如何在R语言自定义函数中新增判断条件适配多场景绘图需求

修改说明 #

• 新增判断分支优先级高于原有两个分支，只要满足存在3行及以上Numbers列取值相同的条件就优先执行对应逻辑
• 默认使用频次统计判断任意3行取值相同的场景，若需要判断连续3行取值相同可替换对应判断逻辑
• 修正新分支中yz的取值逻辑，避免多值冲突

修改后的完整函数代码 #

f1 <- 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

  plot(Numbers ~ Days,  xlim= c(0,45), ylim= c(0,30),
       xaxs='i',data = datas,main = paste0(dmda, "-", CategoryChosse))

  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

  # --------------------------新增判断逻辑开始--------------------------
  # 场景1：判断是否存在任意3行Numbers取值相同（适配你给出的两类测试数据）
  val_freq <- table(datas$Numbers)
  if (any(val_freq >=3)) {
    # 取出现次数≥3的数值，若有多个默认取出现频次最高的
    yz <- as.numeric(names(which.max(val_freq)))
    lines(c(0,datas$Days), c(yz, datas$Numbers), lwd = 2)
    points(0, yz, col = "red", pch = 19, cex = 2, xpd = TRUE)
    text(.1,yz+ .5,round(yz,1), cex=1.1,pos=4,offset =1,col="black")
  # 场景2：如果需要判断连续3行Numbers取值相同，替换上面3行代码为：
  # run_len <- rle(datas$Numbers)$lengths
  # if (any(run_len >=3)) {
  #   yz <- datas$Numbers[which.max(run_len)]
  # --------------------------新增判断逻辑结束--------------------------
  } else if (nrow(datas)<=2){
    abline(h=m,lwd=2) 
    points(0, m, col = "red", pch = 19, cex = 2, xpd = TRUE)
    text(.1,m+ .5, round(m,1), cex=1.1,pos=4,offset =1,col="black")
  } else{
    mod <- nls(Numbers ~ b1*Days^2+b2,start = list(b1 = 0,b2 = 0),data = datas, algorithm = "port")
    new.data <- data.frame(Days = with(datas, seq(min(Days),max(Days),len = 45)))
    new.data <- rbind(0, new.data)
    lines(new.data$Days,predict(mod,newdata = new.data),lwd=2)
    coef<-coef(mod)[2]
    points(0, coef, col="red",pch=19,cex = 2,xpd=TRUE)
    text(.99,coef + 1,max(0, round(coef,1)), cex=1.1,pos=4,offset =1,col="black")
  }
}

内容的提问来源于stack exchange，提问作者user16774617