如何在R中按分组统计数据框列的数值出现次数（排除NA）

Got it, let's sort this out for you! You're looking to count the total number of non-NA values in the Demand column, grouped by Material—not the number of unique values (which is what your current uniqueN() code does).

The Problem with Your Current Code #

Your existing line:

master_data_long_sub[,freq:=uniqueN(Demand),by=.(Material)]

uses uniqueN(), which counts distinct values in each group. For Material 1111, that would return 3 (since the unique non-NA Demand values are 2, 3, 5)—but you need the total count of non-NA entries, which is 4.

The Correct Solution #

To count all non-NA values per group, use sum(!is.na(Demand)) inside your data.table grouping. This checks each entry in Demand:

• is.na(Demand) returns TRUE for NA values, FALSE otherwise
• !is.na(Demand) flips that to TRUE for non-NA values
• sum() adds up all the TRUE values (treated as 1) to get the total count

Here's the full code:

Step 1: Example Data (matching your sample)

library(data.table)
master_data_long_sub <- data.table(
  Material = c(1111, 1111, 1111, 1111, 1111, 2222),
  Demand = c(2, 3, 2, NA, 5, 10)
)

Step 2: Calculate Non-NA Counts per Group

master_data_long_sub[, freq := sum(!is.na(Demand)), by = .(Material)]

Result:

Material Demand freq
1:     1111      2    4
2:     1111      3    4
3:     1111      2    4
4:     1111     NA    4
5:     1111      5    4
6:     2222     10    1

As you can see, Material 1111 gets the desired count of 4, and Material 2222 gets 1 (since it has one non-NA Demand value).

Alternative Approach #

You can also use .N (data.table's built-in count of rows) combined with a filter, though the first method is more explicit:

master_data_long_sub[, freq := .N[!is.na(Demand)], by = .(Material)]

Either way works—pick whichever makes more sense to you!

内容的提问来源于stack exchange，提问作者A380