blocks|key|955167|text|尤里卡!|type|unstyled|depth|inlineStyleRanges|entityRanges|data|955168|xy.list+<-+as.list(as.data.frame(t(xy.df)))|code-block|syntax|javascript|955169|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|$]]

Eureka!

<pre><code>xy.list &lt;- as.list(as.data.frame(t(xy.df)))
</code></pre>

blocks|key|955231|text|如果您想要完全滥用data.frame+(就像我一样)并希望保留$+functionality，一种方法是将您的data.frame拆分成收集在列表中的一行data.frames：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|955232|>+df+=+data.frame(x=c('a','b','c'),+y=3:1)
>+df
++x+y
1+a+3
2+b+2
3+c+1

#+'convert'+into+a+list+of+data.frames
ldf+=+lapply(as.list(1:dim(df)[1]),+function(x)+df[x[1],])

>+ldf
[[1]]
x+y
1+a+3++++
[[2]]
x+y
2+b+2
[[3]]
x+y
3+c+1

#+and+the+'coolest'
>+ldf[[2]]$y
[1]+2|code-block|syntax|javascript|955233|它不仅是智能自慰，还允许将data.frame‘转换’为它的行列表，保留$+indexation，这对进一步使用lapply很有用(假设传递给lapply的函数使用这个$+indexation)|955234|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|$]]

If you want to completely abuse the data.frame (as I do) and like to keep the $ functionality, one way is to split you data.frame into one-line data.frames gathered in a list :

<pre><code>&gt; df = data.frame(x=c('a','b','c'), y=3:1)
&gt; df
 x y
1 a 3
2 b 2
3 c 1

# 'convert' into a list of data.frames
ldf = lapply(as.list(1:dim(df)[1]), function(x) df[x[1],])

&gt; ldf
[[1]]
x y
1 a 3 
[[2]]
x y
2 b 2
[[3]]
x y
3 c 1

# and the 'coolest'
&gt; ldf[[2]]$y
[1] 2
</code></pre>

It is not only intellectual masturbation, but allows to 'transform' the data.frame into a list of its lines, keeping the $ indexation which can be useful for further use with lapply (assuming the function you pass to lapply uses this $ indexation)

blocks|key|1169486|text|如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1169487|xy.list+<-+split(xy.df,+seq(nrow(xy.df)))|code-block|syntax|javascript|1169488|如果您希望xy.df的行名成为输出列表的名称，您可以这样做：|offset|length|style|CODE|1169489|xy.list+<-+setNames(split(xy.df,+seq(nrow(xy.df))),+rownames(xy.df))|1169490|entityMap^0|0|0|5|5|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Like this:

<pre><code>xy.list &lt;- split(xy.df, seq(nrow(xy.df)))
</code></pre>

And if you want the rownames of <code>xy.df</code> to be the names of the output list, you can do:

<pre><code>xy.list &lt;- setNames(split(xy.df, seq(nrow(xy.df))), rownames(xy.df))
</code></pre>

blocks|key|955318|text|另一种方法是将df转换为矩阵，然后在其上应用list+apply+lappy函数：ldf+<-+lapply(as.matrix(myDF),+function(x)x)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|955319|entityMap^0|X|5|15|18|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

An alternative way is to convert the df to a matrix then applying the list apply <code>lappy</code> function over it: <code>ldf &lt;- lapply(as.matrix(myDF), function(x)x)</code>

blocks|key|1169587|text|另一种选择是使用library(purrr)+(在大型data.frames上似乎更快一些)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1169588|flatten(by_row(xy.df,+..f+=+function(x)+flatten_chr(x),+.labels+=+FALSE))|code-block|syntax|javascript|1169589|entityMap^0|8|E|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|$]]

Another alternative using <code>library(purrr)</code> (that seems to be a bit quicker on large data.frames)

