我有一个data. frame，其中前13行包含站点/观察信息。每列代表1个人，但是大多数人都有A和B观察（尽管有些人只有A，而少数人有A、B和C观察）。我想平均每一行每个人，并根据这些信息创建一个新的data.frame。
示例（第1行、第7行、第13行和第56-61行的小子集）：
OriginalID Tree003A Tree003B Tree008B Tree013A
1 Township LY LY LY LY
7 COFECHA ID LY1A003A LY1A003B LY1A008B LY1A013A
13 PathLength 37.5455 54.8963 57.9732 64.0679
56 2006 1.538 1.915 0.827 2.722
57 2007 1.357 1.923 0.854 2.224
58 2008 1.311 2.204 0.669 2.515
59 2009 0.702 1.125 0.382 2.413
60 2010 0.937 1.556 0.907 2.315
61 2011 0.942 1.268 1.514 1.858
我想创建一个新的data. frame来平均每个人的年度观测值，无论他们有A、A和B，还是A B和C观测值。个人的ID在第7行（COFECHAID）：
预期输出：
OriginalID Tree003avg Tree008avg Tree013avg
1 Township LY LY LY
7 COFECHA ID LY1A003avg LY1A008avg LY1A013avg
13 PathLength 46.2209 57.9732 64.0679
56 2006 1.727 0.827 2.722
57 2007 1.640 0.854 2.224
58 2008 1.758 0.669 2.515
59 2009 0.914 0.382 2.413
60 2010 1.247 0.907 2.315
61 2011 1.105 1.514 1.858
关于如何平均列的任何想法都很棒。我一直在尝试修改以下代码，但由于data. frame顶部有13行附加信息，我不知道如何指定仅平均行14:61。
rowMeans（子集（LY011B， select=c（"LY1A003A"，"LY1A003B"）），na.rm=TRUE）
我正在使用的一组更大的数据的代码是：
> dput(LY011B)
structure(list(OriginalTreeID = structure(c(58L, 53L, 57L, 59L,
51L, 61L, 50L, 55L, 56L, 60L, 54L, 49L, 52L, 1L, 2L, 3L, 4L,
5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L,
19L, 20L, 21L, 22L, 23L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 31L,
32L, 33L, 34L, 35L, 36L, 37L, 38L, 39L, 40L, 41L, 42L, 43L, 44L,
45L, 46L, 47L, 48L), .Label = c("1964", "1965", "1966", "1967",
"1968", "1969", "1970", "1971", "1972", "1973", "1974", "1975",
"1976", "1977", "1978", "1979", "1980", "1981", "1982", "1983",
"1984", "1985", "1986", "1987", "1988", "1989", "1990", "1991",
"1992", "1993", "1994", "1995", "1996", "1997", "1998", "1999",
"2000", "2001", "2002", "2003", "2004", "2005", "2006", "2007",
"2008", "2009", "2010", "2011", "AnalysisDateTime", "COFECHA ID",
"CoreLetter", "PathLength", "Plot#", "RingCount", "SiteID", "SP",
"Subplot#", "Township", "Tree#", "YearLastRing", "YearLastWhiteWood"
), class = "factor"), Tree003A = structure(c(35L, 8L, 34L, 7L,
34L, 21L, 36L, 31L, 37L, 30L, 32L, 29L, 33L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 23L, 22L, 25L, 28L, 27L, 24L, 26L, 20L, 16L,
15L, 6L, 18L, 12L, 10L, 3L, 9L, 11L, 19L, 17L, 14L, 13L, 2L,
4L, 5L), .Label = c("", "0.702", "0.803", "0.937", "0.942", "0.961",
"003", "1", "1.09", "1.116", "1.124", "1.224", "1.311", "1.357",
"1.471", "1.509", "1.538", "1.649", "1.679", "1.782", "1999",
"2.084", "2.148", "2.162", "2.214", "2.313", "2.429", "2.848",
"2/19/2014 11:06", "2011", "23017323011sp1", "24", "37.5455",
"A", "LY", "LY1A003A", "sp1"), class = "factor"), Tree003B = structure(c(56L,
19L, 54L, 18L, 55L, 49L, 57L, 51L, 58L, 50L, 52L, 48L, 53L, 1L,
1L, 1L, 1L, 10L, 7L, 8L, 6L, 5L, 4L, 3L, 2L, 11L, 9L, 30L, 15L,
24L, 20L, 23L, 33L, 37L, 42L, 13L, 44L, 36L, 12L, 16L, 21L, 27L,
35L, 41L, 38L, 26L, 40L, 14L, 46L, 32L, 28L, 17L, 31L, 22L, 39L,
43L, 45L, 47L, 25L, 34L, 29L), .Label = c("", "0.073", "0.092",
"0.173", "0.174", "0.358", "0.413", "0.425", "0.58", "0.697",
"0.719", "0.843", "0.883", "0.896", "0.937", "0.941", "0.964",
"003", "1", "1.048", "1.067", "1.075", "1.097", "1.119", "1.125",
"1.176", "1.207", "1.267", "1.268", "1.27", "1.297", "1.402",
"1.429", "1.556", "1.662", "1.693", "1.704", "1.735", "1.76",
"1.792", "1.816", "1.881", "1.915", "1.92", "1.923", "2.155",
"2.204", "2/19/2014 11:06", "2000", "2011", "23017323011sp1",
"48", "54.8963", "A", "B", "LY", "LY1A003B", "sp1"), class = "factor"),
Tree008B = structure(c(59L, 24L, 57L, 23L, 58L, 52L, 60L,
54L, 61L, 53L, 55L, 51L, 56L, 19L, 14L, 13L, 22L, 7L, 8L,
9L, 4L, 6L, 3L, 1L, 2L, 10L, 25L, 47L, 43L, 49L, 46L, 40L,
