data.frame到数组(2列)
data.frame到数组(2列)
提问于 2012-05-24 13:02:45
我想知道如何将data.frame转换为R中的数组，因为我的data.frame由16列和>100行组成。每一行代表一个人(蜗牛)。我给每只蜗牛盖了8个地标。所以现在我有16个坐标/个人。不，我想把data.fram转换成一个数组，每个人都有自己的矩阵，所有的个体都放在一个(大的)数组中。
我的数据如下:有8个地标的6个人，也就是16坐标(X&Y)
X1.Lms Y1.Lms X2.Lms Y2.Lms X3.Lms Y3.Lms X4.Lms Y4.Lms X5.Lms Y5.Lms X6.Lms Y6.Lms X7.Lms Y7.Lms X8.Lms Y8.Lms
1 132500000 132500000 114500000 132000000 150100000 121500000 97600000 123500000 164700000 107800000 77600000 110400000 181600000 81700000 NA NA
2 135700000 150900000 114100000 152000000 147500000 142500000 96900000 143800000 161900000 128600000 78300000 131200000 178000000 100500000 NA NA
3 134100000 136900000 113700000 135400000 148700000 129900000 99600000 127800000 164700000 115900000 76600000 115100000 183600000 87600000 NA NA
4 137400000 147100000 119500000 145600000 149500000 143000000 102400000 136800000 169400000 128200000 84000000 123400000 186700000 106000000 NA NA
5 141300000 144200000 115800000 141300000 154000000 135800000 103900000 132700000 171300000 122400000 83600000 120800000 NA NA NA NA
6 136300000 153700000 118100000 150200000 151700000 146900000 105500000 143700000 168400000 135200000 83600000 132000000 188900000 111600000 NA NA您可以通过输入以下内容将该表输入R：
sample <- structure(list(X1.Lms = c(132500000L, 135700000L, 134100000L, 137400000L, 141300000L, 136300000L), Y1.Lms = c(132500000L, 150900000L, 136900000L, 147100000L, 144200000L, 153700000L), X2.Lms = c(114500000L, 114100000L, 113700000L, 119500000L, 115800000L, 118100000L), Y2.Lms = c(132000000L, 152000000L, 135400000L, 145600000L, 141300000L, 150200000L), X3.Lms = c(150100000L, 147500000L, 148700000L, 149500000L, 154000000L, 151700000L), Y3.Lms = c(121500000L, 142500000L, 129900000L, 143000000L, 135800000L, 146900000L ), X4.Lms = c(97600000L, 96900000L, 99600000L, 102400000L, 103900000L, 105500000L), Y4.Lms = c(123500000L, 143800000L, 127800000L, 136800000L, 132700000L, 143700000L), X5.Lms = c(164700000L, 161900000L, 164700000L, 169400000L, 171300000L, 168400000L ), Y5.Lms = c(107800000L, 128600000L, 115900000L, 128200000L, 122400000L, 135200000L), X6.Lms = c(77600000L, 78300000L, 76600000L, 84000000L, 83600000L, 83600000L), Y6.Lms = c(110400000L, 131200000L, 115100000L, 123400000L, 120800000L, 132000000L ), X7.Lms = c(181600000L, 178000000L, 183600000L, 186700000L, NA, 188900000L), Y7.Lms = c(81700000L, 100500000L, 87600000L, 106000000L, NA, 111600000L), X8.Lms = c(NA_integer_, NA_integer_, NA_integer_, NA_integer_, NA_integer_, NA_integer_), Y8.Lms = c(NA_integer_, NA_integer_, NA_integer_, NA_integer_, NA_integer_, NA_integer_ )), .Names = c("X1.Lms", "Y1.Lms", "X2.Lms", "Y2.Lms", "X3.Lms", "Y3.Lms", "X4.Lms", "Y4.Lms", "X5.Lms", "Y5.Lms", "X6.Lms", "Y6.Lms", "X7.Lms", "Y7.Lms", "X8.Lms", "Y8.Lms"), row.names = c(NA, 6L ), class = "data.frame")或者，我已经将代码巴斯托。
结果应该像“形状”包中大猩猩雌性的样本数据，其中每个坐标(X1.S & Y1.S)都是一个矩阵。我猜维数必须是(8x2xn)。N是原始data.frame中的行数。
install.packages("shapes")
library(shapes)
data(gorf.dat)
gorf.dat数据被组织为一个维数(8x2xn)的数组。
, , 1
[,1] [,2]
[1,] 5 193
[2,] 53 -27
[3,] 0 0
[4,] 0 33
[5,] -2 105
[6,] 18 176
[7,] 72 114
[8,] 92 38
, , 2
[,1] [,2]
[1,] 51 191
[2,] 55 -31
[3,] 0 0
[4,] 0 33
[5,] 25 106
[6,] 56 171
[7,] 98 105
[8,] 99 15
, , 3
[,1] [,2]
[1,] 36 187
[2,] 59 -31
[3,] 0 0
[4,] 0 36
[5,] 12 102
[6,] 38 171
[7,] 91 103
[8,] 100 19等等..。
我试过x <- as.array(样本，dim=.)但到目前为止我还没有找到解决办法。我想我必须尝试类似的东西，thant，在excel女士中的转置功能。
你好，卢克
--编辑--
你的答案帮了很多忙，但我想你错了。第一个人有8个地标，坐标为16 (X1、Y1、X2、Y2、X3、Y3.)。所以要对我想要的结果说清楚：)我的想法应该更像这样：
, , 1
[,1] [,2]
132500000 132500000
114500000 132000000
150100000 121500000
97600000 123500000
164700000 107800000
77600000 110400000
181600000 81700000
NA NA
, , 2
[,1] [,2]
135700000 150900000
114100000 152000000
147500000 142500000
969000000 143800000
161900000 128600000
78300000 131200000
178000000 100500000
NA NA第一行(又名第一例)有8个地标，坐标为16。它们都应该存储在一个矩阵中。在此之后，我希望将所有情况(>100)组合成一个数组，其中每个个体都有自己的矩阵。
谢谢你们的快速反应！
回答 2
Stack Overflow用户
回答已采纳
发布于 2012-05-24 14:58:08
试试这个：
nlandmarks = 8; nind = nrow(sample)
array(t(sample[,c(1:nlandmarks*2 -1, 1:nlandmarks * 2)]),
dim = c(nlandmarks, 2, nind))Stack Overflow用户
发布于 2012-05-24 13:37:37
经修订以处理新的数据结构：
sdat <- apply(sample, 1,matrix, ncol=2, byrow=TRUE ) #unfolds the interleaved columns
dim(sdat) <- c(8,2,6) # expands 16 x 6 into one X column and one Y column per slice
#-----------------
> sdat[, , 1:2]
, , 1
[,1] [,2]
[1,] 132500000 132500000
[2,] 114500000 132000000
[3,] 150100000 121500000
[4,] 97600000 123500000
[5,] 164700000 107800000
[6,] 77600000 110400000
[7,] 181600000 81700000
[8,] NA NA
, , 2
[,1] [,2]
[1,] 135700000 150900000
[2,] 114100000 152000000
[3,] 147500000 142500000
[4,] 96900000 143800000
[5,] 161900000 128600000
[6,] 78300000 131200000
[7,] 178000000 100500000
[8,] NA NA
.... remaining 4 slices omitted页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/10738026
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