R 基于逐月随机抽样从矩阵中提取最大值

我有一个距离矩阵df1，显示了8个位置a:h之间的距离

x <- c("a","b","c","d","e","f","g","h")
df1 <- data.frame(a=c(0,1,2,3,4,5,6,7), b=c(1,0,1,2,3,4,5,6),
                  c=c(2,1,0,1,2,3,4,5), d=c(3,2,1,0,1,2,3,4),
                  e=c(4,3,2,1,0,1,2,3), f=c(5,4,3,2,1,0,1,2),
                  g=c(6,5,4,3,2,1,0,1), h=c(7,6,5,4,3,2,1,0),
                  row.names=x)

> df1
  a b c d e f g h
a 0 1 2 3 4 5 6 7
b 1 0 1 2 3 4 5 6
c 2 1 0 1 2 3 4 5
d 3 2 1 0 1 2 3 4
e 4 3 2 1 0 1 2 3
f 5 4 3 2 1 0 1 2
g 6 5 4 3 2 1 0 1
h 7 6 5 4 3 2 1 0

我还想计算每个月地点之间的累积距离，其输出如下：

  Month Cum.Distance
1    11            5
2    12           11
3     1            6

---

我希望这是有道理的。我曾考虑过使用for循环，但我对R循环的了解有限，因此非常感谢您的帮助。非常感谢

首先，我根据您的示例定义数据帧

df2 <- read.table(text = "
Month Location
1     11        c
2     11        a
3     11        d
4     12        f
5     12        c
6     12        f
7     12        a
8      1        b
9      1        b
10     1        h", h = T)

aggregate(Location ~ Month, df2, function(j) diff(range(sapply(j, function(i) grep(i, letters)))))

  Month Location
1     1        6
2    11        3
3    12        5

x <- c("a","b","c","d","e","f","g","h")
df1 <- data.frame(a=c(0,1,2,3,4,5,6,7), b=c(1,0,1,2,3,4,5,6),
                  c=c(2,1,0,1,2,3,4,5), d=c(3,2,1,0,1,2,3,4),
                  e=c(4,3,2,1,0,1,2,3), f=c(5,4,3,2,1,0,1,2),
                  g=c(6,5,4,3,2,1,0,1), h=c(7,6,5,4,3,2,1,0),
                  row.names=x)

df2 <- data.frame(Month=c(rep(11,3),rep(12,4),rep(1,3)),
                  Location=sample(letters[1:8],10,replace=T))

#    Month Location
# 1     11        d
# 2     11        c
# 3     11        h
# 4     12        e
# 5     12        c
# 6     12        b
# 7     12        h
# 8      1        h
# 9      1        g
# 10     1        b

最后，我将该函数应用于

df2

中的所有月份，并将其作为数据帧重新打包

# Run function on all months
data.frame(month = unique(df2$Month), max_dist = unlist(lapply(unique(df2$Month), max_dist)))

#   month max_dist
# 1    11        5
# 2    12        6
# 3     1        6

---

以下内容提供了总距离：

tot_dist <- function(m){
  tmp <- match(df2$Location[df2$Month == m], rownames(df1))  
  sum(df1[cbind(head(tmp, -1), tail(tmp, -1))])
}

---

谢谢你，Len，你知道如何编辑它来计算每个月每个地点之间的累积距离吗？你到底想要什么？您希望应用的月份顺序是什么？我希望按照每个位置在数据框中出现的顺序对它们之间的距离求和。在你的例子中，对于第11个月，累积距离等于从c到a的距离加上从a到d的距离。月份的顺序并不重要，但如果它们保持原来的顺序会很有用。我现在明白了，我已经把它添加到了我的答案中。这与您的第一个目标的代码几乎相同。Hi Ronak df1沿x=y线对称，因此无论是按行还是按列进行计算都不重要。谢谢你，林巴克这真的很有帮助！我的下一个问题是，是否可以计算每个月的累计距离。例如，对于您的示例中所示的df2，第11个月的累积距离将通过将从d到c的距离与从c到h的距离相加来计算，所以（d:c）+（c:h）=6。@EmilyWinter我在我的解决方案中添加了一个函数，可以做到这一点。这很好，谢谢。然而，有一个问题是，当您在df2的新版本上运行max_dist函数时，每个月只包含一个位置记录，它会在这些月内给出错误的输出。使用df2@EmilyWinter尝试一下，我想我已经修复了它。您好，这段代码一直运行得很好，但我想修改它。是否可以计算每个月地点的最大距离和总距离，以及下个月的第一个地点？因此，在您的示例中，将为位置d、c、h和e计算第11个月的最大距离/总距离。提前感谢你的帮助。

x <- c("a","b","c","d","e","f","g","h")
df1 <- data.frame(a=c(0,1,2,3,4,5,6,7), b=c(1,0,1,2,3,4,5,6),
                  c=c(2,1,0,1,2,3,4,5), d=c(3,2,1,0,1,2,3,4),
                  e=c(4,3,2,1,0,1,2,3), f=c(5,4,3,2,1,0,1,2),
                  g=c(6,5,4,3,2,1,0,1), h=c(7,6,5,4,3,2,1,0),
                  row.names=x)

df2 <- data.frame(Month=c(rep(11,3),rep(12,4),rep(1,3)),
                  Location=sample(letters[1:8],10,replace=T))

#    Month Location
# 1     11        d
# 2     11        c
# 3     11        h
# 4     12        e
# 5     12        c
# 6     12        b
# 7     12        h
# 8      1        h
# 9      1        g
# 10     1        b

# Find maximum distance
max_dist <- function(m){
  # Check if it's just one location
  if(sum(df2$Month == m) == 1)return(0)

  # Get all combinations of locations for given month
  tmp <- t(combn(match(df2$Location[df2$Month == m], rownames(df1)), 2))

  # Get max value from these location combinations
  max(df1[tmp[, 1], tmp[, 2]])
}

# Run function on all months
data.frame(month = unique(df2$Month), max_dist = unlist(lapply(unique(df2$Month), max_dist)))

#   month max_dist
# 1    11        5
# 2    12        6
# 3     1        6

tot_dist <- function(m){
  tmp <- match(df2$Location[df2$Month == m], rownames(df1))  
  sum(df1[cbind(head(tmp, -1), tail(tmp, -1))])
}

# Find maximum distance
max_dist <- function(m){
  # Check if it's just one location
  if(sum(df2$Month == m) == 1)return(0)

  # Get all locations
  locs <- which(df2$Month == m)
  if(tail(which(df2$Month == m), 1) != nrow(df2))locs <- c(locs, tail(which(df2$Month == m), 1) + 1)

  # Get all combinations of locations for given month
  tmp <- t(combn(match(df2$Location[locs], rownames(df1)), 2))

  # Get max value from these location combinations
  max(df1[tmp[, 1], tmp[, 2]])
}

tot_dist <- function(m){
  # Get all locations
  locs <- which(df2$Month == m)
  if(tail(which(df2$Month == m), 1) != nrow(df2))locs <- c(locs, tail(which(df2$Month == m), 1) + 1)

  tmp <- match(df2$Location[locs], rownames(df1))  
  sum(df1[cbind(head(tmp, -1), tail(tmp, -1))])
}