## Tuesday, February 05, 2013

### counts numbers in a interval

Say I have a list of values, and I cut them by some break points, how do I know the number of values in each interval?

We know cut() function in R works for the purpose.  For example,

```tx0 <- c(9, 4, 6, 5, 3, 10, 5, 3, 5)
x <- rep(0:8, tx0)```
```> x
 0 0 0 0 0 0 0 0 0 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 4 4 4 5 5 5 5 5 5 5 5 5 5 6
 6 6 6 6 7 7 7 8 8 8 8 8```
```> table( cut(x, b = 8))

(-0.008,0.994]      (0.994,2]          (2,3]          (3,4]          (4,5]
9              4              6              5             13
(5,6]       (6,7.01]    (7.01,8.01]
5              3              5 ```

In the cut() document, there is a note, saying

```
Instead of `table(cut(x, br))`, `hist(x, br, plot = FALSE)` is more efficient and less memory hungry. Instead of `cut(*, labels = FALSE)`, `findInterval()` is more efficient.

But if you try as it said, you will the counts returned look different:

> hist(x, 8, plot=F)
\$breaks
 0 1 2 3 4 5 6 7 8
\$counts
 13  6  5  3 10  5  3  5

What's wrong?

Nothing is wrong. Just missed argument. "When `breaks` is specified as a single number, the range of the data is divided into `breaks` pieces of equal length, and then the outer limits are moved away by 0.1% of the range to ensure that the extreme values both fall within the break intervals. (If `x` is a constant vector, equal-length intervals are created, one of which includes the single value.)"

The conclusion is:
when breaks is a vector, table( cut(x, b = 0:8,include.lowest = T)) is equal to hist(x, breaks=0:8, plot=F)\$counts; when breaks is a single number, it's not.

```