blocks|key|12721|text|您可以使用top_n(3)和sum对每一列进行转置和map|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|12722|library(tidyverse)

tdf+<-+as.data.frame(t(df))+
var_names+<-+names(tdf)

var_names+%25>%25
++map_dfc(~tdf+%25>%25+select(.x)+%25>%25+top_n(3)+%25>%25+sum())+%25>%25+
++t()

++++[,1]
V1+66.46
V2+56.44
V3+86.30
V4+63.39
V5+90.62
V6+76.47|code-block|syntax|javascript|12723|感觉有点老土，但还是能把工作做好。|12724|更新Moody_Mudskipper的一种更好的方法：|BOLD|12725|var_names+%25>%25+
++map_dfc(~tdf+%25>%25+select(.x)+%25>%25+top_n(3)+%25>%25+summarize_all(sum))+%25>%25+
++gather|12726|数据：|12727|df+<-+read.table(text='"A"+"B"+"C"+"D"+"E"+"F"+"G"+"H"+"I"+"J"+"L"
2.59+23.08+4.61+10.32+5.61+0.05+0.34+24.04+19.34+5.76+4.27
23.31+1.84+15.34+1.23+0+0+6.13+12.88+6.13+17.18+15.95
9.47+0+13.68+9.47+0+0+0+2.11+5.26+6.32+53.68
2.16+29.99+4.58+7.92+0.92+0+0+8.97+16.57+12.05+16.83
5.4+76.35+2.06+8.87+0+0+0.26+0.39+0.9+2.7+3.08
8.24+30+7.65+0+1.18+0+0+4.71+22.94+23.53+1.76',+header=TRUE)|12728|%5E注意:数据与OP略有不同，行号将被删除。|12729|entityMap^0|5|8|E|3|Q|3|0|0|0|0|2|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|18|8|@$9|19|A|1A|B|O]]|D|@]|E|$]]|$1|P|3|Q|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|R|3|S|5|6|7|1C|8|@]|D|@]|E|$]]|$1|T|3|U|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|V|3|W|5|6|7|1E|8|@]|D|@]|E|$]]|$1|X|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|Y|$]]

You can take the transpose, and <code>map</code> over each column with <code>top_n(3)</code> and <code>sum</code>:

<pre><code>library(tidyverse)

tdf &lt;- as.data.frame(t(df)) 
var_names &lt;- names(tdf)

var_names %&gt;%
 map_dfc(~tdf %&gt;% select(.x) %&gt;% top_n(3) %&gt;% sum()) %&gt;% 
 t()

 [,1]
V1 66.46
V2 56.44
V3 86.30
V4 63.39
V5 90.62
V6 76.47
</code></pre>

Feels a bit hacky, but gets the job done.

Update A better way from Moody_Mudskipper:

<pre><code>var_names %&gt;% 
 map_dfc(~tdf %&gt;% select(.x) %&gt;% top_n(3) %&gt;% summarize_all(sum)) %&gt;% 
 gather
</code></pre>

Data:

<pre><code>df &lt;- read.table(text='"A" "B" "C" "D" "E" "F" "G" "H" "I" "J" "L"
2.59 23.08 4.61 10.32 5.61 0.05 0.34 24.04 19.34 5.76 4.27
23.31 1.84 15.34 1.23 0 0 6.13 12.88 6.13 17.18 15.95
9.47 0 13.68 9.47 0 0 0 2.11 5.26 6.32 53.68
2.16 29.99 4.58 7.92 0.92 0 0 8.97 16.57 12.05 16.83
5.4 76.35 2.06 8.87 0 0 0.26 0.39 0.9 2.7 3.08
8.24 30 7.65 0 1.18 0 0 4.71 22.94 23.53 1.76', header=TRUE)
</code></pre>

^ Note: Slightly different data than OP, row numbers are removed.

