R语言中按指定条件从多列筛选值并分配行ID的技术问询

Got it, let's walk through this step by step to get exactly what you need. We'll start by structuring your data, assign the IDs, then implement the conditional checks to filter rows and generate a new column.

Step 1: Set Up the Data Frame with IDs #

First, we'll combine your columns into a data frame and add the ID#1 to ID#5 labels as requested:

# Define your original columns
column1 <- c("rice 2", "apple 4", "melon 6", "blueberry 4", "orange 6")
column2 <- c("rice 8", "blueberry 8", "grape 10", "water 10", "mango 3")
column3 <- c("rice 6", "apple 8", "blueberry 12", "pineapple 8", "mango 3")

# Create data frame with ID column
df <- data.frame(
  ID = paste0("ID#", 1:5),
  column1 = column1,
  column2 = column2,
  column3 = column3,
  stringsAsFactors = FALSE
)

Step 2: Define Conditional Checks #

We need to check if any entry in a row meets one of these criteria:

• rice > 5
• blueberry > 7
• orange > 5

We'll use a helper function to scan each row, split the food-value strings, and validate the conditions. We'll also use tidyverse for cleaner data manipulation (install it with install.packages("tidyverse") if you don't have it):

library(tidyverse)

# Function to check if a row meets any of the conditions
meets_condition <- function(row) {
  # Extract all food-value pairs from the row's columns
  all_pairs <- c(row["column1"], row["column2"], row["column3"])
  # Split each pair into food name and numeric value
  split_pairs <- str_split(all_pairs, " ", simplify = TRUE)
  foods <- split_pairs[, 1]
  values <- as.numeric(split_pairs[, 2])
  
  # Check if any pair matches the conditions
  any(
    (foods == "rice" & values > 5) |
    (foods == "blueberry" & values > 7) |
    (foods == "orange" & values > 5)
  )
}

Step 3: Generate New Column & Filter Rows #

Now we'll add a new column to mark which rows meet the condition, then filter to keep only those rows (which will be ID#1, ID#2, ID#3, ID#5 as expected):

# Add column indicating if row meets conditions
df$meets_condition <- apply(df, 1, meets_condition)

# Filter to keep only qualifying rows
filtered_df <- df %>% filter(meets_condition)

# View the result
print(filtered_df)

Output of Filtered Data Frame #

ID      column1       column2       column3 meets_condition
1 ID#1      rice 2       rice 8       rice 6             TRUE
2 ID#2     apple 4 blueberry 8      apple 8             TRUE
3 ID#3     melon 6      grape 10 blueberry 12             TRUE
4 ID#5    orange 6      mango 3      mango 3             TRUE

Optional: Add a Column with Valid Entries #

If you want a new column that explicitly lists which entries satisfied the conditions (instead of just a TRUE/FALSE flag), use this extended function:

# Function to extract all valid food-value pairs for a row
get_valid_entries <- function(row) {
  all_pairs <- c(row["column1"], row["column2"], row["column3"])
  split_pairs <- str_split(all_pairs, " ", simplify = TRUE)
  foods <- split_pairs[, 1]
  values <- as.numeric(split_pairs[, 2])
  
  # Filter and format valid pairs
  valid_pairs <- paste(foods, values)[
    (foods == "rice" & values > 5) |
    (foods == "blueberry" & values > 7) |
    (foods == "orange" & values > 5)
  ]
  
  paste(valid_pairs, collapse = ", ")
}

# Add the detailed valid entries column
df$valid_entries <- apply(df, 1, get_valid_entries)

# Filter and view the enhanced result
filtered_df <- df %>% filter(meets_condition)
print(filtered_df)

Enhanced Output #

ID      column1       column2       column3 meets_condition       valid_entries
1 ID#1      rice 2       rice 8       rice 6             TRUE rice 8, rice 6
2 ID#2     apple 4 blueberry 8      apple 8             TRUE blueberry 8
3 ID#3     melon 6      grape 10 blueberry 12             TRUE blueberry 12
4 ID#5    orange 6      mango 3      mango 3             TRUE orange 6

内容的提问来源于stack exchange，提问作者Charles