It looks like the main issue here is ensuring your rounded age values are properly included in the correct intervals—especially avoiding cases where the largest rounded value falls outside your break points. Let’s walk through the adjusted code step by step:

Step 1: Round Age Values to Three Decimals #

First, explicitly round your age column to three decimal places to ensure consistency across all calculations:

# Round age values to 3 decimal places
NT$rounded_age <- round(NT$Age, 3)

Step 2: Create Properly Aligned Breaks #

Your original breaks might stop short of covering the maximum rounded age. To fix this, we’ll calculate the upper bound of the breaks by rounding up the max rounded age to the nearest 0.2 increment. This guarantees every rounded value falls into an interval:

# Get the range of rounded ages
rounded_range <- range(NT$rounded_age)
# Calculate upper break: round up max rounded age to the next 0.2
upper_break <- ceiling(rounded_range[2] / 0.2) * 0.2
# Create breaks starting at 1 (as in your original code) and ending at upper_break
breaks_NT <- seq(from = 1, to = upper_break, by = 0.2)

Step 3: Cut Values and Calculate Frequencies #

Now use these adjusted breaks to group your rounded ages, keeping your original interval logic (right=FALSE for left-inclusive intervals):

# Cut rounded ages into defined intervals
freq.cutNT <- cut(NT$rounded_age, 
                  breaks = breaks_NT, 
                  right = FALSE, 
                  include.lowest = TRUE, 
                  ordered_result = TRUE)
# Generate the frequency table
freq.freqNT <- table(freq.cutNT)
# View the final frequency counts
freq.freqNT

Example Clarification #

For your sample value 6.626601 (rounded to 6.627):

• The upper break will be calculated as ceiling(6.627 / 0.2) * 0.2 = 6.8
• This means the interval [6.6, 6.8) will correctly capture 6.627. If you intended for this value to fall into [6.4, 6.6) instead, that would contradict standard interval logic (since 6.627 is greater than 6.6)—but if you need to shift interval boundaries, you can adjust the seq start value or break step size accordingly.

内容的提问来源于stack exchange，提问作者Catarina Ferreira