如何在R的一列中的值序列中找到最大值和最小值？

Question

這個問題可能微不足道，但我發現很難解決。 請指導我。

數據

以下是示例數據：

structure(list(Vehicle.ID2 = c("39-25", "39-25", "39-25", "39-25", 
"39-25", "39-25", "39-25", "39-25", "39-25", "39-25", "39-25", 
"39-25", "39-25", "39-25", "39-25", "39-25", "39-25", "39-25", 
"39-25", "39-25", "39-25", "39-25", "39-25", "39-25", "39-25", 
"39-25", "39-25", "39-25", "39-25", "39-25", "39-25", "39-25", 
"39-25", "39-25", "39-25", "39-25", "39-25", "39-25", "39-25"
), OC_DV = c(".", ".", ".", ".", ".", "CLDV", ".", ".", ".", 
".", ".", ".", ".", ".", ".", "OPDV", ".", ".", ".", ".", ".", 
".", ".", ".", ".", ".", ".", ".", ".", ".", ".", ".", ".", ".", 
".", "CLDV", ".", ".", "."), frspacing = c(35.83373, 35.75742, 
35.70391, 35.67694, 35.67792, 35.70669, 35.7619, 35.84096, 35.93962, 
36.05109, 36.16704, 36.28056, 36.3861, 36.47762, 36.5485, 36.59359, 
36.61402, 36.61791, 36.61383, 36.60651, 36.59694, 36.58372, 36.56525, 
36.54044, 36.50771, 36.46458, 36.40831, 36.33713, 36.25086, 36.15089, 
36.04004, 35.92236, 35.80322, 35.68935, 35.58883, 35.51032, 35.4618, 
35.4492, 35.47479)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
-39L), .Names = c("Vehicle.ID2", "OC_DV", "frspacing"))  

我想做的事

我想找到一組值的最大值和最小值在frspacing標簽之間CLDV和OPDV列OC_DV 。 然后我想找到它們的區別。

期望的輸出

以下是最大和最小：

  Group      Max    Min
1 CLDV-OPDV 36.54   35.70
2 OPDV-CLDV 36.62   35.59  

以下是絕對差異（第一組的最大值-第二組的最小值，反之亦然）：

1 0.95
2 0.92

我沒有任何代碼可以顯示我的嘗試，因為老實說我不知道​​如何解決這個問題。 顯然，簡單的按列的max或min將不起作用。 我正在使用dplyr ，沒有找到任何相關內容。

Answer 1

 library(zoo) # for na.locf
 library(dplyr)

 df[df=="."] = NA
 df$group = paste((na.locf(df$OC_DV, na.rm = FALSE)), lead(na.locf(df$OC_DV, na.rm = FALSE, fromLast = TRUE)), sep = "-")

 df %>% group_by(group) %>% 
   summarise(Max = max(frspacing), Min = min(frspacing)) %>% 
   filter(!grepl("NA",group ))

Source: local data frame [2 x 3]

      group      Max      Min
      (chr)    (dbl)    (dbl)
1 CLDV-OPDV 36.54850 35.70669
2 OPDV-CLDV 36.61791 35.58883

使用多個值時，我將計算更改並將其用作另一個分組變量：（在此示例中，我復制了數據）

df$group2 = NA
df$group2[which(df$group != lag(df$group))] = 1:length(which(df$group != lag(df$group)))
df$group2 = na.locf(df$group2, na.rm = FALSE)

df %>% group_by(group, group2) %>% 
  summarise(Max = max(frspacing), Min = min(frspacing)) %>% 
   filter(!grepl("NA",group ))

Source: local data frame [5 x 4]
Groups: group [3]

      group group2      Max      Min
      (chr)  (int)    (dbl)    (dbl)
1 CLDV-CLDV      3 38.09082 34.30454
2 CLDV-OPDV      1 36.54850 35.70669
3 CLDV-OPDV      4 38.90356 34.08951
4 OPDV-CLDV      2 36.61791 35.58883
5 OPDV-CLDV      5 38.18983 34.27874

但是，如果每個Vehicle.ID2中OC_DV的組合都不同，則只需將ID粘貼到組中即可...

