如何在R中基于年积日(DOY)对多列数据进行每日求和？

Hey Sarah, I see where you hit a snag with the aggregate() function—let’s get this sorted out for you! The issue was your column selection syntax, and we have a couple of straightforward fixes to sum columns 4 through 44 by DOY (or Year+DOY, if you need to account for overlapping DOYs across different years).

Method 1: Using Formula Syntax with Column Subsetting #

First, we’ll explicitly select the grouping variable(s) and the columns you want to sum, then use . in the formula to represent all non-grouping columns. We’ll also add na.rm = TRUE to handle those NaN values in your data (otherwise, your sums would return NaN).

Code for DOY-only grouping: #

# Subset to DOY + columns 4-44, then aggregate
Js_gm2d <- aggregate(. ~ DOY, 
                     data = Js_30min[, c("DOY", 4:44)], 
                     FUN = sum, 
                     na.rm = TRUE)

Code for Year+DOY grouping (to avoid merging same DOYs across different years): #

# Subset to Year + DOY + columns 4-44, then aggregate
Js_gm2d <- aggregate(. ~ Year + DOY, 
                     data = Js_30min[, c("Year", "DOY", 4:44)], 
                     FUN = sum, 
                     na.rm = TRUE)

Method 2: Using x and by Parameters (More Explicit) #

If you prefer a more direct approach without formula syntax, you can specify the columns to aggregate (x) and the grouping variables (by) separately:

Code for DOY-only grouping: #

Js_gm2d <- aggregate(x = Js_30min[, 4:44], 
                     by = list(DOY = Js_30min$DOY), 
                     FUN = sum, 
                     na.rm = TRUE)

Code for Year+DOY grouping: #

Js_gm2d <- aggregate(x = Js_30min[, 4:44], 
                     by = list(Year = Js_30min$Year, DOY = Js_30min$DOY), 
                     FUN = sum, 
                     na.rm = TRUE)

Why Your Original Code Failed #

Your line aggregate(c(,4:44)~DOY,data=Js_30min,FUN=sum) had invalid syntax: c(,4:44) doesn’t make sense to R—you need to explicitly define the columns you want to include, either by position, name, or by subsetting the data first.

Quick Test with Your Sample Data #

Let’s verify with your sample data snippet:

# Sample data
Js_30min <- data.frame(
  Year = rep(2018, 6),
  DOY = rep(157, 6),
  Time = c(0, 30, 100, 130, 200, 230),
  SI1 = c(0, 297.9779, 349.7710, 101.0535, 143.9961, 0),
  SI2 = c(0, 493.8855, 1168.0555, 1279.5865, 1392.9739, 891.1722),
  SI3 = rep(NaN, 6)
)

# Run aggregation
Js_gm2d <- aggregate(. ~ DOY, data = Js_30min[, c("DOY", 4:6)], FUN = sum, na.rm = TRUE)

This will return:

DOY      SI1      SI2 SI3
1 157 892.7985 5225.674   0

Note that SI3 sums to 0 because all values were NaN and we used na.rm = TRUE.

内容的提问来源于stack exchange，提问作者Sarah