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

最直接的方法是求和(numbers == x)。

numbers == x创建一个在x出现的每个位置都为TRUE的逻辑向量，当求和时，逻辑向量被强制转换为numeric，将TRUE转换为1,FALSE转换为0。

但是，请注意，对于浮点数，最好使用如下格式:sum(abs(numbers - x) < 1e-6)。

这里有一个快速而肮脏的方法:

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

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

你可以创建一个函数来得到结果。

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

function1<-function(x){
    if(x==value){return(1)}else{ return(0) }
}

# set your value here
value<-4

# make a vector which return 1 if it equal to your value, 0 else
vector<-sapply(numbers,function(x) function1(x))
sum(vector)

结果:2