基于组分界点的分箱数据分位数划分及按ts_id分组的中位数分箱方法

Hey there! Let's break down your questions one by one, with practical explanations and code for your data.table scenario:

1. Using Group-Specific Cut Points to Split Data into Quantiles #

Here's a straightforward, step-by-step approach to use group-based分界点 (cut points) for quantile binning:

• Step 1: Define your grouping logic
  First, identify which column(s) you’ll use to split your data into groups (e.g., ts_id in your second question).
• Step 2: Calculate group-specific quantile thresholds
  For each group, compute the quantile values you need (like median for 2 bins) using the subset of data relevant to that group (e.g., only rows where ex_id == 1). These values become your group’s cut points.
• Step 3: Assign bins to all data
  Join the computed thresholds back to your full dataset, then use conditional logic to categorize each row into the appropriate bin based on its group’s cut point.

2. Data.Table Implementation for Your Specific Scenario #

Let’s walk through the exact code to solve your problem with the DT data.table:

First, let’s recap the goal: For each ts_id group, calculate the median of s1 where ex_id == 1, then split all rows in that ts_id group into 2 bins using that median.

Step 1: Compute group-specific medians #

First, we calculate the median of s1 for ex_id == 1 rows, grouped by ts_id:

# Load data.table if not already loaded
library(data.table)

# Your original data (I'll set a seed for reproducibility)
set.seed(123)
DT = data.table( ts_id = c(rep(1,15),rep(2,15)), ex_id = c(rep(1,5),rep(2,10),rep(1,5),rep(2,10)), s1 = rnorm(30) )

# Calculate median of s1 (ex_id == 1) per ts_id
group_medians <- DT[ex_id == 1, .(s1_median = median(s1)), by = ts_id]

Step 2: Join medians to full data and assign bins #

We’ll merge the median values back to the original table, then use fifelse() (data.table’s fast if-else) to create the bin column:

# Join median values to DT
DT <- DT[group_medians, on = "ts_id"]

# Assign bins: 1 for s1 <= median, 2 for s1 > median
DT[, bin := fifelse(s1 <= s1_median, 1, 2)]

# Optional: Drop the s1_median column if you don't need it anymore
DT[, s1_median := NULL]

What this does: #

• For each ts_id group, we only use the median from rows where ex_id == 1 as the cut point.
• Every row in the same ts_id group is split into bin 1 (below/equal to the group’s median) or bin 2 (above the median).

You can verify the result by checking the bins alongside ts_id and s1—rows in the same ts_id will be split based on the correct group-specific threshold.

内容的提问来源于stack exchange，提问作者zlu15