50L, 48L, 44L, 17L, 36L, 31L, 27L, 30L, 39L, 37L, 34L, 45L,
38L, 32L, 41L, 29L, 42L, 33L, 28L, 26L, 21L, 11L, 15L, 16L,
18L, 12L, 5L, 20L, 35L), .Label = c("0.302", "0.31", "0.318",
"0.357", "0.382", "0.412", "0.452", "0.476", "0.5", "0.539",
"0.591", "0.669", "0.673", "0.787", "0.79", "0.827", "0.835",
"0.854", "0.879", "0.907", "0.917", "0.967", "008", "1",
"1.027", "1.037", "1.141", "1.152", "1.172", "1.263", "1.383",
"1.411", "1.446", "1.498", "1.514", "1.611", "1.671", "1.685",
"1.695", "1.719", "1.783", "1.879", "1.884", "1.927", "1.97",
"2.019", "2.069", "2.35", "2.696", "2.979", "2/19/2014 11:06",
"2000", "2011", "23017323011sp1", "48", "57.9732", "A", "B",
"LY", "LY1A008B", "sp1"), class = "factor"), Tree013A = structure(c(45L,
6L, 44L, 5L, 44L, 38L, 46L, 40L, 47L, 39L, 42L, 37L, 43L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 10L,
13L, 8L, 22L, 14L, 18L, 24L, 4L, 11L, 25L, 7L, 36L, 41L,
33L, 29L, 17L, 28L, 23L, 21L, 16L, 26L, 15L, 3L, 20L, 12L,
2L, 9L, 34L, 35L, 27L, 32L, 31L, 30L, 19L), .Label = c("",
"0.608", "0.916", "0.945", "013", "1", "1.125", "1.18", "1.388",
"1.423", "1.493", "1.498", "1.554", "1.579", "1.619", "1.629",
"1.719", "1.756", "1.858", "1.867", "1.869", "1.876", "1.9",
"1.916", "2.023", "2.089", "2.224", "2.246", "2.247", "2.315",
"2.413", "2.515", "2.547", "2.645", "2.722", "2.785", "2/19/2014 11:11",
"2002", "2011", "23017323011sp1", "3.375", "34", "64.0679",
"A", "LY", "LY1A013A", "sp1"), class = "factor")), .Names = c("OriginalTreeID",
"Tree003A", "Tree003B", "Tree008B", "Tree013A"), row.names = c(NA,
61L), class = "data.frame")
这是另一种方法，其中大部分工作是通过使用reshape包重新排列数据来完成的。数据被“munging”后，可以使用cast函数将其重新排列成几乎任何你想要的东西。
# I'm used to the transpose
y = t(x)
# Make the first row the column names
# Also get rid of hashes. They make things difficult
library(stringr)
colnames(y) = str_replace( y[1,], "#", "" )
y = data.frame(y[-1,],check.names=FALSE)
# reshape the data by defining the "ID" variables
library(reshape)
z = melt(y,id.vars=c("Township","Plot","Subplot","Tree",
"CoreLetter","COFECHA ID","SiteID","SP","AnalysisDateTime"))
z$value = as.numeric(as.character(z$value))
# Now you can do lots of things!
# All the info you wanted is there, but it's in a different format
# than your "intended output"
cast( z, Tree ~ variable, mean, na.rm=TRUE )
# To get to your "intended output"
out = cast( z, Township + Plot + Subplot + Tree ~ variable, mean, na.rm=TRUE )
out[["COFECHA ID"]] = with(out,paste0(Township,Plot,Subplot,Tree,"avg"))
out2 = out[,c(1,ncol(out),8:(ncol(out)-1))]
out3 = cbind(colnames(out2),t(out2))
colnames(out3) = c("OriginalID",paste0("Tree",out$Tree,"avg"))
# For kicks, here are some other things. Have fun!
cast(z, Tree ~ variable, median, na.rm=TRUE ) # the median instead of the mean
cast(z, Tree + CoreLetter ~ variable ) # back to your original data
cast(z, CoreLetter ~ variable, length ) # How many measurements from each core?
cast(z, CoreLetter ~ variable, mean ) # The average across different cores
为了更多的乐趣！
library(ggplot2)
d = z[-c(1:16), ] # A not so pretty hack
colnames(d)[10] = "Year"
d$Year = as.integer(as.character(d$Year))
ggplot(d, aes(x=Year, y=value, group=Tree, color=Tree, shape=CoreLetter)) +
geom_point() + geom_smooth(method="loess",span=0.3)
这是否意味着21世纪初是干燥的？
试试这个……
d.f <- your data structure...above
子集数据
d.f <- d[-(1:13), -1]
c.n <- colnames(d.f)
构建分组变量
f
f <- d[4, 2:ncol(d)]
将数据帧拆分为子数据帧
d.f <- apply(d.f, 2, as.numeric)
d.f[is.na(d.f)] <- 0
d.f.g <- as.data.frame(t(d.f))
a <- split(d.f.g, f)
将分组平均值计算为colMeans（因为转置）
grp.means <- lapply(a, colMeans)
grp. means是一个数据帧列表，每个数据帧都包含每个grp的日期平均值。根据需要重新形成它，您可能需要再次转置。
