在R [duplicate]中查找向量中最常见的元素

This question already has an answer here:

这个问题已经有了答案:

Fastest way to count occurrences of each unique element 2 answers
计算每个唯一元素2答案出现次数的最快方法
Counting the number of elements with the values of x in a vector 12 answers
用向量12中的x的值来计算元素的个数。

I have a vector say

我有一个向量。

c(1,1,1,1,1,1,2,3,4,5,7,7,5,7,7,7)

How do I find the counts of each element and return the 3 most often occurring elements, i.e. 1, 7, 5?

如何找到每个元素的计数并返回最常发生的3个元素，即1、7、5?

I think this should be really simple but I am having trouble with this.

我认为这应该很简单，但我有问题。

4 个解决方案

#1

I'm sure this is a duplicate, but the answer is simple:

我肯定这是一份副本，但答案很简单:

sort(table(variable),decreasing=TRUE)[1:3]

#2

I don't know if this is better than the table approach, but if your list is already a factor then its summary method will give you frequency counts:

我不知道这是否比表格方法更好，但如果你的列表已经是一个因素，那么它的总结方法将会给你频率计数:

> summary(as.factor(c(1,1,1,1,1,1,2,3,4,5,7,7,5,7,7,7)))
1 2 3 4 5 7 
6 1 1 1 2 5

And then you can get the top 3 most frequent like so:

然后你可以得到前三种最常见的情况如下:

> names(sort(summary(as.factor(c(1,1,1,1,1,1,2,3,4,5,7,7,5,7,7,7))), decreasing=T)[1:3])
[1] "1" "7" "5"

#3

If your vector contains only integers, tabulate will be much faster than anything else. There are a couple of catches to be aware of:

如果你的向量只包含整数，表格将比其他任何东西都快。有几个问题需要注意:

It'll by default return the count for numbers from 1 to N.
默认情况下，它会返回从1到N的计数。
It'll return an unnamed vector.
它会返回一个未命名的向量。

That means, if your x = c(1,1,1,3) then tabulate(x) will return (3, 0, 1). Note that the counts are for 1 to max(x) by default.

这意味着，如果你的x = c(1,1,1,1,3)那么表化(x)将返回(3,0,1)。

How can you use tabulate to make sure that you can pass any numbers?

如何使用表格来确保可以传递任何数字?

set.seed(45)
x <- sample(-5:5, 25, TRUE)
#  [1]  1 -2 -3 -1 -2 -2 -3  1 -3 -5 -1  4 -2  0 -1 -1  5 -4 -1 -3 -4 -2  1  2  4

Just add abs(min(x))+1 when min(x) <= 0 to make sure that the values start from 1. If min(x) > 0, then just use tabulate directly.

只要在min(x) <= 0时添加abs(min(x))+1，确保值从1开始。如果min(x) >，则直接使用表格。

sort(setNames(tabulate(x + ifelse(min(x) <= 0, abs(min(x))+1, 0)), 
      seq(min(x), max(x))), decreasing=TRUE)[1:3]

If your vector does contain NA, then you can use table with useNA="always" parameter.

如果向量确实包含NA，那么可以使用带有useNA="always"参数的表。

#4

you can use table() function to get a tabulation of the frequency of values in an array/vector and then sort this table.

您可以使用table()函数来获得数组/vector中值的频率列表，然后对该表进行排序。

x = c(1, 1, 1, 2, 2)
sort(table(x))
2 1
2 3

#1