<pre><code>flatten(by_row(xy.df, ..f = function(x) flatten_chr(x), .labels = FALSE))
</code></pre>

blocks|key|1169632|text|当前版本的purrr+(0.2.2)包似乎是最快的解决方案：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1169633|by_row(x,+function(v)+list(v)[[1L]],+.collate+=+"list")$.out|code-block|syntax|javascript|1169634|让我们比较一下最有趣的解决方案：|1169635|data("Batting",+package+=+"Lahman")
x+<-+Batting[1:10000,+1:10]
library(benchr)
library(purrr)
benchmark(
++++split+=+split(x,+seq_len(.row_names_info(x,+2L))),
++++mapply+=+.mapply(function(...)+structure(list(...),+class+=+"data.frame",+row.names+=+1L),+x,+NULL),
++++purrr+=+by_row(x,+function(v)+list(v)[[1L]],+.collate+=+"list")$.out
)|1169636|结果：|1169637|Benchmark+summary:
Time+units+:+milliseconds+
++expr+n.eval+++min++lw.qu+median+++mean++up.qu++max++total+relative
+split++++100+983.0+1060.0+1130.0+1130.0+1180.0+1450+113000+++++34.3
mapply++++100+826.0++894.0++963.0++972.0+1030.0+1320++97200+++++29.3
+purrr++++100++24.1+++28.6+++32.9+++44.9+++40.5++183+++4490++++++1.0|1169638|我们也可以用Rcpp得到同样的结果|1169639|#include+<Rcpp.h>
using+namespace+Rcpp;

//+[[Rcpp::export]]
List+df2list(const+DataFrame&+x)+{
++++std::size_t+nrows+=+x.rows();
++++std::size_t+ncols+=+x.cols();
++++CharacterVector+nms+=+x.names();
++++List+res(no_init(nrows));
++++for+(std::size_t+i+=+0;+i+<+nrows;+%2B%2Bi)+{
++++++++List+tmp(no_init(ncols));
++++++++for+(std::size_t+j+=+0;+j+<+ncols;+%2B%2Bj)+{
++++++++++++switch(TYPEOF(x[j]))+{
++++++++++++++++case+INTSXP:+{
++++++++++++++++++++if+(Rf_isFactor(x[j]))+{
++++++++++++++++++++++++IntegerVector+t+=+as<IntegerVector>(x[j]);
++++++++++++++++++++++++RObject+t2+=+wrap(t[i]);
++++++++++++++++++++++++t2.attr("class")+=+"factor";
++++++++++++++++++++++++t2.attr("levels")+=+t.attr("levels");
++++++++++++++++++++++++tmp[j]+=+t2;
++++++++++++++++++++}+else+{
++++++++++++++++++++++++tmp[j]+=+as<IntegerVector>(x[j])[i];
++++++++++++++++++++}
++++++++++++++++++++break;
++++++++++++++++}
++++++++++++++++case+LGLSXP:+{
++++++++++++++++++++tmp[j]+=+as<LogicalVector>(x[j])[i];
++++++++++++++++++++break;
++++++++++++++++}
++++++++++++++++case+CPLXSXP:+{
++++++++++++++++++++tmp[j]+=+as<ComplexVector>(x[j])[i];
++++++++++++++++++++break;
++++++++++++++++}
++++++++++++++++case+REALSXP:+{
++++++++++++++++++++tmp[j]+=+as<NumericVector>(x[j])[i];
++++++++++++++++++++break;
++++++++++++++++}
++++++++++++++++case+STRSXP:+{
++++++++++++++++++++tmp[j]+=+as<std::string>(as<CharacterVector>(x[j])[i]);
++++++++++++++++++++break;
++++++++++++++++}
++++++++++++++++default:+stop("Unsupported+type+'%25s'.",+type2name(x));
++++++++++++}
++++++++}
++++++++tmp.attr("class")+=+"data.frame";
++++++++tmp.attr("row.names")+=+1;
++++++++tmp.attr("names")+=+nms;
++++++++res[i]+=+tmp;
++++}
++++res.attr("names")+=+x.attr("row.names");
++++return+res;
}|1169640|现在与purrr进行比较|1169641|benchmark(
++++purrr+=+by_row(x,+function(v)+list(v)[[1L]],+.collate+=+"list")$.out,
++++rcpp+=+df2list(x)
)|1169642|1169643|Benchmark+summary:
Time+units+:+milliseconds+
+expr+n.eval++min+lw.qu+median+mean+up.qu+++max+total+relative
purrr++++100+25.2++29.8+++37.5+43.4++44.2+159.0++4340++++++1.1
+rcpp++++100+19.0++27.9+++34.3+35.8++37.2++93.8++3580++++++1.0|1169644|entityMap^0|5|5|0|0|0|0|0|0|6|4|0|0|3|5|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|@$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|U|3|V|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|W|3|X|5|6|7|1H|8|@$9|1I|A|1J|B|C]]|D|@]|E|$]]|$1|Y|3|Z|5|H|7|1K|8|@]|D|@]|E|$I|J]]|$1|10|3|P|5|6|7|1L|8|@]|D|@]|E|$]]|$1|11|3|12|5|H|7|1M|8|@]|D|@]|E|$I|J]]|$1|13|3|-4|5|6|7|1N|8|@]|D|@]|E|$]]]|14|$]]