blocks|key|12735|text|df+<-+read.table('"A"+"B"+"C"+"D"+"E"+"F"+"G"+"H"+"I"+"L"+"K"
"1"+2.59+23.08+4.61+10.32+5.61+0.05+0.34+24.04+19.34+5.76+4.27
"2"+23.31+1.84+15.34+1.23+0+0+6.13+12.88+6.13+17.18+15.95
"3"+9.47+0+13.68+9.47+0+0+0+2.11+5.26+6.32+53.68
"4"+2.16+29.99+4.58+7.92+0.92+0+0+8.97+16.57+12.05+16.83
"5"+5.4+76.35+2.06+8.87+0+0+0.26+0.39+0.9+2.7+3.08
"6"+8.24+30+7.65+0+1.18+0+0+4.71+22.94+23.53+1.76')|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|12736|根据@Soheil的建议，一个R+Base解决方案：|unstyled|12737|rowSums(t(apply(df,+1,+FUN+=+function(x)+sort(x,+decreasing+=+TRUE)))[+,+c(1,2,3)])|12738|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>df &lt;- read.table('"A" "B" "C" "D" "E" "F" "G" "H" "I" "L" "K"
"1" 2.59 23.08 4.61 10.32 5.61 0.05 0.34 24.04 19.34 5.76 4.27
"2" 23.31 1.84 15.34 1.23 0 0 6.13 12.88 6.13 17.18 15.95
"3" 9.47 0 13.68 9.47 0 0 0 2.11 5.26 6.32 53.68
"4" 2.16 29.99 4.58 7.92 0.92 0 0 8.97 16.57 12.05 16.83
"5" 5.4 76.35 2.06 8.87 0 0 0.26 0.39 0.9 2.7 3.08
"6" 8.24 30 7.65 0 1.18 0 0 4.71 22.94 23.53 1.76')
</code></pre>

According to @Soheil suggestion, an R Base solution: 

<pre><code>rowSums(t(apply(df, 1, FUN = function(x) sort(x, decreasing = TRUE)))[ , c(1,2,3)])
</code></pre>

blocks|key|2457688|text|如果您有一个向量x，则计算三个最大值之和的函数可能是|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2457689|fun+=+function(x)
++++sum(tail(sort(x),+3))|code-block|syntax|javascript|2457690|您希望将其应用于对象m的每一行|2457691|apply(m,+1,+fun)|2457692|一个稍微快一点(例如40%25)的实现是|2457693|colSums(apply(m,+1,+sort,+decreasing+=+TRUE)[1:3,+])|2457694|或使用部分排序|2457695|colSums(apply(m,+1,+sort.int,+partial+=+9:11)[9:11,+])|2457696|为了提高性能，如果m中有很多行，请避免使用apply()所隐含的行间迭代。一个实现可能是|2457697|library(matrixStats)
rowSums(m+*+(rowRanks(m)+>+ncol(m)+-+3))|2457698|但是，当m行中有ties时，此操作将失败；rowRanks()不支持ties.method+=+"first"。相反，实现我们自己的rowRanks()|2457699|.rowRanks+<-+function(m)+{
++++m[]+=+sort.list(sort.list(m))
++++rowRanks(m)
}
rowSums(m+*+(.rowRanks(m)+>+ncol(m)+-+3))|2457700|对于整齐的解决方案，似乎要从整洁的数据开始|2457701|tbl+<-+df+%25>%25
++rowid_to_column()+%25>%25
++gather("letter","value",-1)+%25>%25
++group_by(rowid)|2457702|使用top_n()的解决方案会生成不正确的结果，因此最好的选择似乎是|2457703|summarize(tbl,+fun(value))|2457704|(fun可以在这里扩展，但这似乎不是一个好主意，因为它使单独修改和测试变得更加困难)。|2457705|比较这些方法|2457706|f0+=+function(m)+apply(m,+1,+fun)
f0a+=+function(m)+colSums(apply(m,+1,+sort,+decreasing=TRUE)[1:3,+])
f0b+=+function(m)+colSums(apply(m,+1,+sort.int,+partial+=+9:11)[9:11,+])
f1+=+function(m)+rowSums(m+*+(.rowRanks(m)+>+ncol(m)+-+3))
f2+=+function(tbl)+summarize(tbl,+fun(value))|2457707|使用|2457708|>+library(microbenchmark)
>+identical(f0(m),+f0a(m))
[1]+TRUE
>+identical(f0(m),+f0b(m))
[1]+TRUE
>+identical(f0(m),+f1(m))
[1]+TRUE
>+identical(f0(m),+f2(tbl)$`fun(value)`)
[1]+TRUE
>+microbenchmark(f0(m),+f0a(m),+f0b(m),+f1(m),+f2(tbl),+times=10)
Unit:+microseconds
++++expr++++++min+++++++lq++++++mean++++median+++++++uq++++++max+neval
+++f0(m)++837.505++860.890++894.9245++905.9880++921.386++948.242++++10
++f0a(m)++594.895++637.258++650.7217++653.1800++673.599++713.167++++10
++f0b(m)++274.925++277.734++305.6551++296.2975++330.482++362.765++++10
+++f1(m)++166.416++169.290++192.8086++189.5945++215.491++219.478++++10
+f2(tbl)+2265.451+2277.599+2425.6083+2327.7015+2359.896+3349.995++++10

