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

我可能会这样做

length(which(numbers==x))

但实际上，更好的方法是

table(numbers)

这里有一个快速而肮脏的方法:

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

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

这是一维原子向量的快速解。它依赖于match()，所以它与NA兼容:

x <- c("a", NA, "a", "c", "a", "b", NA, "c")

fn <- function(x) {
  u <- unique.default(x)
  out <- list(x = u, freq = .Internal(tabulate(match(x, u), length(u))))
  class(out) <- "data.frame"
  attr(out, "row.names") <- seq_along(u)
  out
}

fn(x)

#>      x freq
#> 1    a    3
#> 2 <NA>    2
#> 3    c    2
#> 4    b    1

您还可以调整算法，使其不运行unique()。

fn2 <- function(x) {
  y <- match(x, x)
  out <- list(x = x, freq = .Internal(tabulate(y, length(x)))[y])
  class(out) <- "data.frame"
  attr(out, "row.names") <- seq_along(x)
  out
}

fn2(x)

#>      x freq
#> 1    a    3
#> 2 <NA>    2
#> 3    a    3
#> 4    c    2
#> 5    a    3
#> 6    b    1
#> 7 <NA>    2
#> 8    c    2

在需要该输出的情况下，您甚至可能不需要它来重新返回原始向量，而第二列可能就是您所需要的全部。你可以用pipe在一行中得到:

match(x, x) %>% `[`(tabulate(.), .)

#> [1] 3 2 3 2 3 1 2 2

2021年的基本解决方案

aggregate(numbers, list(num=numbers), length)

       num x
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

tapply(numbers, numbers, length)
  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 

by(numbers, list(num=numbers), length)
num: 4
[1] 2
-------------------------------------- 
num: 5
[1] 1
-------------------------------------- 
num: 23
[1] 2
-------------------------------------- 
num: 34
[1] 2
-------------------------------------- 
num: 43
[1] 1
-------------------------------------- 
num: 54
[1] 1
-------------------------------------- 
num: 56
[1] 2
-------------------------------------- 
num: 65
[1] 1
-------------------------------------- 
num: 67
[1] 2
-------------------------------------- 
num: 324
[1] 1
-------------------------------------- 
num: 435
[1] 3
-------------------------------------- 
num: 453
[1] 1
-------------------------------------- 
num: 456
[1] 1
-------------------------------------- 
num: 567
[1] 1
-------------------------------------- 
num: 657
[1] 1