从绝对数字到两级数据的比例（R！SAC？plyr？）

Question

我将数据嵌套到级别：

L1 L2   x1 x2 x3 x4
A  This 20 14 12 15
A  That 11 NA 8  16
A  Bat  Na 22 13  9
B  This 10  9 11  6
B  That 3   3  1 NA
B  Bat  4  10  2  8

现在我想要一些简单的东西 - 我觉得我上个月才能做到这一点。但是我头脑中已经遗漏了一些东西：我想要百分比（忽略NA），对于L1中的每个变量求和为100

L1 L2   x1  x2   x3   x4
A  This 65% 39%  36%  38%
A  That 35%  0%  24%  40%
A  Bat   0% 61%  40%  22%

我可以用

获得我需要的总数

cast(L1~variable, data=melt(d, na.rm=T),sum)

但是我觉得应该可以做一个能给我想要的功能吗？ 我尝试了各种方法与演员和普莱尔...但它的接缝圣诞已经给我脆弱的大脑带来了许多啤酒。

任何帮助都将受到赞赏 - 任何帮助都不会有任何帮助。

感谢名单

这是我的数据：

d <- structure(list(level1 = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 
2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 
6L, 6L, 7L, 7L, 7L, 7L, 8L, 8L, 8L, 8L, 8L, 10L, 10L, 10L, 10L, 
11L, 11L, 11L, 11L, 11L, 11L, 9L, 9L, 9L, 9L, 9L, 12L, 12L, 12L, 
12L, 13L, 13L, 13L, 13L, 13L, 13L, 14L, 14L, 14L, 14L, 14L, 14L, 
15L, 15L, 15L, 15L, 15L, 16L, 16L, 16L, 16L, 16L, 16L, 17L, 17L, 
17L, 17L, 17L, 18L, 18L, 18L, 18L, 18L, 18L, 19L, 19L, 19L, 19L, 
19L, 19L), .Label = c("a", "b", "c", "d", "e", "f", "g", "h", 
"i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s"), class = "factor"), 
level2 = structure(c(6L, 2L, 1L, 3L, 5L, 6L, 1L, 3L, 5L, 
6L, 1L, 3L, 5L, 6L, 5L, 6L, 1L, 3L, 5L, 4L, 6L, 2L, 1L, 3L, 
5L, 6L, 1L, 5L, 4L, 6L, 2L, 1L, 3L, 5L, 6L, 1L, 3L, 5L, 6L, 
2L, 1L, 3L, 5L, 4L, 6L, 2L, 1L, 3L, 5L, 6L, 1L, 3L, 5L, 6L, 
2L, 1L, 3L, 5L, 4L, 6L, 2L, 1L, 3L, 5L, 4L, 6L, 2L, 1L, 3L, 
5L, 6L, 2L, 1L, 3L, 5L, 4L, 6L, 2L, 1L, 3L, 5L, 6L, 2L, 1L, 
3L, 5L, 4L, 6L, 2L, 1L, 3L, 5L, 4L), .Label = c("This", "That", 
"Phat", "Bat", "Man", "Hat"), class = "factor"), X2002 = c(28L, 
9L, 17L, 8L, 95L, 18L, NA, NA, 36L, 40L, 15L, 10L, 71L, NA, 
14L, 25L, 18L, NA, 56L, 5L, 29L, 5L, 13L, 8L, 65L, 23L, 8L, 
34L, NA, 14L, 5L, 5L, NA, 51L, 18L, NA, 5L, 56L, 30L, 8L, 
9L, 11L, 77L, 5L, 53L, 12L, 16L, 13L, 114L, 30L, 8L, NA, 
52L, 38L, NA, 12L, 5L, 87L, 5L, 35L, NA, 10L, 6L, 92L, 10L, 
41L, NA, 22L, 8L, 115L, 27L, 6L, 9L, NA, 47L, 9L, 29L, 6L, 
11L, NA, 56L, 38L, 7L, 10L, NA, 93L, 6L, 22L, 9L, 9L, NA, 
59L, 5L), X2003 = c(32L, NA, 16L, 9L, 76L, 10L, NA, 5L, 24L, 
22L, 12L, 9L, 63L, 12L, 9L, 36L, 9L, 6L, 83L, 5L, 35L, NA, 
