为所有日期/类别制作系数

Question

你能帮我解决以下问题吗：请注意，此代码根据我选择的日期和类别生成系数，在本例中为 30/06 和 FDE 类别，系数为 4。但是，我想做一个表格使用我拥有的所有日期/类别的系数，这样我就不需要一个一个地继续做。

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

df1 <- structure(
  list(date1= c("2021-06-28","2021-06-28","2021-06-28","2021-06-28"),
       date2 = c("2021-06-30","2021-06-30","2021-07-01","2021-07-01"),
       Category = c("FDE","ABC","FDE","ABC"),
       Week= c("Wednesday","Wednesday","Friday","Friday"),
       DR1 = c(4,1,6,3),
       DR01 = c(4,1,4,3), DR02= c(4,2,6,2),DR03= c(9,5,4,7),
       DR04 = c(5,4,3,2),DR05 = c(5,4,5,4),
       DR06 = c(2,4,3,2),DR07 = c(2,5,4,4),
       DR08 = c(3,4,5,4),DR09 = c(2,3,4,4)),
  class = "data.frame", row.names = c(NA, -4L))

dmda<-"2021-06-30"
CategoryChosse<- "FDE"

  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

      mod <- nls(Numbers ~ b1*Days^2+b2,start = list(b1 = 0,b2 = 0),data = datas, algorithm = "port")
> coef(mod)[2]
b2 
 4

表格的 output 将如下所示：

Answer 1

将代码放入 function 并使用mapply应用它 -

cbind(df1[2:3], coef = mapply(return_coef, df1$date2, df1$Category))

#       date2 Category coef
#1 2021-06-30      FDE    4
#2 2021-06-30      ABC    1
#3 2021-07-01      FDE    6
#4 2021-07-01      ABC    3

其中return_coef是 -

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)])):max(Days)+1) %>%
  ungroup

mod <- nls(Numbers ~ b1*Days^2+b2,start = list(b1 = 0,b2 = 0),data = datas, algorithm = "port")
as.numeric(coef(mod)[2])

}