下面是一种可以用dplyr实现的方法:

library(tidyverse)

numbers <- c(4,23,4,23,5,43,54,56,657,67,67,435,
             453,435,324,34,456,56,567,65,34,435)
ord <- seq(1:(length(numbers)))

df <- data.frame(ord,numbers)

df <- df %>%
  count(numbers)

numbers     n
     <dbl> <int>
 1       4     2
 2       5     1
 3      23     2
 4      34     2
 5      43     1
 6      54     1
 7      56     2
 8      65     1
 9      67     2
10     324     1
11     435     3
12     453     1
13     456     1
14     567     1
15     657     1

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

也有计数(数字)从plyr包。在我看来，比餐桌方便多了。

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
...

一个选项是使用vctrs库中的vec_count()函数:

vec_count(numbers)

   key count
1  435     3
2   67     2
3    4     2
4   34     2
5   56     2
6   23     2
7  456     1
8   43     1
9  453     1
10   5     1
11 657     1
12 324     1
13  54     1
14 567     1
15  65     1

默认顺序将最常用的值放在顶部。如果根据键进行排序(类似table()的输出):

vec_count(numbers, sort = "key")

   key count
1    4     2
2    5     1
3   23     2
4   34     2
5   43     1
6   54     1
7   56     2
8   65     1
9   67     2
10 324     1
11 435     3
12 453     1
13 456     1
14 567     1
15 657     1

这里有一个快速而肮脏的方法:

x <- 23
length(subset(numbers, numbers==x))