如果你想计算随后出现的次数，你可以使用sapply函数:

index<-sapply(1:length(numbers),function(x)sum(numbers[1:x]==numbers[x]))
cbind(numbers, index)

输出:

        numbers index
 [1,]       4     1
 [2,]      23     1
 [3,]       4     2
 [4,]      23     2
 [5,]       5     1
 [6,]      43     1
 [7,]      54     1
 [8,]      56     1
 [9,]     657     1
[10,]      67     1
[11,]      67     2
[12,]     435     1
[13,]     453     1
[14,]     435     2
[15,]     324     1
[16,]      34     1
[17,]     456     1
[18,]      56     2
[19,]     567     1
[20,]      65     1
[21,]      34     2
[22,]     435     3

我可能会这样做

length(which(numbers==x))

但实际上，更好的方法是

table(numbers)

numbers <- c(4,23,4,23,5,43,54,56,657,67,67,435 453,435,324,34,456,56,567,65,34,435)

> length(grep(435, numbers))
[1] 3


> length(which(435 == numbers))
[1] 3


> require(plyr)
> df = count(numbers)
> df[df$x == 435, ] 
     x freq
11 435    3


> sum(435 == numbers)
[1] 3


> sum(grepl(435, numbers))
[1] 3


> sum(435 == numbers)
[1] 3


> tabulate(numbers)[435]
[1] 3


> table(numbers)['435']
435 
  3 


> length(subset(numbers, numbers=='435')) 
[1] 3

你可以使用table():

> a <- table(numbers)
> a
numbers
  4   5  23  34  43  54  56  65  67 324 435 453 456 567 657 
  2   1   2   2   1   1   2   1   2   1   3   1   1   1   1 

然后你可以对它进行子集:

> a[names(a)==435]
435 
  3

或者将它转换为data.frame，如果你更喜欢使用它:

> as.data.frame(table(numbers))
   numbers Freq
1        4    2
2        5    1
3       23    2
4       34    2
...

也有计数(数字)从plyr包。在我看来，比餐桌方便多了。

在R中有一个标准函数

汇总(数字)