R - 行X中的信息的平均列-Java 学习之路

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，而有些人有A，B和C观察） . 我想为每个人平均每一行，并根据这些信息创建一个新的data.frame .

Example (small subset with row 1, row 7, row 13, and row 56-61):

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，平均每个人的年度观察，无论他们是A，A和B，还是A B和C观察 . 个人的ID在第7行（COFECHA ID）：

Intended Output:

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会很棒 . 我一直在尝试修改下面的代码，但由于data.frame顶部有13行附加信息，我不知道如何指定只有平均行14:61 .

rowMeans（子集（LY011B，select = c（“LY1A003A”，“LY1A003B”）），na.rm = TRUE）

The code for a larger set of the data that I'm working with is:

> 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")

2 回答

2
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 < - gsub（“ . ？$”，“”，c.n）
```
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是一个数据框列表，每个数据框包含每个grp的日期平均值 . 根据需要重新形成，你可能想再次转置 .
回复于 2024-05-22T05:16:00+08:00

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 . 在数据为"munged"之后，可以使用 cast 函数将其重新排列为您想要的任何内容 .

# 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)

enter image description here

这是否意味着2000年初是干的？

回复于 2024-05-22T05:16:00+08:00

R - 行X中的信息的平均列

2 回答

相关问题