blocks|key|96213|text|我不确定您想对Event列做什么，但是如果您也想保留它，那么|type|unstyled|depth|inlineStyleRanges|entityRanges|data|96214|isIDmax+<-+with(dd,+ave(Value,+ID,+FUN=function(x)+seq_along(x)==which.max(x)))==1
group[isIDmax,+]

#+++ID+Value+Event
#+3++1+++++5+++++2
#+7++2++++17+++++2
#+9++3+++++5+++++2|code-block|syntax|javascript|96215|在这里，我们使用ave查看每个"ID“的"Value”列。然后我们确定哪个值是最大的，然后将其转换为一个逻辑向量，我们可以使用它来对原始data.frame进行子集。|offset|length|style|CODE|96216|entityMap^0|0|0|8|3|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 wasn't sure what you wanted to do about the Event column, but if you want to keep that as well, how about

<pre><code>isIDmax &lt;- with(dd, ave(Value, ID, FUN=function(x) seq_along(x)==which.max(x)))==1
group[isIDmax, ]

# ID Value Event
# 3 1 5 2
# 7 2 17 2
# 9 3 5 2
</code></pre>

Here we use <code>ave</code> to look at the "Value" column for each "ID". Then we determine which value is the maximal and then turn that into a logical vector we can use to subset the original data.frame.

blocks|key|4602367|text|dplyr解决方案：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4602368|library(dplyr)
ID+<-+c(1,1,1,2,2,2,2,3,3)
Value+<-+c(2,3,5,2,5,8,17,3,5)
Event+<-+c(1,1,2,1,2,1,2,2,2)
group+<-+data.frame(Subject=ID,+pt=Value,+Event=Event)

group+%25>%25
++++group_by(Subject)+%25>%25
++++summarize(max.pt+=+max(pt))|code-block|syntax|javascript|4602369|这将产生以下数据帧：|4602370|++Subject+max.pt
1+++++++1++++++5
2+++++++2+++++17
3+++++++3++++++5|4602371|entityMap^0|0|5|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|$]]

A <code>dplyr</code> solution:

<pre><code>library(dplyr)
ID &lt;- c(1,1,1,2,2,2,2,3,3)
Value &lt;- c(2,3,5,2,5,8,17,3,5)
Event &lt;- c(1,1,2,1,2,1,2,2,2)
group &lt;- data.frame(Subject=ID, pt=Value, Event=Event)

group %&gt;%
 group_by(Subject) %&gt;%
 summarize(max.pt = max(pt))
</code></pre>

This yields the following data frame:

<pre><code> Subject max.pt
1 1 5
2 2 17
3 3 5
</code></pre>

blocks|key|4602413|text|这是一个data.table解决方案：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4602414|require(data.table)+##+1.9.2
group+<-+as.data.table(group)|code-block|syntax|javascript|4602415|如果您想在每个组中保留pt最大值对应的所有条目：|4602416|group[group[,+.I[pt+==+max(pt)],+by=Subject]$V1]
#++++Subject+pt+Event
#+1:+++++++1++5+++++2
#+2:+++++++2+17+++++2
#+3:+++++++3++5+++++2|4602417|如果只想要pt的第一个最大值|4602418|group[group[,+.I[which.max(pt)],+by=Subject]$V1]
#++++Subject+pt+Event
#+1:+++++++1++5+++++2
#+2:+++++++2+17+++++2
#+3:+++++++3++5+++++2|4602419|在这种情况下，这没有什么区别，因为数据中的任何组中都没有多个最大值。|4602420|entityMap^0|4|A|0|0|B|2|0|0|5|2|0|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|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|18|8|@]|D|@]|E|$]]|$1|U|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|V|$]]

Here's a <code>data.table</code> solution:

<pre><code>require(data.table) ## 1.9.2
group &lt;- as.data.table(group)
</code></pre>

If you want to keep all the entries corresponding to max values of <code>pt</code> within each group:

<pre><code>group[group[, .I[pt == max(pt)], by=Subject]$V1]
# Subject pt Event
# 1: 1 5 2
# 2: 2 17 2
# 3: 3 5 2
</code></pre>

If you'd like just the first max value of <code>pt</code>:

