找到一个组内

时间:2015-12-03 19:03:03

标签: r data-manipulation

我希望通过找到组内两个不同变量之间的共同出现来有效地计算共现矩阵,理想情况下不使用迭代所有可能组合的复杂循环。

鉴于我的数据框如下所示:

df = data.frame(group = c(1,1,1,2,2,2),var1 = c(1,2,4,2,2,4),var2 = c(4,1,2,1,3,2))

> df
  group var1 var2
1     1    1    4
2     1    2    1
3     1    4    2
4     2    2    1
5     2    2    3
6     2    4    2

我希望将其变成一个新的共生矩阵,其中行代表var1和列var2。

编辑:对于那些不熟悉共同出现的人,我对组中同时出现的值对感兴趣。例如," 2"的组合和" 1"在组1中发生一次,在组2中发生其他时间,因此暗示2次共同发生。在我的例子中,我将组合接下来两个,但它们可以出现在组内的任何地方。

它应如下所示:

> cooc
  1 2 3 4
1 0 2 0 1
2 2 0 1 2
3 0 1 0 0
4 1 2 0 0

我之前在使用xtabs函数处理组中的一个变量来处理共同出现时已经这样做了,但是不确定如何将它应用于多个列。例如,如果我有兴趣在不同的组中找到var1的共同出现,我会做以下事情:

> td = xtabs(~group + var1,data = df)
> cooc = crossprod(td,td)
> diag(cooc) = 0

1 个答案:

答案 0 :(得分:1)

如果我正确理解你的问题,我相信这应该有效:

# i only use data.table here in case we need to do this "by group"
# but in this solution I do not use it as i did not see the significance
# of grouping
###library(data.table)
###df <-  data.table(df)

# this creates the pair of values "a_b"
df$ID <- paste(df$var1,df$var2,sep="_")
# we enumerate all the unique values that way we can create 
# a map to later match the data and map
uniqval <- sort(unique(c(df$var1,df$var2)))
grid <- expand.grid(uniqval,uniqval)
grid$ID <- paste(grid$Var1,grid$Var2,sep="_")
# match our data to this map
matches <- sort(match(df$ID,grid$ID))
# tabulate our results into a dataframe
tab <- data.frame(table(grid$ID[matches]))
# split up our ID back into values
tab$Var2 <- substr(tab$Var1,3,3)
tab$Var1 <- substr(tab$Var1,1,1)
# create our empty result matrix
cooc <- matrix(0,nrow=length(uniqval),ncol=length(uniqval))
rownames(cooc) <- uniqval
colnames(cooc) <- uniqval

# there are other ways to do this
# but this seemed simple enough of a loop for me
# we just need to replace the tabulation results
# into our desired location in the matrix
# namely, "a_b" frequencies into [a,b] and [b,a] positions
for(m in 1:nrow(tab)){

  i <- tab$Var1[m]
  j <- tab$Var2[m]

# by adding this to the previous value
# we are accounting for "a_b" equiv. to "b_a"
  cooc[i,j] <- cooc[i,j]+tab$Freq[m]
  cooc[j,i] <- cooc[i,j]

}