如何按组计算不同ID随时间的距离？

Question

My (example) data is structured as follows... where the X and Y coordinates of participants, recorded under varying conditions, are collected over time:

    Individ <- data.frame(Participant = c("Bill", "Bill", "Bill", "Bill", "Bill", "Harry", "Harry", "Harry", "Harry","Harry", "Paul", "Paul", "Paul", "Paul", "Paul"),
                          Time = c(0.01, 0.02, 0.03, 0.04, 0.05, 0.01, 0.02, 0.03, 0.04, 0.05, 0.01, 0.02, 0.03, 0.04, 0.05),
                          Condition = c("Expr", "Expr", "Expr", "Expr", "Expr", "Con", "Con", "Con", "Con", "Con", "Nor", "Nor", "Nor", "Nor", "Nor"),
                          X = c(26.07, 26.06, 26.05, 26.09, 26.04, 26.65, 26.64, 26.62, 26.63, 26.62, 27.99, 28.01, 28.01, 28.02, 28.02),
                          Y = c(-5.01, -5.12, -5.14, -5.18, -5.2065, -12.37, 12.36, -12.35, -12.34, 12.33, -5.52, -5.514, -5.51, -5.50, -5.4962))

The X and Y coordinates are captured from the same location. I can calculate the distance covered by each Participant using the following:

require(plyr)
require(dplyr)
DistanceOutput <- Individ %>%
     arrange(Participant, Time, Condition) %>%
     group_by(Participant, Condition) %>%
     mutate( lagX = lag(X, order_by=Time), lagY = lag(Y, order_by=Time)) %>%
     rowwise() %>%
     mutate(Distance = dist( matrix( c(X,Y,lagX,lagY),nrow=2,byrow=TRUE) )) %>%
     select(-lagX, -lagY)

However, how can I calculate the distance between each Participant over Time , according to their Condition . For example, the distance between Bill and Harry, Bill and Paul plus Harry and Paul over Time?

My dataset is 179,800 obs. so ideally, a quick solution is preferred. Thank you!

Answer 1

Here's a way to calculate the distance between each participant at each time point. I doubt it's the most efficient way, but maybe someone else will come along with a more elegant solution.

You said that you'd like to calculate the distance between participants for each Condition . In your sample data, there's only one participant in each condition. However, the solution below can easily be extended to be applied by Condition in addition to Time .

library(reshape2)
library(dplyr)

# Calculate distance matrix for each Time
res = lapply(unique(Individ$Time), function(i) {

  mat = as.matrix(Individ[Individ$Time==i, c("X","Y")])
  rownames(mat) = Individ$Participant[Individ$Time==i]

  # Distance matrix
  d = as.matrix(dist(mat))

  # Keep only lower triangle
  d[upper.tri(d, diag=TRUE)] = NA

  # Return data frame with distances, time and participants
  data.frame(Time=i, d) %>% add_rownames("P1")
})

# Combine all time points into single long data frame of distances
res = bind_rows(res) %>% 
  melt(id.var=c("Time","P1"), variable.name="P2", value.name="Distance") %>%
  filter(!is.na(Distance)) %>% 
  rowwise %>%
  mutate(Pair = paste(sort(c(as.character(P1), as.character(P2))), collapse="-")) %>% 
  select(Pair, Time, Distance) %>%
  arrange(Pair, Time)

res

 Pair Time Distance 1 Bill-Harry 0.01 7.382818 2 Bill-Harry 0.02 17.489620 3 Bill-Harry 0.03 7.232496 4 Bill-Harry 0.04 7.180334 5 Bill-Harry 0.05 17.546089 6 Bill-Paul 0.01 1.986580 7 Bill-Paul 0.02 1.989406 8 Bill-Paul 0.03 1.994618 9 Bill-Paul 0.04 1.956349 10 Bill-Paul 0.05 2.001081 11 Harry-Paul 0.01 6.979835 12 Harry-Paul 0.02 17.926427 13 Harry-Paul 0.03 6.979807 14 Harry-Paul 0.04 6.979807 15 Harry-Paul 0.05 17.881091