请参阅以下数据和代码:
## Dataset pairs of males and females and their colour classes
Pair_Colours <- structure(list(Male = c(7, 6, 4, 6, 8, 8, 5, 6, 6, 8, 6, 6, 5,
7, 9, 5, 8, 7, 5, 5, 4, 6, 7, 7, 3, 6, 5, 4, 7, 4, 3, 9, 4, 4,
4, 4, 9, 6, 6, 6), Female = c(9, 8, 8, 9, 3, 6, 8, 5, 8, 9, 7,
3, 6, 5, 8, 9, 7, 3, 6, 4, 4, 4, 8, 8, 6, 7, 4, 2, 8, 9, 5, 6,
8, 8, 4, 4, 5, 9, 7, 8)), .Names = c("Male", "Female"), class = "data.frame", row.names = c(NA,
40L))
Pair_Colours[] <- as.data.frame(lapply(Pair_Colours, factor, levels=1:9))
## table of pair colour values (colours 1 to 9 - categoricial variable)
table(Pair_Colours$Male, Pair_Colours$Female)
Colour_Count <- as.vector(table(Pair_Colours$Male, Pair_Colours$Female)) #<- the problem occurs here
## plot results to visisually look for possible assortative mating by colour
op<-par(mfrow=c(1,1), oma=c(2,4,0,0), mar=c(4,5,1,2), pty = "s")
plot(1,1, xlim = c(1, 9), ylim = c(1, 9), type="n", xaxt = "n", yaxt = "n", las=1, bty="n", cex.lab = 1.75, cex.axis = 1.5, main = NULL, xlab = "Male Colour", ylab = "Female Colour", pty = "s")
axis(1, at = seq(1, 9, by = 1), labels = T, cex.lab = 1.5, cex.axis = 1.5, tick = TRUE, tck = -0.015, lwd = 1.25, lwd.ticks = 1.25)
axis(2, at = seq(1, 9, by = 1), labels = T, cex.lab = 1.5, cex.axis = 1.5, tick = TRUE, tck = -0.015, lwd = 1.25, lwd.ticks = 1.25, las =2)
points(Pair_Colours$Male, Pair_Colours$Female, pch = 21, cex = Colour_Count, bg = "darkgray", col = "black", lwd = 1)
答案 0 :(得分:4)
您可以使用库plyr的函数ddply()
汇总数据,然后使用此新数据框来绘制数据。计数位于新数据框的V1
列中。
library(plyr)
df<-ddply(Pair_Colours,.(Male,Female),nrow)
df
Male Female V1
1 3 5 1
2 3 6 1
3 4 2 1
4 4 4 3
points(df$Male, df$Female, pch = 21, cex = df$V1,
bg = "darkgray", col = "black", lwd = 1)
其他可能性是使用函数aggregate()
。首先,添加仅包含值1的新列N
。然后为每个aggregate()
和N
组合添加Male
总和Female
值。
Pair_Colours$N<-1
aggregate(N~Male+Female,data=Pair_Colours,FUN=sum)
Male Female N
1 4 2 1
2 6 3 1
3 7 3 1
4 8 3 1
5 4 4 3