>+m+=+m[sample(nrow(m),+1000,+TRUE),]
>+microbenchmark(f0(m),+f0a(m),+f0b(m),+f1(m),+times=10)
Unit:+milliseconds
+++expr++++++++min+++++++++lq+++++++mean+++++median+++++++++uq++++++++max+neval
++f0(m)+137.705781+139.793459+141.658415+141.821540+143.653272+144.428092++++10
+f0a(m)++85.946679++86.663967++88.500392++87.513880++89.634696++94.458554++++10
+f0b(m)++29.762981++30.890124++32.470553++32.649594++33.116767++36.686603++++10
++f1(m)+++2.034407+++2.120689+++2.137723+++2.144328+++2.176306+++2.184712++++10|2457709|entityMap^0|8|1|0|0|A|1|0|0|0|0|0|0|9|1|L|7|0|0|4|1|L|A|Y|L|1U|A|0|0|0|0|2|7|0|0|1|3|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1O|8|@$9|1P|A|1Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1S|8|@$9|1T|A|1U|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1V|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1W|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1X|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1Y|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1Z|8|@]|D|@]|E|$I|J]]|$1|W|3|X|5|6|7|20|8|@$9|21|A|22|B|C]|$9|23|A|24|B|C]]|D|@]|E|$]]|$1|Y|3|Z|5|H|7|25|8|@]|D|@]|E|$I|J]]|$1|10|3|11|5|6|7|26|8|@$9|27|A|28|B|C]|$9|29|A|2A|B|C]|$9|2B|A|2C|B|C]|$9|2D|A|2E|B|C]]|D|@]|E|$]]|$1|12|3|13|5|H|7|2F|8|@]|D|@]|E|$I|J]]|$1|14|3|15|5|6|7|2G|8|@]|D|@]|E|$]]|$1|16|3|17|5|H|7|2H|8|@]|D|@]|E|$I|J]]|$1|18|3|19|5|6|7|2I|8|@$9|2J|A|2K|B|C]]|D|@]|E|$]]|$1|1A|3|1B|5|H|7|2L|8|@]|D|@]|E|$I|J]]|$1|1C|3|1D|5|6|7|2M|8|@$9|2N|A|2O|B|C]]|D|@]|E|$]]|$1|1E|3|1F|5|6|7|2P|8|@]|D|@]|E|$]]|$1|1G|3|1H|5|H|7|2Q|8|@]|D|@]|E|$I|J]]|$1|1I|3|1J|5|6|7|2R|8|@]|D|@]|E|$]]|$1|1K|3|1L|5|H|7|2S|8|@]|D|@]|E|$I|J]]|$1|1M|3|-4|5|6|7|2T|8|@]|D|@]|E|$]]]|1N|$]]

if you had a vector <code>x</code>, a function to calculate the sum of the three largest values might be

