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r - 在循环中轻敲两个变量

问题描述

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", 
"", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", 
"", "", "")))

和 ：

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

我想逐行求和，所以我使用了这段代码：

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

su_test <- do.call(cbind, su_test)

并检查分位数：

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

这是每月一次的观察模拟。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)

我想得到相同的结果，su_test_obs 但除了这些区域的细节之外list(month, area)，我不明白其中的逻辑。请问你有解决办法吗？也许有一个更简单的解决方案dplyr？

谢谢

标签： rdplyr

解决方案

如果将矩阵转换为数据框会简单得多。然后我们可以使用aggregate它可以轻松地应用于多个组

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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