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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
[1] 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
[39] 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)
$breaks [1] 0 1 2 3 4 5 6 7 8$counts
[1] 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.