I have a vector of numbers:

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

How can I have R count the number of times a value *x* appears in the vector?

answered 8 years ago Shane #1

You can just use `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
```

Then you can subset it:

```
> a[names(a)==435]
435
3
```

Or convert it into a data.frame if you're more comfortable working with that:

```
> as.data.frame(table(numbers))
numbers Freq
1 4 2
2 5 1
3 23 2
4 34 2
...
```

answered 8 years ago JD Long #2

here's one fast and dirty way:

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

answered 8 years ago Jesse #3

I would probably do something like this

```
length(which(numbers==x))
```

But really, a better way is

```
table(numbers)
```

answered 8 years ago hadley #4

The most direct way is `sum(numbers == x)`

.

`numbers == x`

creates a logical vector which is TRUE at every location that x occurs, and when `sum`

ing, the logical vector is coerced to numeric which converts TRUE to 1 and FALSE to 0.

However, note that for floating point numbers it's better to use something like: `sum(abs(numbers - x) < 1e-6)`

.

answered 6 years ago Sergej Andrejev #5

There is a standard function in R for that

`tabulate(numbers)`

answered 5 years ago JBecker #6

My preferred solution uses `rle`

, which will return a value (the label, `x`

in your example) and a length, which represents how many times that value appeared in sequence.

By combining `rle`

with `sort`

, you have an extremely fast way to count the number of times any value appeared. This can be helpful with more complex problems.

Example:

```
> numbers <- c(4,23,4,23,5,43,54,56,657,67,67,435,453,435,324,34,456,56,567,65,34,435)
> a <- rle(sort(numbers))
> a
Run Length Encoding
lengths: int [1:15] 2 1 2 2 1 1 2 1 2 1 ...
values : num [1:15] 4 5 23 34 43 54 56 65 67 324 ...
```

If the value you want doesn't show up, or you need to store that value for later, make `a`

a `data.frame`

.

```
> b <- data.frame(number=a$values, n=a$lengths)
> b
values n
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
```

I find it is rare that I want to know the frequency of one value and not all of the values, and rle seems to be the quickest way to get count and store them all.

answered 5 years ago geotheory #7

There is also `count(numbers)`

from `plyr`

package. Much more convenient than `table`

in my opinion.

answered 3 years ago Akash #8

One more way i find convenient is:

```
numbers <- c(4,23,4,23,5,43,54,56,657,67,67,435,453,435,324,34,456,56,567,65,34,435)
(s<-summary (as.factor(numbers)))
```

This converts the dataset to factor, and then summary() gives us the control totals (counts of the unique values).

Output is:

```
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
```

This can be stored as dataframe if preferred.

as.data.frame(cbind(Number = names(s),Freq = s), stringsAsFactors=F, row.names = 1:length(s))

here row.names has been used to rename row names. without using row.names, column names in s are used as row names in new dataframe

Output is:

```
Number Freq
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
```

answered 3 years ago pomber #9

Using table but without comparing with `names`

:

```
numbers <- c(4,23,4,23,5,43,54,56,657,67,67,435)
x <- 67
numbertable <- table(numbers)
numbertable[as.character(x)]
#67
# 2
```

`table`

is useful when you are using the counts of different elements several times. If you need only one count, use `sum(numbers == x)`

answered 3 years ago Berny #10

If you want to count the number of appearances subsequently, you can make use of the `sapply`

function:

```
index<-sapply(1:length(numbers),function(x)sum(numbers[1:x]==numbers[x]))
cbind(numbers, index)
```

Output:

```
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
```

answered 2 years ago uttkarsh dharmadhikari #11

You can change the number to whatever you wish in following line

```
length(which(numbers == 4))
```

answered 9 months ago ishandutta2007 #12

```
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
```