【问题标题】:Adjust condition when all columns have 0 values当所有列都有 0 值时调整条件
【发布时间】:2021-10-13 20:06:57
【问题描述】:

下面的代码根据我在date2 上选择的日期/类别生成图表。这几天是 30/06、01/07 和 02/07。对于 30/06 和 01/07,我可以正常生成,如您在附图中看到的那样,但不能用于 02/07。这是因为我所有的列都有 0 值,最终会在datas 中产生问题。所以我需要如果所有列都是 0,我希望我的图表考虑代码的这种情况:

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")}

所以我的图表没有点,只有m 中的线。

下面的可执行代码

library(dplyr)

df1 <- structure(
  list(date1= c("2021-06-28","2021-06-28","2021-06-28"),
       date2 = c("2021-06-30","2021-07-01","2021-07-02"),
       Category = c("ABC","ABC","ABC"),
       Week= c("Wednesday","Wednesday","Wednesday"),
       DR1 = c(4,1,0),
       DR01 = c(4,1,0), DR02= c(4,2,0),DR03= c(9,5,0),
       DR04 = c(5,4,0),DR05 = c(5,4,0)),
  class = "data.frame", row.names = c(NA, -3L))


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)])):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
  
  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 if(any(table(datas$Numbers) >= 3) & length(unique(datas$Numbers)) == 1){
    yz <- unique(datas$Numbers)
    lines(c(0,datas$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")}
  
  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")
  }
}


f1("2021-06-30", "ABC")
f1("2021-07-01", "ABC")
f1("2021-07-02", "ABC")

【问题讨论】:

  • 在调用points 之前,您正在创建plot。当您有 0 行的 datas 时,步骤 plot(Numbers ~ Days, xlim= c(0,45), ylim= c(0,30), xaxs='i',data = datas,main = paste0(dmda, "-", CategoryChosse)) 不起作用,并且错误发生在该步骤之前
  • Akrun,是否可以在这个问题中做一个 `if`:stackoverflow.com/questions/69535009/…
  • 我会说,对于第三种情况,列“DR0”将被删除,因为该列中的所有值都是 0 SPV &lt;- SPV %&gt;% filter(date2 == dmda, Category == CategoryChosse) %&gt;% select(-any_of(dropnames))。如果没有 DRO 列,我们可能需要一个条件
  • 如果没有 DR0 列,summarize(across(starts_with("DR0"), sum)) 步骤将失败。在这种情况下你想要什么作为价值
  • 所以,我是这样想的:如果给定日期的 DR0 列等于 0 或没有信息,即 NA,则将绘制一个图,该行将是我提到的那个条件.问题是如何生成这个图,如果datas 给出问题,对吗?

标签: r


【解决方案1】:

DR0 列在最后一种情况下被删除,这会导致错误,因为 summarise 正在循环通过这些列 summarize(across(starts_with("DR0"), sum))。一种选择是创建条件检查,即如果没有 DR0 列,则将这些列添加为 NA 并且它应该可以正常工作

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))
  
  if(length(grep("DR0", names(SPV))) == 0) {
    SPV[mat1] <- NA_real_
  }
 
  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)])):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
 
 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 if(any(table(datas$Numbers) >= 3) & length(unique(datas$Numbers)) == 1){
   yz <- unique(datas$Numbers)
   lines(c(0,datas$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")}
 
 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")
 }
 
}

-测试

f1("2021-07-02", "ABC")

-输出

【讨论】:

  • Akrun,你插入:` if(length(grep("DR0", names(SPV))) == 0) { SPV[mat1]
  • @JVieira 基于df1,即m&lt;-df1 %&gt;% group_by(Category,Week) %&gt;% summarize(across(starts_with("DR1"), mean)),与数据无关
  • 那么当nrow小于等于2时满足条件即m&lt;-subset(m, Week == df1$Week[match(ymd(dmda), ymd(df1$date2))] &amp; Category == CategoryChosse)$DR1 if (nrow(datas)&lt;=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")}
  • @JVieira 如果长度大于 10 左右,您可能可以对 DR0 的数量进行子集化,即您可以选择 SPV[head(mat1, 10)] &lt;- NA_real_
  • Akun,我使用了SPV[head(mat1, 10)] &lt;- NA_real,这对我的情况有好处。现在我觉得有必要调整一下这个问题的DR0列没有值,也就是NA。如果你愿意,我可以问一个新问题
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