我是R世界的新手，我有一个包含这样的行系列的文件：
"0000010000010000000101000001000000011000000001
0000000000000000000000010001000001001000110001
0000100000000000000000010000000000000000010100
0100000001100000000001001001100000010000000001
0001000000000100010000010000000000010000000000"
我想从这个字符串开始构建一个矩阵。从现在开始我写了这段代码：
for(line in readLines(ff)){
line <- as.numeric(substring(line, seq(1,nchar(line),1), seq(1,nchar(line),1)))
}
但它只从文件中提取行，如何使用line向量构建矩阵？
答案 0 :(得分：6)
编辑：感谢Ananda Matho和agstudy的建议，这里有一个更好的方法来自动处理width参数。如果您的数据位于名为test.txt的文件中，则可以执行以下操作：
width <- nchar(readLines("test.txt", n=1))
m <- as.matrix(read.fwf("test.txt", widths=rep(1,width)))
我假设每个0/1都是一个不同的值。在这种情况下，您可以使用read.fwf，它允许通过指定每个字段的宽度来读取数据：
text <- "0000010000010000000101000001000000011000000001
0000000000000000000000010001000001001000110001
0000100000000000000000010000000000000000010100
0100000001100000000001001001100000010000000001
0001000000000100010000010000000000010000000000"
m <- as.matrix(read.fwf(textConnection(text), widths=rep(1,46)))
给出了：
R> m
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19
[1,] 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[3,] 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[4,] 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0
[5,] 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0
V20 V21 V22 V23 V24 V25 V26 V27 V28 V29 V30 V31 V32 V33 V34 V35 V36
[1,] 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1
[2,] 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0
[3,] 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0
[4,] 0 0 1 0 0 1 0 0 1 1 0 0 0 0 0 0 1
[5,] 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1
V37 V38 V39 V40 V41 V42 V43 V44 V45 V46
[1,] 1 0 0 0 0 0 0 0 0 1
[2,] 1 0 0 0 1 1 0 0 0 1
[3,] 0 0 0 0 0 1 0 1 0 0
[4,] 0 0 0 0 0 0 0 0 0 1
[5,] 0 0 0 0 0 0 0 0 0 0
在您的情况下，您将使用您的文件名替换textConnection(text))，并将rep(1,46)中的值46修改为矩阵每行中的值数。
答案 1 :(得分：6)
你也可以使用：
t <- readLines(textConnection("0000010000010000000101000001000000011000000001
0000000000000000000000010001000001001000110001
0000100000000000000000010000000000000000010100
0100000001100000000001001001100000010000000001
0001000000000100010000010000000000010000000000"))
do.call("rbind", lapply(strsplit(t, ""), as.numeric))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[1,] 0 0 0 0 0 1 0 0 0 0 0 1 0 0
[2,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[3,] 0 0 0 0 1 0 0 0 0 0 0 0 0 0
[4,] 0 1 0 0 0 0 0 0 0 1 1 0 0 0
[5,] 0 0 0 1 0 0 0 0 0 0 0 0 0 1
[,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26]
[1,] 0 0 0 0 0 1 0 1 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 1 0 0
[3,] 0 0 0 0 0 0 0 0 0 1 0 0
[4,] 0 0 0 0 0 0 0 1 0 0 1 0
[5,] 0 0 0 1 0 0 0 0 0 1 0 0
[,27] [,28] [,29] [,30] [,31] [,32] [,33] [,34] [,35] [,36] [,37] [,38]
[1,] 0 1 0 0 0 0 0 0 0 1 1 0
[2,] 0 1 0 0 0 0 0 1 0 0 1 0
[3,] 0 0 0 0 0 0 0 0 0 0 0 0
[4,] 0 1 1 0 0 0 0 0 0 1 0 0
[5,] 0 0 0 0 0 0 0 0 0 1 0 0
[,39] [,40] [,41] [,42] [,43] [,44] [,45] [,46]
[1,] 0 0 0 0 0 0 0 1
[2,] 0 0 1 1 0 0 0 1
[3,] 0 0 0 1 0 1 0 0
[4,] 0 0 0 0 0 0 0 1
[5,] 0 0 0 0 0 0 0 0