我有一个数字向量:
numbers <- c(4,23,4,23,5,43,54,56,657,67,67,435,
453,435,324,34,456,56,567,65,34,435)
我如何让R计算x值在向量中出现的次数?
当前回答
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
其他回答
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
...
这里有一个快速而肮脏的方法:
x <- 23
length(subset(numbers, numbers==x))
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
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
