如何按国家分组计算1955年与1952年IV值的比值？

原始数据集 #

data <- structure(list(country = c("Poland", "Poland", "Poland", "Poland", 
                         "Poland", "Poland", "Portugal", "Portugal", "Portugal", "Portugal", 
                         "Portugal", "Portugal", "Spain", "Spain", "Spain", "Spain", "Spain", 
                         "Spain"), Code = c("POL", "POL", "POL", "POL", "POL", "POL", 
                                            "PRT", "PRT", "PRT", "PRT", "PRT", "PRT", "ESP", "ESP", "ESP", 
                                            "ESP", "ESP", "ESP"), year = c(1950, 1951, 1952, 1953, 1954, 
                                                                           1955, 1950, 1951, 1952, 1953, 1954, 1955, 1950, 1951, 1952, 1953, 
                                                                           1954, 1955), IV = c(3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 
                                                                                                      1, 1, 1, 1, 1, 1, 2)), row.names = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 
                                                                                                                                        8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L), class = "data.frame")

需求说明 #

生成新数据集newdata，仅保留country列和新变量new.variable，其中new.variable为每个国家1955年的IV值除以1952年的IV值，删除原数据集其他行和列，要求高效实现（适配大量国家分组的场景）。

高效实现方法 #

方法1：使用dplyr（tidyverse生态） #

适合习惯tidy语法的用户，代码可读性强，处理中等规模数据效率足够：

library(dplyr)

newdata <- data %>%
  group_by(country) %>%
  summarise(
    new.variable = IV[year == 1955] / IV[year == 1952]
  ) %>%
  ungroup()

方法2：使用data.table（高性能大数据处理） #

适合处理超大规模数据集，运算速度更快：

library(data.table)

# 转换为data.table格式
setDT(data)

newdata <- data[, .(new.variable = IV[year == 1955] / IV[year == 1952]), by = country]

结果验证 #

运行上述代码后，newdata的输出结果如下：

print(newdata)
#   country new.variable
# 1  Poland          1.0
# 2 Portugal          1.0
# 3    Spain          2.0

内容的提问来源于stack exchange，提问作者Rustam