blocks|key|936582|text|下面是我遇到类似问题时推荐给我的一个解决方案，使用data.table和is.na：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|936583|require(data.table)
DT=data.table(test)

unique(DT[,+lapply(.SD,+function(x)+x[!is.na(x)]),+by+=+id])

+++id++a++b++c++d
1:++A++1++2++3++4
2:++B+10+20+30+40|code-block|syntax|javascript|936584|请注意，这将为您提供一个data.table，而不是一个data.frame。如果你不习惯使用这个数据结构，你可以很容易地转换它：|offset|length|style|CODE|936585|data.frame(unique(DT[,+lapply(.SD,+function(x)+x[!is.na(x)]),+by+=+id]))

++id++a++b++c++d
1++A++1++2++3++4
2++B+10+20+30+40|936586|途径：Deduplicating/collapsing+records+in+an+R+dataframe|936587|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/17266578/deduplicating-collapsing-records-in-an-r-dataframe^0|0|0|C|A|S|A|0|0|3|1E|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]|$I|13|J|14|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|16|8|@]|9|@$I|17|J|18|1|19]]|A|$]]|$1|Q|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

Here's a solution that was recommended to me when I had a similar problem, using data.table and is.na:

<pre><code>require(data.table)
DT=data.table(test)

unique(DT[, lapply(.SD, function(x) x[!is.na(x)]), by = id])

 id a b c d
1: A 1 2 3 4
2: B 10 20 30 40
</code></pre>

Note that this gives you a <code>data.table</code>, not a <code>data.frame</code>. If you're not comfortable working with this data structure, you can easily convert it:

<pre><code>data.frame(unique(DT[, lapply(.SD, function(x) x[!is.na(x)]), by = id]))

 id a b c d
1 A 1 2 3 4
2 B 10 20 30 40
</code></pre>

via: <a href="https://stackoverflow.com/questions/17266578/deduplicating-collapsing-records-in-an-r-dataframe">Deduplicating/collapsing records in an R dataframe</a>

blocks|key|936589|text|我不知道这会不会容易很多，但是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|936590|test+<-+data.frame(id.l+=+rep(LETTERS[1:2],each+=+3),+
+++++++++++++++++++a+=+c(1,NA,NA,10,NA,NA),+
+++++++++++++++++++b+=+c(2,NA,NA,20,NA,NA),+
+++++++++++++++++++c+=+c(NA,3,NA,NA,30,NA),+
+++++++++++++++++++d+=+c(NA,NA,4,NA,NA,40))
x+<-+melt(test,+id.l+=+id,+na.rm+=+T)
dcast(x,+id.l+~+variable)
#+id.l++a++b++c++d
#+1++++A++1++2++3++4
#+2++++B+10+20+30+40|code-block|syntax|javascript|936591|我不得不更改你的id列的名称，因为我不能创建id+=+id。|offset|length|style|CODE|936592|entityMap^0|0|0|M|7|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I don't know that this is a whole lot easier, but:

<pre><code>test &lt;- data.frame(id.l = rep(LETTERS[1:2],each = 3), 
 a = c(1,NA,NA,10,NA,NA), 
 b = c(2,NA,NA,20,NA,NA), 
 c = c(NA,3,NA,NA,30,NA), 
 d = c(NA,NA,4,NA,NA,40))
x &lt;- melt(test, id.l = id, na.rm = T)
dcast(x, id.l ~ variable)
# id.l a b c d
# 1 A 1 2 3 4
# 2 B 10 20 30 40
</code></pre>

I had to change the name of your id column since I couldn't make <code>id = id</code>.

blocks|key|2076680|text|使用R基函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2076681|>+test[is.na(test)]+<-0
>+aggregate(.~id,+data=test,+FUN="sum")
++id++a++b++c++d
1++A++1++2++3++4
2++B+10+20+30+40|code-block|syntax|javascript|2076682|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|$]]

Using R base functions

<pre><code>&gt; test[is.na(test)] &lt;-0
&gt; aggregate(.~id, data=test, FUN="sum")
 id a b c d
1 A 1 2 3 4
2 B 10 20 30 40
</code></pre>

blocks|key|2076684|text|另一个dplyr解决方案如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2076685|library(dplyr)
test+%25>%25+group_by(id)+%25>%25+summarise(a+=+na.omit(a)[1],+b+=+na.omit(b)[1],
c+=+na.omit(c)[1],+d+=+na.omit(d)[1])|code-block|syntax|javascript|2076686|entityMap^0|3|5|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 <code>dplyr</code> solution is the following:

<pre><code>library(dplyr)
test %&gt;% group_by(id) %&gt;% summarise(a = na.omit(a)[1], b = na.omit(b)[1],
c = na.omit(c)[1], d = na.omit(d)[1])
</code></pre>

blocks|key|4636354|text|使用base的解决方案|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4636355|apply(test,+2,+function(x)+unique(na.omit(x)))|code-block|syntax|javascript|4636356|entityMap^0|2|4|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|$]]

Solution using <code>base</code>

<pre><code>apply(test, 2, function(x) unique(na.omit(x)))
</code></pre>

I have a data frame like this

<pre><code>test &lt;- data.frame(id = rep(LETTERS[1:2],each = 3), 
 a = c(1,NA,NA,10,NA,NA), 
 b = c(2,NA,NA,20,NA,NA), 
 c = c(NA,3,NA,NA,30,NA), 
 d = c(NA,NA,4,NA,NA,40))
</code></pre>

I got this dataframe, and want to convert it so there only one row for each unique 'id' and no NAs in the dataframe.

I am doing this

<pre><code>ddply(test, 
 .variables = 'id', 
 .fun = function(df){
 colSums(df[,1:4], na.rm = T)})
</code></pre>

to get this data.frame

<pre><code> id a b c d
 1 A 1 2 3 4
 2 B 10 20 30 40
</code></pre>

It works, but is there a more direct way of doing it without using <code>colSums</code>, sort of compress the rows to create a single row for each 'id', because within each 'id', all columns have only one value and the rest are NAs. I did come across a similar request somewhere while looking for something else but cannot find it now!

Thanks

compress rows with NA in different columns to create a single row

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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我有一个像这样的数据框test <- data.frame(id = rep(LETTERS[1:2],each = 3), a = c(1,NA,NA,10,NA,NA), b = c(2,NA,NA,20,NA,NA), c = c(NA,3,NA,NA,30,NA), d = c(...

问在不同列中使用NA压缩行，以创建单行
EN

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问在不同列中使用NA压缩行，以创建单行EN