<pre><code>group[group[, .I[which.max(pt)], by=Subject]$V1]
# Subject pt Event
# 1: 1 5 2
# 2: 2 17 2
# 3: 3 5 2
</code></pre>

In this case, it doesn't make a difference, as there aren't multiple maximum values within any group in your data.

blocks|key|96313|text|使用data.table的更短的解决方案|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|96314|setDT(group)[,+.SD[which.max(pt)],+by=Subject]
#++++Subject+pt+Event
#+1:+++++++1++5+++++2
#+2:+++++++2+17+++++2
#+3:+++++++3++5+++++2|code-block|syntax|javascript|96315|entityMap^0|2|A|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 shorter solution using <code>data.table</code>:

<pre><code>setDT(group)[, .SD[which.max(pt)], by=Subject]
# Subject pt Event
# 1: 1 5 2
# 2: 2 17 2
# 3: 3 5 2
</code></pre>

blocks|key|96347|text|最直观的方法是在dplyr中使用group_by和top_n函数|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|96348|++++group+%25>%25+group_by(Subject)+%25>%25+top_n(1,+pt)|code-block|syntax|javascript|96349|你得到的结果是|96350|++++Source:+local+data+frame+[3+x+3]
++++Groups:+Subject+[3]

++++++Subject++++pt+Event
++++++++(dbl)+(dbl)+(dbl)
++++1+++++++1+++++5+++++2
++++2+++++++2++++17+++++2
++++3+++++++3+++++5+++++2|96351|entityMap^0|8|5|G|8|P|5|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

The most intuitive method is to use <code>group_by</code> and <code>top_n</code> function in <code>dplyr</code>
<pre><code> group %&gt;% group_by(Subject) %&gt;% top_n(1, pt)
</code></pre>
The result you get is
<pre><code> Source: local data frame [3 x 3]
 Groups: Subject [3]

 Subject pt Event
 (dbl) (dbl) (dbl)
 1 1 5 2
 2 2 17 2
 3 3 5 2
</code></pre>

blocks|key|4602468|text|另一个选项是slice|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4602469|library(dplyr)
group+%25>%25
+++++group_by(Subject)+%25>%25
+++++slice(which.max(pt))
#++++Subject++++pt+Event
#++++<dbl>+<dbl>+<dbl>
#1+++++++1+++++5+++++2
#2+++++++2++++17+++++2
#3+++++++3+++++5+++++2|code-block|syntax|javascript|4602470|entityMap^0|6|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 option is <code>slice</code>

<pre><code>library(dplyr)
group %&gt;%
 group_by(Subject) %&gt;%
 slice(which.max(pt))
# Subject pt Event
# &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt;
#1 1 5 2
#2 2 17 2
#3 3 5 2
</code></pre>

blocks|key|4602499|text|do.call(rbind,+lapply(split(group,as.factor(group$Subject)),+function(x)+{return(x[which.max(x$pt),])}))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4602500|使用基本R|unstyled|offset|length|style|CODE|4602501|entityMap^0|0|4|1|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>do.call(rbind, lapply(split(group,as.factor(group$Subject)), function(x) {return(x[which.max(x$pt),])}))
</code></pre>

Using Base <code>R</code>

blocks|key|4602550|text|如果你想要一个主题的最大pt值，你可以简单地使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4602551|+++pt_max+=+as.data.frame(aggregate(pt~Subject,+group,+max))|code-block|syntax|javascript|4602552|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|$]]

If you want the biggest pt value for a subject, you could simply use:

<pre><code> pt_max = as.data.frame(aggregate(pt~Subject, group, max))
</code></pre>

blocks|key|96437|text|这是另一个data.table解决方案，因为which.max不能处理字符|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|96438|library(data.table)
group+<-+data.table(Subject=ID,+pt=Value,+Event=Event)

group[,+.SD[order(pt,+decreasing+=+TRUE)+==+1],+by+=+Subject]|code-block|syntax|javascript|96439|entityMap^0|5|A|M|9|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Here's another <code>data.table</code> solution, since <code>which.max</code> does not work on characters

<pre><code>library(data.table)
group &lt;- data.table(Subject=ID, pt=Value, Event=Event)

group[, .SD[order(pt, decreasing = TRUE) == 1], by = Subject]
</code></pre>

blocks|key|4602625|text|另一个基础解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4602626|group_sorted+<-+group[order(group$Subject,+-group$pt),]
group_sorted[!duplicated(group_sorted$Subject),]