12L, 8L, 82L, 19L, 5L, 53L, 5L, 10L, NA, 7L, NA, 35L, 15L, 
6L, 6L, 40L, 30L, NA, 10L, 8L, 85L, 9L, 46L, NA, 14L, 9L, 
106L, 24L, 6L, 7L, 56L, 33L, NA, 12L, 9L, 106L, NA, 37L, 
7L, 11L, 8L, 79L, 5L, 54L, 5L, 10L, 6L, 100L, 25L, 9L, 5L, 
6L, 49L, NA, 31L, NA, 13L, 10L, 79L, 46L, NA, 14L, NA, 82L, 
5L, 21L, 7L, 11L, NA, 69L, NA), X2004 = c(35L, 6L, 13L, 8L, 
82L, 12L, 5L, NA, 35L, 34L, 5L, 6L, 75L, 9L, 9L, 40L, 13L, 
9L, 70L, NA, 41L, NA, 17L, 10L, 83L, 10L, 6L, 40L, NA, 18L, 
NA, 6L, NA, 34L, 10L, NA, NA, 45L, 38L, 6L, 11L, NA, 74L, 
NA, 45L, 5L, 12L, 9L, 131L, 34L, NA, NA, 64L, 28L, 5L, NA, 
NA, 93L, NA, 32L, NA, 9L, 11L, 99L, NA, 40L, NA, 18L, 8L, 
104L, 14L, NA, 13L, 6L, 67L, NA, 23L, NA, 6L, 8L, 85L, 49L, 
NA, 19L, 7L, 102L, NA, 28L, 5L, 7L, 7L, 74L, NA), X2005 = c(36L, 
NA, 20L, 10L, 93L, 22L, NA, NA, 35L, 38L, 13L, 9L, 99L, NA, 
14L, 48L, 17L, 7L, 70L, NA, 35L, NA, 13L, 9L, 103L, 16L, 
5L, 49L, NA, 12L, NA, 5L, 8L, 51L, 15L, 7L, 5L, 45L, 40L, 
NA, 12L, 5L, 102L, NA, 40L, NA, 21L, 16L, 141L, 25L, 9L, 
10L, 70L, 41L, NA, 10L, NA, 111L, NA, 37L, NA, 10L, 9L, 124L, 
NA, 37L, NA, 12L, 12L, 124L, 32L, NA, 16L, 6L, 45L, NA, 33L, 
NA, 8L, NA, 101L, 51L, NA, 19L, 5L, 117L, NA, 17L, NA, 11L, 
5L, 73L, NA), X2006 = c(38L, NA, 22L, 13L, 103L, 15L, NA, 
7L, 44L, 39L, 11L, 6L, 95L, NA, 15L, 53L, 16L, 9L, 89L, NA, 
41L, NA, 12L, 13L, 87L, 30L, 6L, 43L, NA, 14L, NA, 6L, 5L, 
50L, 19L, 5L, NA, 63L, 23L, NA, 6L, NA, 75L, NA, 38L, NA, 
12L, 19L, 142L, 32L, 7L, 7L, 64L, 49L, NA, 13L, 12L, 114L, 
NA, 48L, NA, 23L, 5L, 136L, NA, 52L, NA, 15L, 16L, 127L, 
24L, NA, 6L, NA, 57L, NA, 32L, NA, NA, 13L, 96L, 20L, NA, 
10L, 21L, 102L, NA, 31L, NA, 5L, 12L, 93L, NA)), .Names = c("level1", 
"level2", "X2002", "X2003", "X2004", "X2005", "X2006"), row.names = c(NA, 
-93L), class = "data.frame")

Answer 1

这应该是我认为的伎俩：

by(d, d$level1, function(x) cbind(x[,1:2], t(t(x[,-1:-2]) / colSums(x[,-1:-2], na.rm=TRUE))))

如果你想要一个数据框中的所有东西，你可以在那里运行 do.call（rbind，...）。

Answer 2

据我了解这个问题，你有总计，使用：

totals <- cast(level1 ~ variable, data=melt(d, na.rm=T),sum)

...并且您希望将它们转换为百分比。 （请注意，您在问题文本中调用了第一列“L1”，但数据结构调用了列“level1”。）

从总数到百分比比你想象的要简单得多。

prc <- 100 * totals[,-1] / colSums(totals[,-1])
rownames(prc) <- totals[,1]