Answer 2

d <- your_dput
# Build your subsetted dataframes
e <- d[grep("CLDV", d$OC_DV)[1]: grep("OPDV", d$OC_DV),]
f <- d[(grep("OPDV", d$OC_DV): grep("CLDV", d$OC_DV)[2]),]
# Make the diff() calls
diff(c(max(e$frspacing), min(f$frspacing)))
diff(c(max(f$frspacing), min(e$frspacing)))

我的值與您的值不一樣，您可以根據要處理邊界包含/排除的方式手動調整grep值。

Answer 3

以下是基本的R解決方案：

MaxMinSeq <- function(df) {
    myInd <- which(df$OC_DV != ".")
    myVals <- df$frspacing
    myTitles <- df$OC_DV[myInd]
    myLen <- length(myInd)-1L
    NewDf <- as.data.frame(t(sapply(1:myLen, function(x) {
               list(Group = paste(c(myTitles[x],"-",myTitles[x+1L]), collapse = ""),
                   Max = max(myVals[myInd[x]:(myInd[x+1L]-1L)]),
                   Min = min(myVals[myInd[x]:(myInd[x+1L]-1L)]))})))
    for (i in 1:3) {NewDf[,i] <- unlist(NewDf[,i])}
    NewDf
}

df2 <- MaxMinSeq(df)
df2
      Group      Max      Min
1 CLDV-OPDV 36.54850 35.70669
2 OPDV-CLDV 36.61791 35.58883

這比上面發布的dplyr解決方案快了dplyr 。 注意：

TestDplyr <- function(df) {
    df[df=="."] <- NA
    df$group <- paste((na.locf(df$OC_DV, na.rm = FALSE)), lead(na.locf(df$OC_DV, na.rm = FALSE, fromLast = TRUE)), sep = "-")

    df$group2 <- NA
    df$group2[which(df$group != lag(df$group))] <- 1:length(which(df$group != lag(df$group)))
    df$group2 <- na.locf(df$group2, na.rm = FALSE)

    df %>% group_by(group, group2) %>% 
        summarise(Max = max(frspacing), Min = min(frspacing)) %>% 
        filter(!grepl("NA",group ))
}

microbenchmark(Joseph = MaxMinSeq(df), Cabana = TestDplyr(df))
Unit: microseconds
expr      min        lq      mean    median       uq      max neval
Joseph  338.671  377.6695  405.0257  405.9945  429.188  496.718   100
Cabana 2622.336 2698.2810 2890.5430 2765.6045 2977.427 7772.180   100

這是一個非常重要的例子：

myDfs <- lapply(1:10000, function(x) df)
bigDf <- do.call(rbind, myDfs)
bigDf$frspacing[40:nrow(bigDf)] <- runif((nrow(bigDf)-39), 10, 100)

a <- MaxMinSeq(bigDf)
b <- TestDplyr(bigDf)
b <- b[order(b$group2),]

identical(a$Max, b$Max)
[1] TRUE
identical(a$Min, b$Min)
[1] TRUE

system.time(TestDplyr(bigDf))
 user  system elapsed 
 1.54    0.00    1.54 
system.time(MaxMinSeq(bigDf))
 user  system elapsed 
  0.3     0.0     0.3

至於問題的第二部分，我不確定OP希望答案有多一般，尤其是當有兩個以上不同的最終配對時。 例如，OP是否要查找一行的最大值並將其與所有行的最小值進行比較，還是只是比較鄰居？ 下面的函數采用第一種方法（即通用方法）。

GetDiff <- function(df) {
    df2 <- cbind(df, t(sapply(1:nrow(df), function(x) {
                        c(rowMin = min(df[x,2:3]),
                          rowMax = max(df[x,2:3]))})))
    myRows <- 1:nrow(df)
    sapply(myRows, function(x) df2$rowMax[x] - min(df2$rowMin[-x]))
}

GetDiff(df2)   ## df2 comes from above
[1] 0.95967 0.91122