#+Subject+pt+Event
#+++++++1++5+++++2
#+++++++2+17+++++2
#+++++++3++5+++++2|code-block|syntax|javascript|4602627|按pt+(降序)对数据框进行排序，然后移除Subject中重复的行|offset|length|style|CODE|4602628|entityMap^0|0|0|1|2|L|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]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

Another base solution

<pre><code>group_sorted &lt;- group[order(group$Subject, -group$pt),]
group_sorted[!duplicated(group_sorted$Subject),]

# Subject pt Event
# 1 5 2
# 2 17 2
# 3 5 2
</code></pre>

Order the data frame by <code>pt</code> (descending) and then remove rows duplicated in <code>Subject</code>

blocks|key|96490|text|又一个基础R解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|96491|merge(aggregate(pt+~+Subject,+max,+data+=+group),+group)

++Subject+pt+Event
1+++++++1++5+++++2
2+++++++2+17+++++2
3+++++++3++5+++++2|code-block|syntax|javascript|96492|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|$]]

One more base R solution:

<pre><code>merge(aggregate(pt ~ Subject, max, data = group), group)

 Subject pt Event
1 1 5 2
2 2 17 2
3 3 5 2
</code></pre>

blocks|key|96518|text|另一个data.table选项：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|96519|library(data.table)
setDT(group)
group[group[order(-pt),+.I[1L],+Subject]$V1]|code-block|syntax|javascript|96520|或者另一个(可读性较差，但速度稍快)：|96521|group[group[,+rn+:=+.I][order(Subject,+-pt),+{
++++rn[c(1L,+1L+%2B+which(diff(Subject)>0L))]
}]]|96522|计时码：|96523|library(data.table)
nr+<-+1e7L
ng+<-+nr/4L
set.seed(0L)
DT+<-+data.table(Subject=sample(ng,+nr,+TRUE),+pt=1:nr)#rnorm(nr))
DT2+<-+copy(DT)


microbenchmark::microbenchmark(times=3L,
++++mtd0+=+{a0+<-+DT[DT[,+.I[which.max(pt)],+by=Subject]$V1]},
++++mtd1+=+{a1+<-+DT[DT[order(-pt),+.I[1L],+Subject]$V1]},
++++mtd2+=+{a2+<-+DT2[DT2[,+rn+:=+.I][
++++++++order(Subject,+-pt),+rn[c(TRUE,+diff(Subject)>0L)]
++++]]},
++++mtd3+=+{a3+<-+unique(DT[order(Subject,+-pt)],+by="Subject")}
)
fsetequal(a0[order(Subject)],+a1[order(Subject)])
#[1]+TRUE
fsetequal(a0[order(Subject)],+a2[,+rn+:=+NULL][order(Subject)])
#[1]+TRUE
fsetequal(a0[order(Subject)],+a3[order(Subject)])
#[1]+TRUE|96524|计时：|96525|Unit:+seconds
+expr++++++min+++++++lq+++++mean+++median+++++++uq++++++max+neval
+mtd0+3.256322+3.335412+3.371439+3.414502+3.428998+3.443493+++++3
+mtd1+1.733162+1.748538+1.786033+1.763915+1.812468+1.861022+++++3
+mtd2+1.136307+1.159606+1.207009+1.182905+1.242359+1.301814+++++3
+mtd3+1.123064+1.166161+1.228058+1.209257+1.280554+1.351851+++++3|96526|entityMap^0|3|A|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|14|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|16|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|W|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|X|$]]

Another <code>data.table</code> option:

<pre><code>library(data.table)
setDT(group)
group[group[order(-pt), .I[1L], Subject]$V1]
</code></pre>

Or another (less readable but slightly faster):

<pre><code>group[group[, rn := .I][order(Subject, -pt), {
 rn[c(1L, 1L + which(diff(Subject)&gt;0L))]
}]]
</code></pre>

timing code:

