R t循环中两个变量的组合

我的数据：

TEST <- structure(c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 
2, 2, 2, 2, 2, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 
0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 3, 1, 0, 0, 
0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 1, 2, 0, 0, 1, 0, 0, 1, 1, 1, 
0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 
0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 2, 
0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0, 0, 
1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 
0, 0, 0, 3, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 
0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 
3, 0, 0, 2, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 
1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 
0, 0, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
3, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 0, 0, 0, 
0, 1, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 
0, 0, 3, 0, 0, 1, 0, 0, 2, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 
0, 0, 0, 1, 0, 0, 2, 0, 1, 2, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 
0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 3, 1, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 
1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 1, 1, 0, 0, 1, 
0, 0, 0), .Dim = c(22L, 20L), .Dimnames = list(NULL, c("month", 
"", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", 
"", "", "")))

以及：

month <- c(1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2)

我想按行求和，所以我使用了以下代码：

su_test <- list()
for (i in 1:ncol(TEST)){
    su_test[[i]] <- tapply(TEST[,i], month, sum)
}

su_test <- do.call(cbind, su_test)

检查分位数：

su_test_obs <- apply(su_test,1,quantile,c(0.1,0.9))

这是每月的观察模拟。但是，我也有按区域划分的详细信息：

TEST2 <- structure(c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 
2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 
1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 3, 1, 0, 
0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 1, 2, 0, 0, 1, 0, 0, 1, 1, 
1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 
0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 
2, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0, 
0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 
0, 0, 0, 0, 3, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 
0, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 
0, 3, 0, 0, 2, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 3, 
0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 
0, 0, 0, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 0, 0, 
0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 
0, 0, 0, 3, 0, 0, 1, 0, 0, 2, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 
0, 0, 0, 0, 1, 0, 0, 2, 0, 1, 2, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 
1, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 3, 1, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 
0, 1, 1), .Dim = c(22L, 20L), .Dimnames = list(NULL, c("month", 
"area", "", "", "", "", "", "", "", "", "", "", "", "", "", "", 
"", "", "", "")))

area <- c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)

我想得到同样的su_测试结果 但是除了像listmonth这样的区域细节之外，我不理解该区域的逻辑。 你能想出一个解决办法吗？也许有一个更简单的dplyr解决方案

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谢谢

如果您将矩阵转换为数据帧，会简单得多。然后，我们可以使用可以轻松应用于多个组的聚合

df <- data.frame(TEST2)
apply(aggregate(.~month + area, df, sum)[-c(1, 2)], 1, quantile, c(0.1,0.9))

#    [,1] [,2]
#10%    2  1.7
#90%    6  5.3

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谢谢，这工作做得很好。但我意识到我需要一个像su_test这样的对象来制作模拟ggplot类型模拟数据的直方图分布。frame x=su_test[1，]，aes x=x+geom_histogram bins=30，col='黑色'，fill='白色'。你认为有可能用我的两个变量重新建立su_测试吗？感谢you@thomasleonsu_测试是聚合的。~month+面积，df，sum，而su_测试obs是聚合的。~month+面积，df，sum[-c1,2]，1，分位数，c0.1,0.9