<pre><code>fun = function(x)
 sum(tail(sort(x), 3))
</code></pre>

You'd like to apply this to each row of your object <code>m</code>

<pre><code>apply(m, 1, fun)
</code></pre>

A somewhat faster (e.g., 40%) implementation is

<pre><code>colSums(apply(m, 1, sort, decreasing = TRUE)[1:3, ])
</code></pre>

Or using partial sorting

<pre><code>colSums(apply(m, 1, sort.int, partial = 9:11)[9:11, ])
</code></pre>

For performance, if there are many rows in <code>m</code>, avoid the iteration over rows implied by use of <code>apply()</code>. An implementation might be

<pre><code>library(matrixStats)
rowSums(m * (rowRanks(m) &gt; ncol(m) - 3))
</code></pre>

but this fails when there are ties in a row of <code>m</code>; <code>rowRanks()</code> does not support <code>ties.method = "first"</code>. Instead, implement our own <code>rowRanks()</code>

<pre><code>.rowRanks &lt;- function(m) {
 m[] = sort.list(sort.list(m))
 rowRanks(m)
}
rowSums(m * (.rowRanks(m) &gt; ncol(m) - 3))
</code></pre>

For a tidyverse solution, it seems one wants to start with tidy data

<pre><code>tbl &lt;- df %&gt;%
 rowid_to_column() %&gt;%
 gather("letter","value",-1) %&gt;%
 group_by(rowid)
</code></pre>

Solutions using <code>top_n()</code> generate incorrect results, so the best bet would seem to be

<pre><code>summarize(tbl, fun(value))
</code></pre>

(<code>fun</code> could be expanded here, but that doesn't really seem to be a good idea because it makes it harder to separately modify and test).

Comparing the methods

<pre><code>f0 = function(m) apply(m, 1, fun)
f0a = function(m) colSums(apply(m, 1, sort, decreasing=TRUE)[1:3, ])
f0b = function(m) colSums(apply(m, 1, sort.int, partial = 9:11)[9:11, ])
f1 = function(m) rowSums(m * (.rowRanks(m) &gt; ncol(m) - 3))
f2 = function(tbl) summarize(tbl, fun(value))
</code></pre>

with

<pre><code>&gt; library(microbenchmark)
&gt; identical(f0(m), f0a(m))
[1] TRUE
&gt; identical(f0(m), f0b(m))
[1] TRUE
&gt; identical(f0(m), f1(m))
[1] TRUE
&gt; identical(f0(m), f2(tbl)$`fun(value)`)
[1] TRUE
&gt; microbenchmark(f0(m), f0a(m), f0b(m), f1(m), f2(tbl), times=10)
Unit: microseconds
 expr min lq mean median uq max neval
 f0(m) 837.505 860.890 894.9245 905.9880 921.386 948.242 10
 f0a(m) 594.895 637.258 650.7217 653.1800 673.599 713.167 10
 f0b(m) 274.925 277.734 305.6551 296.2975 330.482 362.765 10
 f1(m) 166.416 169.290 192.8086 189.5945 215.491 219.478 10
 f2(tbl) 2265.451 2277.599 2425.6083 2327.7015 2359.896 3349.995 10

&gt; m = m[sample(nrow(m), 1000, TRUE),]
&gt; microbenchmark(f0(m), f0a(m), f0b(m), f1(m), times=10)
Unit: milliseconds
 expr min lq mean median uq max neval
 f0(m) 137.705781 139.793459 141.658415 141.821540 143.653272 144.428092 10
 f0a(m) 85.946679 86.663967 88.500392 87.513880 89.634696 94.458554 10
 f0b(m) 29.762981 30.890124 32.470553 32.649594 33.116767 36.686603 10
 f1(m) 2.034407 2.120689 2.137723 2.144328 2.176306 2.184712 10
</code></pre>

blocks|key|2667101|text|使用tidyverse|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2667102|df+%25>%25
++rowid_to_column()+%25>%25
++gather("letter","value",-1)+%25>%25
++group_by(rowid)+%25>%25
++arrange(desc(value))+%25>%25
++slice(1:3)+%25>%25
++summarize(value=+sum(value))