Seems a current version of the <code>purrr</code> (0.2.2) package is the fastest solution:

<pre class="lang-r prettyprint-override"><code>by_row(x, function(v) list(v)[[1L]], .collate = "list")$.out
</code></pre>

Let's compare the most interesting solutions:

<pre class="lang-r prettyprint-override"><code>data("Batting", package = "Lahman")
x &lt;- Batting[1:10000, 1:10]
library(benchr)
library(purrr)
benchmark(
 split = split(x, seq_len(.row_names_info(x, 2L))),
 mapply = .mapply(function(...) structure(list(...), class = "data.frame", row.names = 1L), x, NULL),
 purrr = by_row(x, function(v) list(v)[[1L]], .collate = "list")$.out
)
</code></pre>

Rsults:

<pre><code>Benchmark summary:
Time units : milliseconds 
 expr n.eval min lw.qu median mean up.qu max total relative
 split 100 983.0 1060.0 1130.0 1130.0 1180.0 1450 113000 34.3
mapply 100 826.0 894.0 963.0 972.0 1030.0 1320 97200 29.3
 purrr 100 24.1 28.6 32.9 44.9 40.5 183 4490 1.0
</code></pre>

Also we can get the same result with <code>Rcpp</code>:

<pre><code>#include &lt;Rcpp.h&gt;
using namespace Rcpp;

// [[Rcpp::export]]
List df2list(const DataFrame&amp; x) {
 std::size_t nrows = x.rows();
 std::size_t ncols = x.cols();
 CharacterVector nms = x.names();
 List res(no_init(nrows));
 for (std::size_t i = 0; i &lt; nrows; ++i) {
 List tmp(no_init(ncols));
 for (std::size_t j = 0; j &lt; ncols; ++j) {
 switch(TYPEOF(x[j])) {
 case INTSXP: {
 if (Rf_isFactor(x[j])) {
 IntegerVector t = as&lt;IntegerVector&gt;(x[j]);
 RObject t2 = wrap(t[i]);
 t2.attr("class") = "factor";
 t2.attr("levels") = t.attr("levels");
 tmp[j] = t2;
 } else {
 tmp[j] = as&lt;IntegerVector&gt;(x[j])[i];
 }
 break;
 }
 case LGLSXP: {
 tmp[j] = as&lt;LogicalVector&gt;(x[j])[i];
 break;
 }
 case CPLXSXP: {
 tmp[j] = as&lt;ComplexVector&gt;(x[j])[i];
 break;
 }
 case REALSXP: {
 tmp[j] = as&lt;NumericVector&gt;(x[j])[i];
 break;
 }
 case STRSXP: {
 tmp[j] = as&lt;std::string&gt;(as&lt;CharacterVector&gt;(x[j])[i]);
 break;
 }
 default: stop("Unsupported type '%s'.", type2name(x));
 }
 }
 tmp.attr("class") = "data.frame";
 tmp.attr("row.names") = 1;
 tmp.attr("names") = nms;
 res[i] = tmp;
 }
 res.attr("names") = x.attr("row.names");
 return res;
}
</code></pre>

