如何在R表中创建缺失值？

Question

我有40对鸟儿，每对男性和女性都为他们的颜色评分。颜色分数是一个分类变量，取值范围为1到9.我想创建一个表格，其中包含每个组合的编号（1/1，1/2，1/3，... 9/7，9/8,9/9）。我的问题是，当我尝试创建表格时，有一些组合在我的数据中不存在（在这些情况下，我会为缺失值设置零）。以下是数据和示例代码。我确信答案在于使用'expand.grid（）'命令，例如请参阅post，但我不确定如何实施它。有什么建议么？

## 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$Male <- as.factor(Pair_Colours$Male) 
Pair_Colours$Female <- as.factor(Pair_Colours$Female) 

## table of pair colour values (colours 1 to 9 - categoricial variable) 
table(Pair_Colours$Male, Pair_Colours$Female) 

## my attempt to create a table with a count of each possible value for pairs 
Colour_Male <- rep(seq(1, 9, by = 1), each = 9) 
Colour_Female <- rep(seq(1, 9, by = 1), times = 9) 
Colour_Count <- as.vector(table(Pair_Colours$Male, Pair_Colours$Female)) # <- the problem occurs here 
Pairs_Colour_Table <- as.data.frame(cbind(cbind(Colour_Male, Colour_Female), Colour_Count)) 

## 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 = Pairs_Colour_Table$Colour_Count, bg = "darkgray", col = "black", lwd = 1) 

Answer 1

你只需要调用table之前转换您Pair_Colours为factor所有要求的水平：

# Convert each column to factor with levels 1 to 9 
Pair_Colours[] <- lapply(Pair_Colours, factor, levels=1:9) 
table(Pair_Colours$Male, Pair_Colours$Female) 
#  1 2 3 4 5 6 7 8 9 
# 1 0 0 0 0 0 0 0 0 0 
# 2 0 0 0 0 0 0 0 0 0 
# 3 0 0 0 0 1 1 0 0 0 
# 4 0 1 0 3 0 0 0 3 1 
# 5 0 0 0 2 0 2 0 1 1 
# 6 0 0 1 1 1 0 3 3 2 
# 7 0 0 1 0 1 0 0 3 1 
# 8 0 0 1 0 0 1 1 0 1 
# 9 0 0 0 0 1 1 0 1 0 

您可以用as.data.frame转换，如果你想要的格式是“combn1，combn2，频率”。