r - 使用 R 创建起点-终点矩阵

Question

我的数据框由个人和他们在某个时间点居住的城市组成。我想为每年生成一个起点-终点矩阵，记录从一个城市到另一个城市的移动次数。我想知道:

如何自动生成数据集中每一年的起点-终点表？

如何以相同的 5x5 格式生成所有表格，5 是我示例中的城市数量？

有没有比我在下面提出的更有效的代码？我打算在一个非常大的数据集上运行它。

考虑以下示例:

#An example dataframe
id=sample(1:5,50,T)
year=sample(2005:2010,50,T)
city=sample(paste(rep("City",5),1:5,sep=""),50,T)
df=as.data.frame(cbind(id,year,city),stringsAsFactors=F)
df$year=as.numeric(df$year)
df=df[order(df$id,df$year),]
rm(id,year,city)

我最好的尝试

#Creating variables
for(i in 1:length(df$id)){
  df$origin[i]=df$city[i]
  df$destination[i]=df$city[i+1]
  df$move[i]=ifelse(df$orig[i]!=df$dest[i] & df$id[i]==df$id[i+1],1,0) #Checking whether a move has taken place and whether its the same person
  df$year_move[i]=ceiling((df$year[i]+df$year[i+1])/2) #I consider that the person has moved exactly between the two dates at which its location was recorded
}
df=df[df$move!=0,c("origin","destination","year_move")]    

创建 2007 年的起点-终点表

yr07=df[df$year_move==2007,]
table(yr07$origin,yr07$destination)

结果

        City1 City2 City3 City5
  City1     0     0     1     2
  City2     2     0     0     0
  City5     1     1     0     0

Answer 1

您可以按 id 拆分数据，对特定于 id 的数据框执行必要的计算以获取该人的所有 Action ，然后重新组合:

spl <- split(df, df$id)
move.spl <- lapply(spl, function(x) {
  ret <- data.frame(from=head(x$city, -1), to=tail(x$city, -1),
                    year=ceiling((head(x$year, -1)+tail(x$year, -1))/2),
                    stringsAsFactors=FALSE)
  ret[ret$from != ret$to,]
})
(moves <- do.call(rbind, move.spl))
#       from    to year
# 1.1  City4 City2 2007
# 1.2  City2 City1 2008
# 1.3  City1 City5 2009
# 1.4  City5 City4 2009
# 1.5  City4 City2 2009
# ...

因为此代码对每个 id 使用矢量化计算，所以它应该比在提供的代码中循环遍历数据帧的每一行要快得多。

现在您可以使用 split 获取特定年份的 5x5 移动矩阵和 table :

moves$from <- factor(moves$from)
moves$to <- factor(moves$to)
lapply(split(moves, moves$year), function(x) table(x$from, x$to))
# $`2005`
#        
#         City1 City2 City3 City4 City5
#   City1     0     0     0     0     1
#   City2     0     0     0     0     0
#   City3     0     0     0     0     0
#   City4     0     0     0     0     0
#   City5     0     0     1     0     0
# 
# $`2006`
#        
#         City1 City2 City3 City4 City5
#   City1     0     0     0     1     0
#   City2     0     0     0     0     0
#   City3     1     0     0     1     0
#   City4     0     0     0     0     0
#   City5     2     0     0     0     0
# ...