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r - r 平均列表的元素

问题描述

假设我有一个包含三个对象的列表，如下所示

[[1]]
     yeargp gender   Estimate   ci.lower  ci.upper
1 1991-1995      M  0.8757711 -0.8407402  2.592282
2 1991-1995      F  0.0000000  0.0000000  0.000000
3 1996-2000      M  2.2119671 -0.8536629  5.277597
4 1996-2000      F  2.8254349 -0.3718457  6.022715
5 2001-2005      M  7.7695653  2.6460791 12.893051
6 2001-2005      F  2.2710074 -0.3108077  4.852822
7 2006-2010      M 12.1639403  6.1435827 18.184298
8 2006-2010      F  6.3637686  2.5667028 10.160834

[[2]]
     yeargp gender  Estimate   ci.lower  ci.upper
1 1991-1995      M  0.000000  0.0000000  0.000000
2 1991-1995      F  0.000000  0.0000000  0.000000
3 1996-2000      M  2.211967 -0.8536629  5.277597
4 1996-2000      F  2.825435 -0.3718457  6.022715
5 2001-2005      M  8.599076  3.2238115 13.974341
6 2001-2005      F  1.517900 -0.6003366  3.636137
7 2006-2010      M 13.485237  7.1911854 19.779289
8 2006-2010      F  5.991342  2.2651006  9.717582

[[3]]
     yeargp gender  Estimate   ci.lower  ci.upper
1 1991-1995      M  0.000000  0.0000000  0.000000
2 1991-1995      F  0.000000  0.0000000  0.000000
3 1996-2000      M  3.317951 -0.4366640  7.072565
4 1996-2000      F  1.883623 -0.7269454  4.494192
5 2001-2005      M  7.643263  2.6144621 12.672065
6 2001-2005      F  2.366219 -0.3266446  5.059082
7 2006-2010      M 13.637280  7.2795528 19.995008
8 2006-2010      F  5.991342  2.2651006  9.717582

计算 column3-column5 (Estimate, ci.lower, ci.upper) 中所有元素的平均值的有效方法是什么？

这是我期望实现的。

year      Gender Estimate     L.C.L          U.C.L
1991-1995   M    0.2919237   -0.280246733   0.864094
1991-1995   F    0            0             0
1996-2000   M    2.580628367 -0.714663267   5.875919667
1996-2000   F    2.511497633 -0.490212267   5.513207333
2001-2005   M    8.0039681    2.828117567   13.179819
2001-2005   F    2.0517088   -0.4125963     4.516013667
2006-2010   M    13.09548577  6.8714403     19.31953167
2006-2010   F    6.1154842    2.365634667   9.865332667

非常感谢任何建议。下面是我列表中 dput 函数的输出。

templist <- list(structure(list(yeargp = structure(c(1L, 1L, 2L, 2L, 3L, 
3L, 4L, 4L), .Label = c("1991-1995", "1996-2000", "2001-2005", 
"2006-2010"), class = "factor"), gender = structure(c(1L, 2L, 1L, 
2L, 1L, 2L, 1L, 2L), .Label = c("M", "F"), class = "factor"), 
    Estimate = c(0.875771052955988, 0, 2.2119670520759, 2.82543488793347, 
    7.76956525829443, 2.27100738124732, 12.1639402903974, 6.36376856610303
    ), ci.lower = c(-0.840740210837749, 0, -0.853662876400907, 
    -0.371845674593782, 2.64607905876294, -0.310807679155956, 
    6.14358267928312, 2.56670275678554), ci.upper = c(2.59228231674973, 
    0, 5.2775969805527, 6.02271545046073, 12.8930514578259, 4.85282244165059, 
    18.1842979015118, 10.1608343754205)), .Names = c("yeargp", 
"gender", "Estimate", "ci.lower", "ci.upper"), row.names = c(NA, 
-8L), class = "data.frame"), structure(list(yeargp = structure(c(1L, 
1L, 2L, 2L, 3L, 3L, 4L, 4L), .Label = c("1991-1995", "1996-2000", 
"2001-2005", "2006-2010"), class = "factor"), gender = structure(c(1L, 
2L, 1L, 2L, 1L, 2L, 1L, 2L), .Label = c("M", "F"), class = "factor"), 
    Estimate = c(0, 0, 2.2119670520759, 2.82543488793347, 8.59907630432197, 
    1.51790034439859, 13.4852371898016, 5.9913415231189), ci.lower = c(0, 
    0, -0.853662876400907, -0.371845674593782, 3.2238114821611, 
    -0.600336642205772, 7.19118540022504, 2.26510058455415), 
    ci.upper = c(0, 0, 5.2775969805527, 6.02271545046073, 13.9743411264828, 
    3.63613733100296, 19.7792889793781, 9.71758246168364)), .Names = c("yeargp", 
"gender", "Estimate", "ci.lower", "ci.upper"), row.names = c(NA, 
-8L), class = "data.frame"), structure(list(yeargp = structure(c(1L, 
1L, 2L, 2L, 3L, 3L, 4L, 4L), .Label = c("1991-1995", "1996-2000", 
"2001-2005", "2006-2010"), class = "factor"), gender = structure(c(1L, 
2L, 1L, 2L, 1L, 2L, 1L, 2L), .Label = c("M", "F"), class = "factor"), 
    Estimate = c(0, 0, 3.31795057811384, 1.88362325862232, 
    7.6432632822894, 2.36621893284824, 13.6372803202135, 5.9913415231189
    ), ci.lower = c(0, 0, -0.436663954372684, -0.726945388947865, 
    2.6144620600312, -0.32664459212626, 7.27955279689059, 2.26510058455415
    ), ci.upper = c(0, 0, 7.07256511060037, 4.4941919061925, 
    12.6720645045476, 5.05908245782275, 19.9950078435365, 9.71758246168364
    )), .Names = c("yeargp", "gender", "Estimate", "ci.lower", 
"ci.upper"), row.names = c(NA, -8L), class = "data.frame"))

标签： rlist

解决方案

简短而甜蜜：

df <- do.call(rbind, templist)
aggregate(df[3:5], df[1:2], mean)

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