_38237805" title="here">here</a>

<pre><code>library(dplyr)
df &lt;- tibble(x = 1:2, y = c("a", "b"))
df %&gt;% slice(rep(1:n(), each = 2))
</code></pre>

blocks|key|2916454|text|有一个可爱的矢量化解决方案，它只重复某些行n次，例如可以通过向数据框添加ntimes列：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2916455|++A+B+++C+ntimes
1+j+i+100++++++2
2+K+P+101++++++4
3+Z+Z+102++++++1|code-block|syntax|javascript|2916456|方法：|2916457|df+<-+data.frame(A=c("j","K","Z"),+B=c("i","P","Z"),+C=c(100,101,102),+ntimes=c(2,4,1))
df+<-+as.data.frame(lapply(df,+rep,+df$ntimes))|2916458|结果：|2916459|++A+B+++C+ntimes
1+Z+Z+102++++++1
2+j+i+100++++++2
3+j+i+100++++++2
4+K+P+101++++++4
5+K+P+101++++++4
6+K+P+101++++++4
7+K+P+101++++++4|2916460|这与Josh+O‘’Brien和Mark+Miller的方法非常相似：|2916461|df[rep(seq_len(nrow(df)),+df$ntimes),]|2916462|然而，这种方法看起来要慢得多：|2916463|df+<-+data.frame(A=c("j","K","Z"),+B=c("i","P","Z"),+C=c(100,101,102),+ntimes=c(2000,3000,4000))

microbenchmark::microbenchmark(
++df[rep(seq_len(nrow(df)),+df$ntimes),],
++as.data.frame(lapply(df,+rep,+df$ntimes)),
++times+=+10
)|2916464|2916465|Unit:+microseconds
++++++++++++++++++++++++++++++++++++++expr++++++min+++++++lq++++++mean+++median+++++++uq++++++max+neval
+++df[rep(seq_len(nrow(df)),+df$ntimes),+]+3563.113+3586.873+3683.7790+3613.702+3657.063+4326.757++++10
+as.data.frame(lapply(df,+rep,+df$ntimes))++625.552++654.638++676.4067++668.094++681.929++799.893++++10|2916466|entityMap^0|10|6|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|19|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1B|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1D|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1E|8|@]|D|@]|E|$I|J]]|$1|W|3|X|5|6|7|1F|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|10|3|P|5|6|7|1H|8|@]|D|@]|E|$]]|$1|11|3|12|5|H|7|1I|8|@]|D|@]|E|$I|J]]|$1|13|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|14|$]]

There is a lovely vectorized solution that repeats only certain rows n-times each, possible for example by adding an <code>ntimes</code> column to your data frame:

<pre><code> A B C ntimes
1 j i 100 2
2 K P 101 4
3 Z Z 102 1
</code></pre>

Method:

<pre><code>df &lt;- data.frame(A=c("j","K","Z"), B=c("i","P","Z"), C=c(100,101,102), ntimes=c(2,4,1))
df &lt;- as.data.frame(lapply(df, rep, df$ntimes))
</code></pre>

Result:

<pre><code> A B C ntimes
1 Z Z 102 1
2 j i 100 2
3 j i 100 2
4 K P 101 4
5 K P 101 4
6 K P 101 4
7 K P 101 4
</code></pre>

This is very similar to Josh O'Brien and Mark Miller's method:

<pre><code>df[rep(seq_len(nrow(df)), df$ntimes),]
</code></pre>

However, that method appears quite a bit slower:

<pre><code>df &lt;- data.frame(A=c("j","K","Z"), B=c("i","P","Z"), C=c(100,101,102), ntimes=c(2000,3000,4000))

microbenchmark::microbenchmark(
 df[rep(seq_len(nrow(df)), df$ntimes),],
 as.data.frame(lapply(df, rep, df$ntimes)),
 times = 10
)
</code></pre>

Result:

<pre><code>Unit: microseconds
 expr min lq mean median uq max neval
 df[rep(seq_len(nrow(df)), df$ntimes), ] 3563.113 3586.873 3683.7790 3613.702 3657.063 4326.757 10
 as.data.frame(lapply(df, rep, df$ntimes)) 625.552 654.638 676.4067 668.094 681.929 799.893 10
</code></pre>

I want to repeat the rows of a data.frame, each <code>N</code> times. The result should be a new <code>data.frame</code> (with <code>nrow(new.df) == nrow(old.df) * N</code>) keeping the data types of the columns.

Example for N = 2:

<pre><code> A B C
 A B C 1 j i 100
1 j i 100 --&gt; 2 j i 100
2 K P 101 3 K P 101
 4 K P 101
</code></pre>

So, each row is repeated 2 times and characters remain characters, factors remain factors, numerics remain numerics, ...

My first attempt used apply: <code>apply(old.df, 2, function(co) rep(co, each = N))</code>, but this one transforms my values to characters and I get:

<pre><code> A B C 
[1,] "j" "i" "100"
[2,] "j" "i" "100"
[3,] "K" "P" "101"
[4,] "K" "P" "101"
</code></pre>

Repeat rows of a data.frame

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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我想在每次N的时候重复data.frame的行数。结果应该是一个新的data.frame (带有nrow(new.df) == nrow(old.df) * N)，它保留了列的数据类型。N=2的示例：                        A B   C  A B   C             1 j i 1...

问重复data.frame的行
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问重复data.frame的行EN