Now caompare with <code>purrr</code>:

<pre class="lang-r prettyprint-override"><code>benchmark(
 purrr = by_row(x, function(v) list(v)[[1L]], .collate = "list")$.out,
 rcpp = df2list(x)
)
</code></pre>

Results:

<pre><code>Benchmark summary:
Time units : milliseconds 
 expr n.eval min lw.qu median mean up.qu max total relative
purrr 100 25.2 29.8 37.5 43.4 44.2 159.0 4340 1.1
 rcpp 100 19.0 27.9 34.3 35.8 37.2 93.8 3580 1.0
</code></pre>

blocks|key|955445|text|purrrlyr包中的by_row函数将为您完成此操作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|955446|这个例子演示了|955447|myfn+<-+function(row)+{
++#row+is+a+tibble+with+one+row,+and+the+same+number+of+columns+as+the+original+df
++l+<-+as.list(row)
++return(l)
}

list_of_lists+<-+purrrlyr::by_row(df,+myfn,+.labels=FALSE)$.out|code-block|syntax|javascript|955448|默认情况下，从myfn返回的值被放入名为.out的df中的一个新list+column中。上述语句末尾的$.out会立即选择此列，返回一个列表列表。|955449|entityMap|0|LINK|mutability|MUTABLE|url|http://r4ds.had.co.nz/many-models.html#list-columns-1^0|0|8|B|6|0|0|0|7|4|K|4|1G|5|W|B|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@$9|1A|A|1B|1|1C]]|E|$]]|$1|O|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

The <code>by_row</code> function from the <code>purrrlyr</code> package will do this for you. 

This example demonstrates

<pre><code>myfn &lt;- function(row) {
 #row is a tibble with one row, and the same number of columns as the original df
 l &lt;- as.list(row)
 return(l)
}

list_of_lists &lt;- purrrlyr::by_row(df, myfn, .labels=FALSE)$.out
</code></pre>

By default, the returned value from <code>myfn</code> is put into a new <a href="http://r4ds.had.co.nz/many-models.html#list-columns-1" rel="nofollow noreferrer">list column</a> in the df called <code>.out</code>. The <code>$.out</code> at the end of the above statement immediately selects this column, returning a list of lists.

blocks|key|955451|text|对我来说最好的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|955452|示例数据：|955453|Var1<-c("X1",X2","X3")
Var2<-c("X1",X2","X3")
Var3<-c("X1",X2","X3")

Data<-cbind(Var1,Var2,Var3)

ID++++Var1+++Var2++Var3+
1++++++X1+++++X2++++X3
2++++++X4+++++X5++++X6
3++++++X7+++++X8++++X9|code-block|syntax|javascript|955454|我们调用BBmisc库|offset|length|style|CODE|955455|library(BBmisc)

data$lists<-convertRowsToList(data[,2:4])|955456|结果将是：|955457|ID++++Var1+++Var2++Var3++lists
1++++++X1+++++X2++++X3+++list("X1",+"X2",+X3")+
2++++++X4+++++X5++++X6+++list("X4","X5",+"X6")+
3++++++X7+++++X8++++X9+++list("X7,"X8,"X9)+|955458|entityMap^0|0|0|0|4|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@$K|10|L|11|M|N]]|9|@]|A|$]]|$1|O|3|P|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

The best way for me was:

Example data:

<pre><code>Var1&lt;-c("X1",X2","X3")
Var2&lt;-c("X1",X2","X3")
Var3&lt;-c("X1",X2","X3")

