blocks|key|872806|text|也许桌子就是你想要的？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|872807|dummyData+=+rep(c(1,2,+2,+2),+25)

table(dummyData)
#+dummyData
#++1++2+
#+25+75

##+or+another+presentation+of+the+same+data
as.data.frame(table(dummyData))
#++++dummyData+Freq
#++1+++++++++1+++25
#++2+++++++++2+++75|code-block|syntax|javascript|872808|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|$]]

Perhaps table is what you are after?

<pre><code>dummyData = rep(c(1,2, 2, 2), 25)

table(dummyData)
# dummyData
# 1 2 
# 25 75

## or another presentation of the same data
as.data.frame(table(dummyData))
# dummyData Freq
# 1 1 25
# 2 2 75
</code></pre>

blocks|key|1087991|text|如果您有多个因子(=多维数据框)，则可以使用dplyr包来计算每个因子组合中的唯一值：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1087992|library("dplyr")
data+%25>%25+group_by(factor1,+factor2)+%25>%25+summarize(count=n())|code-block|syntax|javascript|1087993|它使用管道操作符%25>%25来链接数据帧data上的方法调用。|1087994|entityMap^0|M|5|0|0|8|3|H|4|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

If you have multiple factors (= a multi-dimensional data frame), you can use the <code>dplyr</code> package to count unique values in each combination of factors:

<pre><code>library("dplyr")
data %&gt;% group_by(factor1, factor2) %&gt;% summarize(count=n())
</code></pre>

It uses the pipe operator <code>%&gt;%</code> to chain method calls on the data frame <code>data</code>.

blocks|key|1087788|text|它是使用aggregate的单行方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1087789|>+aggregate(data.frame(count+=+v),+list(value+=+v),+length)

++value+count
1+++++1++++25
2+++++2++++75|code-block|syntax|javascript|1087790|entityMap^0|4|9|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|$]]

It is a one-line approach by using <code>aggregate</code>. 

<pre><code>&gt; aggregate(data.frame(count = v), list(value = v), length)

 value count
1 1 25
2 2 75
</code></pre>

blocks|key|1087935|text|正如Chase建议的那样，table()函数是一种很好的方法。如果您正在分析大型数据集，另一种方法是使用datatable包中的.N函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1087936|确保您通过以下方式安装了数据表包|1087937|install.packages("data.table")|code-block|syntax|javascript|1087938|代码：|1087939|#+Import+the+data.table+package
library(data.table)

#+Generate+a+data+table+object,+which+draws+a+number+10%5E7+times++
#+from+1+to+10+with+replacement
DT<-data.table(x=sample(1:10,1E7,TRUE))

#+Count+Frequency+of+each+factor+level
DT[,.N,by=x]|1087940|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/415635/chase^0|2|5|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|6|7|10|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|11|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|C|$]]|$1|M|3|N|5|H|7|13|8|@]|9|@]|C|$I|J]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

table() function is a good way to go, as <a href="https://stackoverflow.com/users/415635/chase">Chase</a> suggested.
If you are analyzing a large dataset, an alternative way is to use .N function in datatable package.

Make sure you installed the data table package by 

<pre><code>install.packages("data.table")
</code></pre>

Code:

<pre><code># Import the data.table package
library(data.table)

# Generate a data table object, which draws a number 10^7 times 
# from 1 to 10 with replacement
DT&lt;-data.table(x=sample(1:10,1E7,TRUE))

# Count Frequency of each factor level
DT[,.N,by=x]
</code></pre>

blocks|key|1087623|text|要获得包含唯一值计数的无量纲整数向量，请使用c()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1087624|dummyData+=+rep(c(1,+2,+2,+2),+25)+#+Chase's+reproducible+data
c(table(dummyData))+#+get+un-dimensioned+integer+vector
+1++2+
25+75

str(c(table(dummyData))+)+#+confirm+structure
+Named+int+[1:2]+25+75
+-+attr(*,+"names")=+chr+[1:2]+"1"+"2"|code-block|syntax|javascript|1087625|如果您需要将唯一值的计数提供给另一个函数，这可能会很有用，而且比Chase的答案的评论中发布的t(as.data.frame(table(dummyData))[,2]更简短、更惯用。感谢Ricardo+Saporta向我指出了这一点，here。|1087626|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/15720885/1036500^0|M|3|0|0|1B|11|3A|4|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|M|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

To get an un-dimensioned integer vector that contains the count of unique values, use <code>c()</code>. 

