调用函数以生成值但不绘制图形答案

【问题标题】：Call function to generate the value but without plotting the graph调用函数以生成值但不绘制图形
【发布时间】：2021-12-28 23:58:30
【问题描述】：

我有一个可以生成地图和coef_val 值的函数，但我想知道是否可以调用相同的函数并仅获取生成的值，而不绘制图形？我知道存在的一种可能性是创建一个新函数，例如f2，没有绘图部分，但我不想这样做。还有其他方法吗？

library(dplyr)
library(tidyverse)
library(lubridate)


Test <- structure(
  list(date1= c("2021-06-28","2021-06-28"),
       date2 = c("2021-07-01","2021-07-01"),
       Category = c("FDE","ABC"),
       Week= c("Friday","Monday"),
       DR1 = c(14,11),
       DR01 = c(14,12), DR02= c(14,12),DR03= c(19,15),
       DR04 = c(15,14),DR05 = c(15,14),
       DR06 = c(12,14)),
  class = "data.frame", row.names = c(NA, -2L))


f1 <- function(df1, 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),.groups = 'drop')
  
  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"))
  
  datas<-SPV %>%
    filter(date2 == ymd(dmda)) %>%
    group_by(Category) %>%
    summarize(across(starts_with("DR0"), sum),.groups = 'drop') %>%
    pivot_longer(cols= -Category, names_pattern = "DR0(.+)", values_to = "val") %>%
    mutate(name = readr::parse_number(name))
  colnames(datas)[-1]<-c("Days","Numbers")
  
  if(as.Date(dmda) < min(as.Date(df1$date1))){
    datas <- datas %>% 
      group_by(Category) %>% 
      slice(1:max(Days)+1) %>%
      ungroup
  }else{
    datas <- datas %>% 
      group_by(Category) %>% 
      slice((as.Date(dmda) - min(as.Date(df1$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))
  
  model <- 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(model,newdata = new.data),lwd=2)
  coef_val<-coef(model)[2]
  points(0, coef_val, col="red",pch=19,cex = 2,xpd=TRUE)
  return(coef_val)
}

f1(Test, "2021-07-01", "ABC")
b2 
12.5

【问题讨论】：

是的 - f <- function(x, plot=TRUE) {if(plot) plot(x); return(x)}
我不明白在哪里插入这个。您不必选择一天中的部分时间和类别吗？例如：f1(Test, "2021-07-01", "ABC")
这只是一个示例，表明您可以返回一个值并选择是否需要图表。将所有 plot points 等位包装在 if(plot) {...} 中，然后在函数中添加 plot= 参数以选择是否发生。
具体来说，您需要将条件句括在plot、lines 和points 周围。
感谢您的回答！您能否以答案的形式留下，以便我更好地测试它？

标签： r plot

【解决方案1】：

您可以在下面提供一个参数，允许用户选择是否绘图，默认情况下FALSE。然后只需使用 if 语句包装对plot() 的调用，如果参数是TRUE，则执行该语句。

library(dplyr)
library(tidyverse)
library(lubridate)


Test <- structure(
  list(date1= c("2021-06-28","2021-06-28"),
       date2 = c("2021-07-01","2021-07-01"),
       Category = c("FDE","ABC"),
       Week= c("Friday","Monday"),
       DR1 = c(14,11),
       DR01 = c(14,12), DR02= c(14,12),DR03= c(19,15),
       DR04 = c(15,14),DR05 = c(15,14),
       DR06 = c(12,14)),
  class = "data.frame", row.names = c(NA, -2L))


f1 <- function(df1, dmda, CategoryChosse, plot = FALSE) {
  
  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),.groups = 'drop')
  
  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"))
  
  datas<-SPV %>%
    filter(date2 == ymd(dmda)) %>%
    group_by(Category) %>%
    summarize(across(starts_with("DR0"), sum),.groups = 'drop') %>%
    pivot_longer(cols= -Category, names_pattern = "DR0(.+)", values_to = "val") %>%
    mutate(name = readr::parse_number(name))
  colnames(datas)[-1]<-c("Days","Numbers")
  
  if(as.Date(dmda) < min(as.Date(df1$date1))){
    datas <- datas %>% 
      group_by(Category) %>% 
      slice(1:max(Days)+1) %>%
      ungroup
  }else{
    datas <- datas %>% 
      group_by(Category) %>% 
      slice((as.Date(dmda) - min(as.Date(df1$date1) [
        df1$Category == first(Category)])):max(Days)+1) %>%
      ungroup
  }
  
  model <- nls(Numbers ~ b1*Days^2+b2,start = list(b1 = 0,b2 = 0),data = datas, algorithm = "port")
  coef_val<-coef(model)[2]

  if(plot){
    new.data <- data.frame(Days = with(datas, seq(min(Days),max(Days),len = 45)))
    new.data <- rbind(0, new.data)
    
    plot(Numbers ~ Days,  xlim= c(0,45), ylim= c(0,30),
           xaxs='i',data = datas,main = paste0(dmda, "-", CategoryChosse))
    lines(new.data$Days,predict(model,newdata = new.data),lwd=2)
    points(0, coef_val, col="red",pch=19,cex = 2,xpd=TRUE)
  }

  return(coef_val)
}

在使用中演示：


f1(Test, "2021-07-01", "ABC")
f1(Test, "2021-07-01", "ABC", plot = TRUE)

【讨论】：

如果这回答了您的问题，请告诉我:) 如果是，请单击它旁边的复选标记以接受它。否则，如果需要，我很乐意澄清/编辑。
抱歉耽搁了，感谢@BHudson 的回复。当我执行f1(Test, "2021-07-01", "ABC") 时，会出现以下错误：Error in plot.xy(xy.coords(x, y), type = type, ...) : plot.new has not been called yet
@JVieira ;那是因为以下绘图命令lines(new.data$Days,predict(model,newdata = new.data),lwd=2) 和points(0, coef_val, col="red",pch=19,cex = 2,xpd=TRUE) 也需要包含在if(plot) 语句中。（所以需要在 nls 调用之后移动 if/plot 语句
@JVieira;我更新了 BHudson 的答案，所以你不应该得到你提到的错误 here 。（希望没关系，BHudson——我只是稍微移动了一下以避免情节错误）
现在，它可以工作了！感谢@user20650 和@BHudson！ =)

【解决方案2】：

您可以将绘图重定向到NULL 文件。来自help(pdf)的相关部分：

file：给出文件路径的字符串。 [...]如果它是'NULL'，则不会创建外部文件（实际上，不会发生绘图）[...]

所以，为了不产生任何情节：

pdf(NULL)
f1(Test, "2021-07-01", "ABC")
dev.off()

【讨论】：