Data&lt;-cbind(Var1,Var2,Var3)

ID Var1 Var2 Var3 
1 X1 X2 X3
2 X4 X5 X6
3 X7 X8 X9
</code></pre>

We call the <code>BBmisc</code> library

<pre><code>library(BBmisc)

data$lists&lt;-convertRowsToList(data[,2:4])
</code></pre>

And the result will be:

<pre><code>ID Var1 Var2 Var3 lists
1 X1 X2 X3 list("X1", "X2", X3") 
2 X4 X5 X6 list("X4","X5", "X6") 
3 X7 X8 X9 list("X7,"X8,"X9) 
</code></pre>

blocks|key|955493|text|更现代的解决方案只使用purrr::transpose：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|955494|library(purrr)
iris[1:2,]+%25>%25+purrr::transpose()
#>+[[1]]
#>+[[1]]$Sepal.Length
#>+[1]+5.1
#>+
#>+[[1]]$Sepal.Width
#>+[1]+3.5
#>+
#>+[[1]]$Petal.Length
#>+[1]+1.4
#>+
#>+[[1]]$Petal.Width
#>+[1]+0.2
#>+
#>+[[1]]$Species
#>+[1]+1
#>+
#>+
#>+[[2]]
#>+[[2]]$Sepal.Length
#>+[1]+4.9
#>+
#>+[[2]]$Sepal.Width
#>+[1]+3
#>+
#>+[[2]]$Petal.Length
#>+[1]+1.4
#>+
#>+[[2]]$Petal.Width
#>+[1]+0.2
#>+
#>+[[2]]$Species
#>+[1]+1|code-block|syntax|javascript|955495|entityMap^0|B|G|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|$]]

A more modern solution uses only <code>purrr::transpose</code>:

 

<pre class="lang-r prettyprint-override"><code>library(purrr)
iris[1:2,] %&gt;% purrr::transpose()
#&gt; [[1]]
#&gt; [[1]]$Sepal.Length
#&gt; [1] 5.1
#&gt; 
#&gt; [[1]]$Sepal.Width
#&gt; [1] 3.5
#&gt; 
#&gt; [[1]]$Petal.Length
#&gt; [1] 1.4
#&gt; 
#&gt; [[1]]$Petal.Width
#&gt; [1] 0.2
#&gt; 
#&gt; [[1]]$Species
#&gt; [1] 1
#&gt; 
#&gt; 
#&gt; [[2]]
#&gt; [[2]]$Sepal.Length
#&gt; [1] 4.9
#&gt; 
#&gt; [[2]]$Sepal.Width
#&gt; [1] 3
#&gt; 
#&gt; [[2]]$Petal.Length
#&gt; [1] 1.4
#&gt; 
#&gt; [[2]]$Petal.Width
#&gt; [1] 0.2
#&gt; 
#&gt; [[2]]$Species
#&gt; [1] 1
</code></pre>

I have a data.frame which I would like to convert to a list by rows, meaning each row would correspond to its own list elements. In other words, I would like a list that is as long as the data.frame has rows.

So far, I've tackled this problem in the following manner, but I was wondering if there's a better way to approach this.

<pre><code>xy.df &lt;- data.frame(x = runif(10), y = runif(10))

# pre-allocate a list and fill it with a loop
xy.list &lt;- vector("list", nrow(xy.df))
for (i in 1:nrow(xy.df)) {
 xy.list[[i]] &lt;- xy.df[i,]
}
</code></pre>

data.frame rows to a list

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。

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我有一个data.frame，我想按行将其转换为列表，这意味着每一行都对应于它自己的列表元素。换句话说，我想要一个和data.frame有多少行一样长的列表。到目前为止，我已经用以下方式解决了这个问题，但我想知道是否有更好的方法来解决这个问题。xy.df <- data.frame(x = runif(10), y ...

问将行data.frame到列表中
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