<pre><code>dummyData = rep(c(1, 2, 2, 2), 25) # Chase's reproducible data
c(table(dummyData)) # get un-dimensioned integer vector
 1 2 
25 75

str(c(table(dummyData)) ) # confirm structure
 Named int [1:2] 25 75
 - attr(*, "names")= chr [1:2] "1" "2"
</code></pre>

This may be useful if you need to feed the counts of unique values into another function, and is shorter and more idiomatic than the <code>t(as.data.frame(table(dummyData))[,2]</code> posted in a comment to Chase's answer. Thanks to Ricardo Saporta who pointed this out to me <a href="https://stackoverflow.com/a/15720885/1036500">here</a>.

blocks|key|873432|text|length(unique(df$col))是我能看到的最简单的方式。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|873433|entityMap^0|0|M|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

<code>length(unique(df$col))</code> is the most simple way I can see.

blocks|key|1088034|text|这对我很有效。取你的向量v|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1088035|length(summary(as.factor(v),maxsum=50000))|1088036|备注:将maxsum设置为足够大，以捕获唯一值的数量|1088037|或者使用magrittr包|1088038|v+%25>%25+as.factor+%25>%25+summary(maxsum=50000)+%25>%25+length|1088039|entityMap^0|C|1|0|0|16|0|0|4|8|0|0|1G|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|V|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|N|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|O|$]]

This works for me. Take your vector <code>v</code>

<code>length(summary(as.factor(v),maxsum=50000))</code>

Comment: set maxsum to be large enough to capture the number of unique values

or with the <code>magrittr</code> package

<code>v %&gt;% as.factor %&gt;% summary(maxsum=50000) %&gt;% length</code>

blocks|key|872982|text|如果您需要将唯一值的数量作为包含您的值的数据框中的附加列(例如，可能代表样本大小的列)，plyr提供了一种简洁的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|872983|data_frame+<-+data.frame(v+=+rep(c(1,2,+2,+2),+25))

library("plyr")
data_frame+<-+ddply(data_frame,+.(v),+transform,+n+=+length(v))|code-block|syntax|javascript|872984|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 need to have the number of unique values as an additional column in the data frame containing your values (a column which may represent sample size for example), plyr provides a neat way:

<pre><code>data_frame &lt;- data.frame(v = rep(c(1,2, 2, 2), 25))

library("plyr")
data_frame &lt;- ddply(data_frame, .(v), transform, n = length(v))
</code></pre>

blocks|key|873400|text|您也可以尝试使用tidyverse|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|873401|library(tidyverse)+
dummyData+%25>%25+
++++as.tibble()+%25>%25+
++++count(value)
#+A+tibble:+2+x+2
++value+++++n
++<dbl>+<int>
1+++++1++++25
2+++++2++++75|code-block|syntax|javascript|873402|entityMap^0|8|9|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|$]]

You can try also a <code>tidyverse</code>

<pre><code>library(tidyverse) 
dummyData %&gt;% 
 as.tibble() %&gt;% 
 count(value)
# A tibble: 2 x 2
 value n
 &lt;dbl&gt; &lt;int&gt;
1 1 25
2 2 75
</code></pre>

blocks|key|1088075|text|此外，将值分类并调用summary()也是可行的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1088076|>+v+=+rep(as.factor(c(1,2,+2,+2)),+25)
>+summary(v)
+1++2+
25+75+|code-block|syntax|javascript|1088077|entityMap^0|A|9|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|$]]

Also making the values categorical and calling <code>summary()</code> would work.

<pre><code>&gt; v = rep(as.factor(c(1,2, 2, 2)), 25)
&gt; summary(v)
 1 2 
25 75 
</code></pre>

