R语言使用dplyr计算系数时如何将负值自动替换为0

解决方法 #

你可以通过两种方式实现负值自动替换为0的需求：

方案1：修改return_coef函数，从源头限制返回值非负 #

直接在函数的两个返回结果处用pmax()将输出值和0取最大值，负数会自动被替换为0，修改后的函数如下：

return_coef <- function(dmda, CategoryChosse) {
  x<-df1 %>% select(starts_with("DR0"))
  
  x<-cbind(df1, setNames(df1$DR1 - x, paste0(names(x), "_PV")))
  PV<-select(x, date2,Week, Category, DR1, ends_with("PV"))
  
  med<-PV %>%
    group_by(Category,Week) %>%
    summarize(across(ends_with("PV"), median))
  
  SPV<-df1%>%
    inner_join(med, by = c('Category', 'Week')) %>%
    mutate(across(matches("^DR0\\d+$"), ~.x + 
                    get(paste0(cur_column(), '_PV')),
                  .names = '{col}_{col}_PV')) %>%
    select(date1:Category, DR01_DR01_PV:last_col())
  
  SPV<-data.frame(SPV)
  
  mat1 <- df1 %>%
    filter(date2 == dmda, Category == CategoryChosse) %>%
    select(starts_with("DR0")) %>%
    pivot_longer(cols = everything()) %>%
    arrange(desc(row_number())) %>%
    mutate(cs = cumsum(value)) %>%
    filter(cs == 0) %>%
    pull(name)
  
  (dropnames <- paste0(mat1,"_",mat1, "_PV"))
  
  SPV <- SPV %>%
    filter(date2 == dmda, Category == CategoryChosse) %>%
    select(-any_of(dropnames))
  
  datas<-SPV %>%
    filter(date2 == ymd(dmda)) %>%
    group_by(Category) %>%
    summarize(across(starts_with("DR0"), sum)) %>%
    pivot_longer(cols= -Category, names_pattern = "DR0(.+)", values_to = "val") %>%
    mutate(name = readr::parse_number(name))
  colnames(datas)[-1]<-c("Days","Numbers")
  
  datas <- datas %>% 
    group_by(Category) %>% 
    slice((as.Date(dmda) - min(as.Date(df1$date1) [
      df1$Category == first(Category)])-2):max(Days)+1) %>%
    ungroup
  
  m<-df1 %>%
    group_by(Category,Week) %>%
    summarize(across(starts_with("DR1"), mean))
  
  m<-subset(m, Week == df1$Week[match(ymd(dmda), ymd(df1$date2))] & Category == CategoryChosse)$DR1
  
  if (nrow(datas)<=2){
    pmax(as.numeric(m), 0)
  }
  else{
    mod <- nls(Numbers ~ b1*Days^2+b2,start = list(b1 = 0,b2 = 0),data = datas, algorithm = "port")
    pmax(as.numeric(coef(mod)[2]), 0)
  }
}

运行原有生成结果的代码即可得到替换后的结果：

cbind(df1 %>% select(date2, Category), coef = mapply(return_coef, df1$date2, df1$Category))

输出结果：

date2 Category     coef
1 2021-06-29      FDE 5.347892
2 2021-06-29      ABC 1.369478
3 2021-07-06      FDE 0.000000
4 2021-07-06      ABC 0.000000
5 2021-07-06      DDE 2.000000

方案2：不修改原有函数，最后生成结果时统一替换 #

如果你不想改动原有函数逻辑，也可以在生成最终结果后对coef列做处理：

library(dplyr)
result <- cbind(df1 %>% select(date2, Category), coef = mapply(return_coef, df1$date2, df1$Category))
result <- result %>% mutate(coef = pmax(coef, 0))

效果和方案1完全一致。

内容的提问来源于stack exchange，提问作者Antonio