R语言：按ID计算指定日期差值并新增数据列的技术求助

Hey there! Let's work through this R problem together. I'll start by addressing the date difference calculation you need, then we can circle back to the 'Performed' column once you share a bit more about what that should track.

First, let's assume your data looks something like this (I made a sample to demonstrate):

# Sample data to mimic your structure
df <- data.frame(
  ID = c(0, 1, 0, 2, 2, 0, 3),
  Date = as.Date(c("2023-01-01", "2023-01-05", "2023-01-10", "2023-01-15", "2023-01-17", "2023-01-20", "2023-01-25"))
)

Using dplyr (tidyverse) for clean, readable code #

I recommend using the dplyr package here—it makes this kind of row-wise and grouping logic straightforward. Here's how to compute the date difference and set up the columns:

library(dplyr)

df_processed <- df %>%
  # Step 1: Capture dates where ID = 0, leave others as NA
  mutate(prev_zero_date = ifelse(ID == 0, Date, NA)) %>%
  # Step 2: Fill those NA values with the most recent previous 0 date
  fill(prev_zero_date, .direction = "down") %>%
  # Step 3: Calculate difference only for non-0 IDs; set 0 IDs to NA
  mutate(Difference = ifelse(ID != 0, Date - prev_zero_date, NA)) %>%
  # Step 4: Example for Performed column (adjust based on your actual need!)
  # I'm assuming this marks if a difference was calculated—tweak as needed
  mutate(Performed = ifelse(ID != 0, TRUE, FALSE))

# Check the result
print(df_processed)

What this does: #

• prev_zero_date: Creates a column that holds the date of the last preceding row with ID = 0 for every row. The fill() function ensures we carry that date forward until the next ID = 0 comes along.
• Difference: Computes the number of days between the current row's date and the last 0 date—only for rows where ID isn't 0 (you can adjust this if you want 0 rows to show 0 instead of NA).
• Performed: Since your original request was incomplete, I added a simple example where it flags rows where we calculated a difference. If this needs to track something else (like whether a task was performed), just share more details and I'll adjust the code!

If you prefer base R (no packages needed): #

If you don't want to use dplyr, here's a base R approach that does the same thing:

# Base R solution
df$prev_zero_date <- NA
zero_rows <- which(df$ID == 0)

for (i in 1:nrow(df)) {
  # Find the last 0 row before the current one
  last_zero <- max(zero_rows[zero_rows < i])
  df$prev_zero_date[i] <- df$Date[last_zero]
}

# Calculate difference and Performed column
df$Difference <- ifelse(df$ID != 0, df$Date - df$prev_zero_date, NA)
df$Performed <- ifelse(df$ID != 0, TRUE, FALSE)

Quick note: Make sure your Date column is formatted as a proper R Date object first! If it's stored as text, convert it with df$Date <- as.Date(df$Date).

Let me know if you need to adjust the logic (like handling edge cases, different date units, or clarifying what the 'Performed' column should do)!

内容的提问来源于stack exchange，提问作者Arjan