#+#+A+tibble:+6+x+2
#+++rowid+value
#+++<int>+<dbl>
#+1+++++1++66.5
#+2+++++2++56.4
#+3+++++3++76.8
#+4+++++4++63.4
#+5+++++5++90.6
#+6+++++6++76.5|code-block|syntax|javascript|2667103|替代解决方案，灵感来自@andrew_reece的解决方案：|2667104|tdf+<-+setNames(as.data.frame(t(df)),seq(nrow(df)))
tdf+%25>%25
++summarize_all(~sum(tail(sort(.),3)))+%25>%25
++gather(rowid,value)

#+++rowid+value
#+1+++++1+66.46
#+2+++++2+56.44
#+3+++++3+76.83
#+4+++++4+63.39
#+5+++++5+90.62
#+6+++++6+76.47++++|2667105|注意：|2667106|执行setNames(as.data.frame(t(df)),seq(nrow(df)))的真正tidyverse方法如下：|2667107|df+%25>%25
++rowid_to_column+%25>%25
++gather("letter","value",-1)+%25>%25
++spread("rowid","value")|2667108|entityMap^0|2|9|0|0|0|0|0|2|18|1D|9|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|12|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|S|3|T|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|V|$]]

using <code>tidyverse</code>:

<pre><code>df %&gt;%
 rowid_to_column() %&gt;%
 gather("letter","value",-1) %&gt;%
 group_by(rowid) %&gt;%
 arrange(desc(value)) %&gt;%
 slice(1:3) %&gt;%
 summarize(value= sum(value))

# # A tibble: 6 x 2
# rowid value
# &lt;int&gt; &lt;dbl&gt;
# 1 1 66.5
# 2 2 56.4
# 3 3 76.8
# 4 4 63.4
# 5 5 90.6
# 6 6 76.5
</code></pre>

Alternate solution, inspired by @andrew_reece's solution:

<pre><code>tdf &lt;- setNames(as.data.frame(t(df)),seq(nrow(df)))
tdf %&gt;%
 summarize_all(~sum(tail(sort(.),3))) %&gt;%
 gather(rowid,value)

# rowid value
# 1 1 66.46
# 2 2 56.44
# 3 3 76.83
# 4 4 63.39
# 5 5 90.62
# 6 6 76.47 
</code></pre>

note:

the real <code>tidyverse</code> way to do <code>setNames(as.data.frame(t(df)),seq(nrow(df)))</code> would be as follows:

<pre><code>df %&gt;%
 rowid_to_column %&gt;%
 gather("letter","value",-1) %&gt;%
 spread("rowid","value")
</code></pre>

Considering the following sample data:

<pre><code>"A" "B" "C" "D" "E" "F" "G" "H" "I" "L" "K"
"1" 2.59 23.08 4.61 10.32 5.61 0.05 0.34 24.04 19.34 5.76 4.27
"2" 23.31 1.84 15.34 1.23 0 0 6.13 12.88 6.13 17.18 15.95
"3" 9.47 0 13.68 9.47 0 0 0 2.11 5.26 6.32 53.68
"4" 2.16 29.99 4.58 7.92 0.92 0 0 8.97 16.57 12.05 16.83
"5" 5.4 76.35 2.06 8.87 0 0 0.26 0.39 0.9 2.7 3.08
"6" 8.24 30 7.65 0 1.18 0 0 4.71 22.94 23.53 1.76
</code></pre>

I am struggling to find a solution that row-wise find and sum the three highest values.

Any idea?

Row-wise sum of first, second and third highest values

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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