使用plyr包dlply/ldply后，如何按Subject ID生成起始为1的序列列

Hey there! Let's walk through this with a concrete example so you can see exactly how to add that incrementing column you need, while keeping your existing workflow with plyr's dlply and ldply.

First, let's start with a reproducible dataset to mimic your scenario:

library(plyr)

# Create sample data matching your structure
set.seed(123)
df <- data.frame(
  `Subject IDs` = rep(c(10, 11, 12), each = 3),
  time_seq = rep(c(1, 2, 3), 3),
  measurement = rnorm(9) # Random values to represent your data
)

Step 1: Add the incrementing column first (simplest approach) #

Before splitting your data, you can generate the group-specific incrementing column using ddply to group by Subject IDs and add a sequence number for each row in the group:

# Add a new column "group_seq" that starts at 1 for each Subject ID
df_with_seq <- ddply(df, .(`Subject IDs`), transform, 
                     group_seq = seq_along(`Subject IDs`))

The seq_along(Subject IDs) part will generate a sequence from 1 to the number of rows in each Subject group—perfect for your "start at 1 and increment" requirement.

Step 2: Split into named lists (as you originally did) #

Now split the modified data frame by Subject IDs and time_seq to get your desired named list (like 10.1, 10.2):

split_list <- dlply(df_with_seq, .(`Subject IDs`, time_seq), identity)

# Check the list names to confirm they match your format
names(split_list)
# Output: "10.1" "10.2" "10.3" "11.1" "11.2" "11.3" "12.1" "12.2" "12.3"

Step 3: Merge back to a data frame with ldply #

Finally, use ldply to combine the list back into a single data frame, which will include your new group_seq column:

final_df <- ldply(split_list, identity)

# View the result
head(final_df)

This will give you output like this:

Subject IDs time_seq measurement group_seq
1          10        1  -0.5604756         1
2          10        2  -0.2301775         2
3          10        3   1.5587083         3
4          11        1   0.0705084         1
5          11        2   0.1292877         2
6          11        3   1.7150650         3

If you need to add the column after splitting #

If you must add the column post-split (e.g., if your list is already generated), you can adjust the workflow to first split by Subject IDs, add the sequence to each group, then split further by time_seq:

# Split by Subject IDs first
split_by_subject <- dlply(df, .(`Subject IDs`), function(subject_data) {
  # Add the incrementing column to each subject's data
  subject_data$group_seq <- seq_len(nrow(subject_data))
  subject_data
})

# Split each subject's data by time_seq to get the ID.time names
final_split_list <- ldply(split_by_subject, function(x) dlply(x, .(time_seq), identity)) %>%
  unlist(recursive = FALSE)

# Fix the names to match ID.time format
names(final_split_list) <- sapply(strsplit(names(final_split_list), "\\."), function(parts) paste(parts[2], parts[1], sep = "."))

# Merge back to data frame
final_df <- ldply(final_split_list, identity)

Either way, you'll end up with the incrementing column grouped by Subject IDs that you need.

内容的提问来源于stack exchange，提问作者Frederick