blocks|key|1087747|text|如果您想在data.frame+(例如，train.data)上运行唯一的，并获得计数(可以用作分类器中的权重)，您可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1087748|unique.count+=+function(train.data,+all.numeric=FALSE)+{+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++#+first+convert+each+row+in+the+data.frame+to+a+string++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++train.data.str+=+apply(train.data,+1,+function(x)+paste(x,+collapse=','))+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++#+use+table+to+index+and+count+the+strings++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++train.data.str.t+=+table(train.data.str)++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++#+get+the+unique+data+string+from+the+row.names+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++train.data.str.uniq+=+row.names(train.data.str.t)+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++weight+=+as.numeric(train.data.str.t)+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++#+convert+the+unique+data+string+to+data.frame
++if+(all.numeric)+{
++++train.data.uniq+=+as.data.frame(t(apply(cbind(train.data.str.uniq),+1,+
++++++function(x)+as.numeric(unlist(strsplit(x,+split=","))))))++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++}+else+{
++++train.data.uniq+=+as.data.frame(t(apply(cbind(train.data.str.uniq),+1,+
++++++function(x)+unlist(strsplit(x,+split=",")))))++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++}
++names(train.data.uniq)+=+names(train.data)++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++list(data=train.data.uniq,+weight=weight)+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
}++|code-block|syntax|javascript|1087749|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 to run unique on a data.frame (e.g., train.data), and also get the counts (which can be used as the weight in classifiers), you can do the following:

<pre><code>unique.count = function(train.data, all.numeric=FALSE) { 
 # first convert each row in the data.frame to a string 
 train.data.str = apply(train.data, 1, function(x) paste(x, collapse=',')) 
 # use table to index and count the strings 
 train.data.str.t = table(train.data.str) 
 # get the unique data string from the row.names 
 train.data.str.uniq = row.names(train.data.str.t) 
 weight = as.numeric(train.data.str.t) 
 # convert the unique data string to data.frame
 if (all.numeric) {
 train.data.uniq = as.data.frame(t(apply(cbind(train.data.str.uniq), 1, 
 function(x) as.numeric(unlist(strsplit(x, split=",")))))) 
 } else {
 train.data.uniq = as.data.frame(t(apply(cbind(train.data.str.uniq), 1, 
 function(x) unlist(strsplit(x, split=","))))) 
 }
 names(train.data.uniq) = names(train.data) 
 list(data=train.data.uniq, weight=weight) 
} 
</code></pre>

blocks|key|1087706|text|count_unique_words+<-function(wlist)+{
ucountlist+=+list()
unamelist+=+c()
for+(i+in+wlist)
{
if+(is.element(i,+unamelist))
++++ucountlist[[i]]+<-+ucountlist[[i]]+%2B1
else
++++{
++++listlen+<-+length(ucountlist)
++++ucountlist[[i]]+<-+1
++++unamelist+<-+c(unamelist,+i)
++++}
}
ucountlist
}

expt_counts+<-+count_unique_words(population)
for(i+in+names(expt_counts))
++++cat(i,+expt_counts[[i]],+"\n")|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1087707|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>count_unique_words &lt;-function(wlist) {
ucountlist = list()
unamelist = c()
for (i in wlist)
{
if (is.element(i, unamelist))
 ucountlist[[i]] &lt;- ucountlist[[i]] +1
else
 {
 listlen &lt;- length(ucountlist)
 ucountlist[[i]] &lt;- 1
 unamelist &lt;- c(unamelist, i)
 }
}
ucountlist
}

expt_counts &lt;- count_unique_words(population)
for(i in names(expt_counts))
 cat(i, expt_counts[[i]], "\n")
</code></pre>

Let's say I have:

<pre><code>v = rep(c(1,2, 2, 2), 25)
</code></pre>

Now, I want to count the number of times each unique value appears. <code>unique(v)</code> returns what the unique values are, but not how many they are. 

<pre><code>&gt; unique(v)
[1] 1 2
</code></pre>

I want something that gives me 

<pre><code>length(v[v==1])
[1] 25
length(v[v==2])
[1] 75
</code></pre>

but as a more general one-liner :) Something close (but not quite) like this:

<pre><code>#&lt;doesn't work right&gt; length(v[v==unique(v)])
</code></pre>

Count number of occurences for each unique value

假设我有：v = rep(c(1,2, 2, 2), 25)现在，我想计算每个唯一值出现的次数。unique(v)返回唯一值是什么，但不返回它们有多少。> unique(v)[1] 1 2我想要能给我的东西length(v[v==1])[1] 25length(v[v==2])[1] 75但作为更一般的一行代码:)类似如下的内容(但不完全是)：#<doesn't work right> lengt

问计数每个唯一值的出现次数
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回答 12

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问计数每个唯一值的出现次数
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