<pre><code>library(data.table)
nr &lt;- 1e7L
ng &lt;- nr/4L
set.seed(0L)
DT &lt;- data.table(Subject=sample(ng, nr, TRUE), pt=1:nr)#rnorm(nr))
DT2 &lt;- copy(DT)


microbenchmark::microbenchmark(times=3L,
 mtd0 = {a0 &lt;- DT[DT[, .I[which.max(pt)], by=Subject]$V1]},
 mtd1 = {a1 &lt;- DT[DT[order(-pt), .I[1L], Subject]$V1]},
 mtd2 = {a2 &lt;- DT2[DT2[, rn := .I][
 order(Subject, -pt), rn[c(TRUE, diff(Subject)&gt;0L)]
 ]]},
 mtd3 = {a3 &lt;- unique(DT[order(Subject, -pt)], by="Subject")}
)
fsetequal(a0[order(Subject)], a1[order(Subject)])
#[1] TRUE
fsetequal(a0[order(Subject)], a2[, rn := NULL][order(Subject)])
#[1] TRUE
fsetequal(a0[order(Subject)], a3[order(Subject)])
#[1] TRUE
</code></pre>

timings:

<pre><code>Unit: seconds
 expr min lq mean median uq max neval
 mtd0 3.256322 3.335412 3.371439 3.414502 3.428998 3.443493 3
 mtd1 1.733162 1.748538 1.786033 1.763915 1.812468 1.861022 3
 mtd2 1.136307 1.159606 1.207009 1.182905 1.242359 1.301814 3
 mtd3 1.123064 1.166161 1.228058 1.209257 1.280554 1.351851 3
</code></pre>

blocks|key|4602720|text|另一个data.table解决方案：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4602721|library(data.table)
setDT(group)[,+head(.SD[order(-pt)],+1),+by+=+.(Subject)]|code-block|syntax|javascript|4602722|entityMap^0|3|A|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>data.table</code> solution:

<pre><code>library(data.table)
setDT(group)[, head(.SD[order(-pt)], 1), by = .(Subject)]
</code></pre>

blocks|key|4602759|text|by是用于数据帧的tapply版本：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4602760|res+<-+by(group,+group$Subject,+FUN=function(df)+df[which.max(df$pt),])|code-block|syntax|javascript|4602761|它返回一个by类的对象，因此我们将其转换为data+frame：|4602762|do.call(rbind,+b)
++Subject+pt+Event
1+++++++1++5+++++2
2+++++++2+17+++++2
3+++++++3++5+++++2|4602763|entityMap^0|0|2|9|6|0|0|5|2|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

<code>by</code> is a version of <code>tapply</code> for data frames:

<pre><code>res &lt;- by(group, group$Subject, FUN=function(df) df[which.max(df$pt),])
</code></pre>

It returns an object of class <code>by</code> so we convert it to data frame:

<pre><code>do.call(rbind, b)
 Subject pt Event
1 1 5 2
2 2 17 2
3 3 5 2
</code></pre>

In a dataset with multiple observations for each subject. For each subject I want to select the row which have the maximum value of 'pt'. For example, with a following dataset:
<pre><code>ID &lt;- c(1,1,1,2,2,2,2,3,3)
Value &lt;- c(2,3,5,2,5,8,17,3,5)
Event &lt;- c(1,1,2,1,2,1,2,2,2)

group &lt;- data.frame(Subject=ID, pt=Value, Event=Event)
# Subject pt Event
# 1 1 2 1
# 2 1 3 1
# 3 1 5 2 # max 'pt' for Subject 1
# 4 2 2 1
# 5 2 5 2
# 6 2 8 1
# 7 2 17 2 # max 'pt' for Subject 2
# 8 3 3 2
# 9 3 5 2 # max 'pt' for Subject 3
</code></pre>
Subject 1, 2, and 3 have the biggest pt value of 5, 17, and 5 respectively.
How could I first find the biggest pt value for each subject, and then, put this observation in another data frame? The resulting data frame should only have the biggest pt values for each subject.

Select the row with the maximum value in each group

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在具有每个对象的多个观察值的数据集中。对于每个主题，我希望选择具有最大值'pt‘的行。例如，使用以下数据集：ID <- c(1,1,1,2,2,2,2,3,3)Value <- c(2,3,5,2,5,8,17,3,5)Event <- c(1,1,2,1,2,1,2,2,2)group <- data.fram...

问选择每组中具有最大值的行
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