将向量的最小值排列并匹配到下一个向量中的对应值

Question

我有一个dataframe，它是根据每个id（从高到低）的最小值排列的。我要实现的是每个id的最小值与下一个id中对应值的位置重合，依此类推。创建了一个新向量 Time，即 1:nrow(df)。

我已经手动完成了这个，它看起来像这样：

我想知道如何使这个过程自动化？

这是我的数据示例：

MA_vol <- c(0.2486667, 0.2463333, 0.2426667, 0.2423333, 0.2376667, 0.2323333, 0.2270000, 0.2246667, 0.2216667, 0.2203333, 0.2183333, 0.2126667, 0.2076667, 0.2060000)
R_id <- rep(15, length(MA_vol))
df1 <- data.frame(R_id, MA_vol)

MA_vol <- c(0.2073333, 0.2053333, 0.2013333, 0.1993333, 0.1973333, 0.1970000, 0.1966667, 0.1946667, 0.1920000, 0.1890000, 0.1883333, 0.1866667, 0.1843333, 0.1823333, 0.1810000)
R_id <- rep(13, length(MA_vol))
df2 <- data.frame(R_id, MA_vol)

MA_vol <- c(0.2016667, 0.1996667, 0.1980000, 0.1970000, 0.1963333, 0.1956667, 0.1930000, 0.1913333, 0.1900000, 0.1893333, 0.1890000, 0.1863333, 0.1853333, 0.1820000, 0.1800000, 0.1780000, 0.1763333)
R_id <- rep(4, length(MA_vol))
df3 <- data.frame(R_id, MA_vol)

MA_vol <- c(0.2180000, 0.2146667, 0.2126667, 0.2103333, 0.2070000, 0.2040000, 0.2010000, 0.1993333, 0.1956667, 0.1950000, 0.1926667, 0.1920000, 0.1896667, 0.1890000, 0.1856667, 0.1830000, 0.1786667, 0.1763333, 0.1733333, 0.1720000, 0.1700000, 0.1686667, 0.1670000)
R_id <- rep(8, length(MA_vol))
df4 <- data.frame(R_id, MA_vol)

MA_vol <- c(0.2096667, 0.2063333, 0.2030000, 0.1993333, 0.1953333, 0.1916667, 0.1880000, 0.1870000, 0.1850000, 0.1830000, 0.1783333, 0.1753333, 0.1726667, 0.1716667, 0.1673333, 0.1666667, 0.1656667) 
R_id <- rep(2, length(MA_vol))
df5 <- data.frame(R_id, MA_vol)

df <- bind_rows(df1, df2, df3, df4, df5)

感谢您的帮助！

Answer 1

这可行，但可能有更好的方法。由于每个组的大小都不同，这使事情变得复杂。

library(tidyverse)

# first made a list of data frames
df1 <- df %>% arrange(MA_vol) %>% 
  group_by(R_id) %>% 
  group_nest() 

# then created a graph for each data frame in the list
df2 <- map(1:nrow(df1), 
           ~ggplot(df1$data[[.x]], 
                   aes(x = 1:nrow(df1$data[[.x]]),
                       y = MA_vol)) +
             geom_path() +
             # label the x-axis the group name
             xlab(paste0("Group ", 
                         as.character(df1$R_id[[.x]]))))

# then stacked the graphs for comparison
cowplot::plot_grid(df2[[1]], df2[[2]], df2[[3]],
                   df2[[4]], df2[[5]], ncol = 1)

Answer 2

# Order based on each min value (high to low)
R_minvalues <- df %>%
  group_by(R_id) %>%                # group by recession id
  slice(which.min(MA_vol)) %>%      # extract min volume values for each recession
  select(R_id, MA_vol)

x <- R_minvalues[with(R_minvalues, order(-MA_vol)), ]     # order by MA-vol min value (high to low)
R_id_order <- as.numeric(x$R_id)


# Reorder dataframe based on R_minvalues (high to low)
MRC_DF <- df %>%
  arrange(match(R_id, R_id_order)) %>%       # match R_id rows with R_id_order
  transform(t = 1:nrow(df)) %>%     # create t (time) column the length of the df
  select(t, R_id, MA_vol)                    # select columns

R_order_chr <- as.character(R_id_order)     # convert R_id_order to character so can rearrange columns

MRC_DF_wide <- dcast(setDT(MRC_DF), t ~ R_id, value.var = "MA_vol") %>%     # convert df to wide format 
  select(all_of(R_order_chr))       # rearrange column order

colnames(MRC_DF_wide)[1:ncol(MRC_DF_wide)] <-
  paste("R", colnames(MRC_DF_wide)[1:ncol(MRC_DF_wide)], sep = "")     # add "R_" to start of numbers so syntax is correct

matching.strip.fn <- function(.x, .y) {
  .y <- .y[!is.na(.y)]
  pos <- which.min(.x) - which.min(abs(min(.x, na.rm = T) - .y))
  c(rep(NA, pos), .y, rep(NA, length(.x) - pos - length(.y)))
}

MRC_DF_wide <- MRC_DF_wide %>% 
  accumulate(matching.strip.fn) %>%
  set_names(names(MRC_DF_wide)) %>%
  as.data.frame()

MRC_DF_wide <-
  MRC_DF_wide[rowSums(is.na(MRC_DF_wide)) != ncol(MRC_DF_wide),]

MRC_DF_wide <- MRC_DF_wide %>%
  transform(t = 1:nrow(MRC_DF_wide)) %>%
  select(t, everything())

将向量的最小值排列并匹配到下一个向量中的对应值

Arrange and match minimum value of vector to corresponding value in the next vector

sorting

r

match

reshape

dataframe