groupby + row与dplyr的平均值

Question

I have a large data frame, which includes impoT and nlc as key (ignore t all together), and other columns which each hold a number. I want to find, for each impoT and nlc pair, the average of all the other columns, or basically a rowMean. A subset of my data, which only includes one nlc, is given at the end. The last thing I tried was :

avg <- data.frame(a %>% group_by(impoT, nlc) %>% select(-c(1:3)) %>% mutate(r= rowMeans(.) ))
stds = (a %>% group_by(impoT, nlc) %>% select(-c(1:3)) %>% apply( 1, sd)) #wrong

dput(a)
    structure(list(impoT = 1:18, nlc = c(669L, 669L, 669L, 669L, 
669L, 669L, 669L, 669L, 669L, 669L, 669L, 669L, 669L, 669L, 669L, 
669L, 669L, 669L), t = c(102L, 118L, 134L, 150L, 166L, 182L, 
198L, 214L, 230L, 246L, 262L, 278L, 294L, 310L, 326L, 342L, 358L, 
374L), X11950 = c(6, 14, 40, 53, 59, 70, 118, 119, 111, 114, 
103, 220, 278, 94, 28, 13, 5, 8), X11951 = c(4, 18, 41, 64, 78, 
87, 140, 112, 113, 129, 112, 245, 322, 102, 52, 20, 15, 7), X11952 = c(8, 
13, 30, 42, 52, 86, 126, 118, 52, 87, 116, 251, 262, 101, 35, 
21, 15, 21), X11955 = c(9, 11, 47, 38, 39, 70, 95, 82, 80, 77, 
77, 142, 192, 78, 13, 13, 5, 0), X11956 = c(14, 13, 44, 65, 65, 
72, 125, 138, 117, 111, 104, 175, 282, 93, 28, 14, 8, 4), X11957 = c(10, 
7, 45, 42, 50, 83, 123, 102, 104, 82, 102, 234, 265, 101, 23, 
13, 7, 6), X11958 = c(10, 13, 42, 60, 68, 69, 106, 125, 104, 
103, 112, 233, 310, 128, 50, 22, 10, 5), X11959 = c(7, 11, 32, 
45, 63, 74, 119, 87, 121, 108, 104, 229, 266, 111, 46, 26, 22, 
11), X11962 = c(8, 12, 38, 35, 49, 58, 96, 66, 73, 109, 82, 161, 
192, 75, 22, 4, 2, 3), X11963 = c(8, 9, 39, 40, 56, 50, 142, 
98, 102, 78, 79, 220, 229, 87, 25, 5, 7, 2), X11964 = c(10, 9, 
42, 60, 53, 52, 105, 114, 96, 94, 95, 180, 268, 114, 23, 10, 
7, 10), X11965 = c(9, 9, 41, 40, 61, 81, 150, 102, 102, 121, 
125, 222, 347, 116, 37, 18, 3, 4), X11966 = c(10, 9, 34, 43, 
49, 73, 112, 123, 102, 92, 107, 207, 239, 115, 60, 18, 15, 5), 
    X11969 = c(8, 9, 31, 34, 41, 51, 93, 92, 68, 103, 76, 166, 
    182, 63, 24, 14, 6, 4), X11970 = c(7, 12, 33, 48, 56, 59, 
    102, 88, 99, 86, 103, 194, 233, 90, 25, 13, 7, 3), X11971 = c(9, 
    16, 37, 60, 78, 62, 114, 106, 129, 107, 91, 212, 272, 88, 
    31, 10, 3, 3), X12088 = c(6, 11, 41, 44, 56, 70, 106, 97, 
    64, 73, 75, 161, 186, 76, 17, 8, 2, 2), X12089 = c(0, 11, 
    53, 59, 62, 64, 114, 109, 109, 100, 66, 222, 241, 88, 19, 
    8, 8, 3), X12090 = c(4, 12, 57, 52, 65, 73, 132, 109, 120, 
    101, 104, 227, 238, 99, 17, 8, 10, 8), X12091 = c(4, 16, 
    54, 167, 74, 62, 111, 95, 120, 102, 92, 227, 317, 106, 44, 
    16, 10, 4), X12092 = c(9, 10, 50, 55, 63, 64, 130, 103, 98, 
    116, 83, 249, 279, 88, 35, 36, 22, 15), X12095 = c(5, 15, 
    39, 44, 53, 58, 95, 92, 67, 63, 69, 163, 182, 69, 20, 8, 
    4, 2), X12096 = c(3, 14, 49, 53, 71, 70, 107, 130, 90, 89, 
    101, 214, 253, 100, 30, 10, 3, 3), X12097 = c(2, 16, 53, 
    61, 82, 83, 123, 124, 125, 98, 89, 220, 274, 107, 20, 17, 
    7, 5), X12098 = c(6, 17, 56, 59, 51, 77, 102, 115, 93, 98, 
    83, 221, 288, 97, 36, 16, 9, 10), X12099 = c(2, 16, 39, 49, 
    60, 84, 112, 91, 102, 103, 108, 246, 261, 131, 49, 24, 18, 
    14), X12102 = c(4, 12, 29, 47, 64, 69, 104, 111, 92, 72, 
    105, 174, 179, 64, 16, 10, 2, 1)), .Names = c("impoT", "nlc", 
"t", "X11950", "X11951", "X11952", "X11955", "X11956", "X11957", 
"X11958", "X11959", "X11962", "X11963", "X11964", "X11965", "X11966", 
"X11969", "X11970", "X11971", "X12088", "X12089", "X12090", "X12091", 
"X12092", "X12095", "X12096", "X12097", "X12098", "X12099", "X12102"
), row.names = c(NA, -18L), class = "data.frame")

Answer 1

It's easiest if you split your means into two steps, as you're actually taking the mean of irregular groups: first each row, and second each group. This means you're taking the means of means, but given each of the row means is of the same amount of numbers, they should be fine that way, although you should consider that the grouping means may be means of different amounts of rows.

You also need to nest the select so you don't lose your grouping variables, and use summarise to collapse the groups. All told,

a %>% mutate(r = rowMeans(select(a, -c(1:3)))) %>% 
    group_by(impoT, nlc) %>% summarise(r = mean(r))

produces

Source: local data frame [18 x 3]
Groups: impoT [?]

   impoT   nlc          r
   (int) (int)      (dbl)
1      1   669   6.740741
2      2   669  12.407407
3      3   669  42.074074
4      4   669  54.037037
5      5   669  59.925926
6      6   669  69.296296
7      7   669 114.888889
8      8   669 105.481481
9      9   669  98.259259
10    10   669  96.888889
11    11   669  94.925926
12    12   669 207.962963
13    13   669 253.222222
14    14   669  95.592593
15    15   669  30.555556
16    16   669  14.629630
17    17   669   8.592593
18    18   669   6.037037