基于特定值条件对成对列向量计算行求和(R语言)
问题:按条件配对列求和生成
overall_total列 原始数据集
df <- structure(list(datA = c(1L, NA, 5L, 3L, 8L, NA), datA_total = c(20L, 30L, 40L, 15L, 10L, NA), datB = c(5L, 5L, NA, 6L, 1L, NA), datB_total = c(80L, 10L, 10L, 5L, 4L, NA), datC = c(NA, 4L, 1L, NA, 3L, NA), datC_total = c(NA, 10L, 15L, NA, 20L, NA)), class = "data.frame", row.names = c(NA, -6L))
打印后结果:
# datA datA_total datB datB_total datC datC_total #1 1 20 5 80 NA NA #2 NA 30 5 10 4 10 #3 5 40 NA 10 1 15 #4 3 15 6 5 NA NA #5 8 10 1 4 3 20 #6 NA NA NA NA NA NA
需求
创建overall_total列,仅当类型列(datA/datB/datC)的值在1-5范围内时,将对应总计列(datA_total/datB_total/datC_total)的值纳入行求和;全NA的行求和为0。期望结果:
# datA datA_total datB datB_total datC datC_total overall_total #1 1 20 5 80 NA NA 100 #2 NA 30 5 10 4 10 20 #3 5 40 NA 10 1 15 55 #4 3 15 6 5 NA NA 15 #5 8 10 1 4 3 20 24 #6 NA NA NA NA NA NA 0
错误代码分析
你尝试的代码逻辑完全错误:
type_vars <- c("datA", "datB", "datC") type_scores <- c("1", "2", "3", "4", "5") type_visits <- c("datA_total", "datB_total", "datC_total") df <- df %>% mutate(overall_total = rowSums(all_of(type_visits[type_vars %in% type_scores])))
type_vars是列名字符向量,type_scores是数字字符串,两者用%in%比较毫无意义,导致type_visits[type_vars %in% type_scores]是空向量,rowSums无法计算- 没有实现按行判断每个类型列的值是否符合条件,再对应取总计列求和的核心逻辑
正确解法
方法1:用rowwise+purrr批量处理配对列
适合列数较多的场景,扩展性强:
library(dplyr) library(purrr) # 定义类型列与总计列的配对关系 type_pairs <- list( c("datA", "datA_total"), c("datB", "datB_total"), c("datC", "datC_total") ) df <- df %>% rowwise() %>% mutate( overall_total = sum( map_dbl(type_pairs, ~{ # 取出当前行的类型列值和总计列值 type_val <- pick(all_of(.x[1]))[[1]] total_val <- pick(all_of(.x[2]))[[1]] # 符合条件则取总计值,否则取0 if (!is.na(type_val) && type_val %in% 1:5) total_val else 0 }), na.rm = TRUE ) ) %>% ungroup()
方法2:直接对每对列做条件判断(简洁直观)
适合列数较少的场景:
library(dplyr) df <- df %>% mutate( overall_total = # 对每对列分别判断,符合条件则加对应总计,否则加0 (ifelse(between(datA, 1, 5) & !is.na(datA), datA_total, 0) + ifelse(between(datB, 1, 5) & !is.na(datB), datB_total, 0) + ifelse(between(datC, 1, 5) & !is.na(datC), datC_total, 0)) %>% # 把全NA的行结果替换为0 replace_na(0) )
两种方法都能得到你期望的结果,其中方法2代码更简洁,方法1适合后续新增配对列的情况,只需在type_pairs里添加新的配对即可。
内容的提问来源于stack exchange,提问作者Monarch
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