关于R语言base包.bincode()函数示例的结果困惑

Let's break down why you're seeing that unexpected output—your confusion comes from assuming .bincode() automatically removes duplicate values in breaks, but that's not the case. Here's the step-by-step explanation:

1. How .bincode() Handles Your Input Breaks #

Your breaks vector b <- c(0, 0, 1, 1) is non-decreasing, which meets .bincode()'s requirement for valid breaks. The function does not deduplicate or modify the breaks sequence in this scenario—it uses the exact sequence you provided.

With length(b) = 4, this creates length(b)-1 = 3 intervals. Since you set right = FALSE, each interval follows the left-closed, right-open format [breaks[i], breaks[i+1)).

2. The Generated Intervals #

Here are the actual intervals created from your breaks:

• Interval 1: [0, 0) → An empty interval (no value can satisfy 0 ≤ x < 0)
• Interval 2: [0, 1) → All values where 0 ≤ x < 1 fall here
• Interval 3: [1, 1) → Another empty interval

3. Matching Your x Values to Intervals #

Now let's map each element in x <- c(0, 0.01, 0.5, 0.99, 1):

• x=0: Doesn't fit Interval 1, but fits Interval 2 → returns code 2
• x=0.01, x=0.5, x=0.99: All fall within Interval 2 → return code 2
• x=1: Per .bincode() rules, when right = FALSE, any value greater than or equal to the last break value (here, 1) is considered outside the range of breaks → returns NA

4. Getting Your Expected Output (If Needed) #

If you want .bincode() to use deduplicated breaks, you'll need to clean the breaks manually first:

x <- c(0, 0.01, 0.5, 0.99, 1)
b <- c(0, 0, 1, 1)
clean_breaks <- sort(unique(b)) # Results in c(0, 1)

# This will return [1, 1, 1, 1, NA]
.bincode(x, clean_breaks, right = FALSE, include.lowest = FALSE)

To get x=1 to map to a valid code, you'd need to extend your breaks to include a value above 1 (e.g., clean_breaks <- c(0,1,2)), which would create a second interval [1,2) and return 2 for x=1.

内容的提问来源于stack exchange，提问作者Angus Zhang