基本上，我有几个频率表d1和d2。假设我有：
UPDATE2：d1的实际结构是表。因此d1由d1 <- table(datavector)获得，类似于d2。
d1
Value 0 1 2 3 4 9
Freq 25 30 100 10 10 10
d2
Value 0 1 3 5 7 11 13
Freq 25 30 100 10 10 10 12
问题：我想生成一个矩阵，其行对应于d1和d2，列对应于d1和d2中所见的所有不同“值”。所以我想生成一个行和列的矩阵，如下所示：
[,"0"] [,"1"] [,"2"] [,"3"] [,"4"] [,"5"] [,"7"] [,"9"] [,"11"] [,"13"]
[1,] 25 30 100 10 10 0 0 10 0 0
[2,] 25 30 0 100 0 10 10 0 10 12
请注意，没有列号6,8和10，因为它们没有出现在频率表中。最后，我试图将此矩阵放入函数image.plot()。
更新1：我想我可以允许列号6,8和10出现在矩阵中，但最终我必须编写一个for循环来消除仅包含零个列的列。
更新3：请注意，我实际上使用250个数据向量，因此有250个表（每个表具有不同的长度/维度）。所以，我正在寻找一个有效的解决方案
更新4：请将上述内容视为我想要实现的内容的摘要。真实数据集如下：
> dput(head(get.dist(fnn[1])))
structure(c(0.999214894571557, 0.000134589502018843, 4.48631673396142e-05,
2.24315836698071e-05, 6.72947510094213e-05, 8.97263346792284e-05,
2.24315836698071e-05, 4.48631673396142e-05, 4.48631673396142e-05,
2.24315836698071e-05, 2.24315836698071e-05, 6.72947510094213e-05,
2.24315836698071e-05, 2.24315836698071e-05, 4.48631673396142e-05,
2.24315836698071e-05, 6.72947510094213e-05, 2.24315836698071e-05
), class = "table", .Dim = 18L, .Dimnames = structure(list(d = c("0",
"1", "2", "3", "4", "5", "8", "9", "11", "12", "15", "16", "17",
"18", "20", "22", "24", "31")), .Names = "d"))
> dput(head(get.dist(fnn[2])))
structure(c(0.71161956034096, 0.199147599820547, 0.0644010767160162,
0.0147599820547331, 0.00327501121579183, 0.000807537012113055,
6.72947510094213e-05, 0.000785105428443248, 0.000179452669358457,
0.000134589502018843, 0.000112157918349035, 4.48631673396142e-05,
6.72947510094213e-05, 0.00307312696276357, 0.00107671601615074,
0.000336473755047106, 6.72947510094213e-05, 2.24315836698071e-05,
2.24315836698071e-05), class = "table", .Dim = 19L, .Dimnames = structure(list(
d = c("0", "1", "2", "3", "4", "5", "6", "9", "10", "11",
"35", "36", "37", "38", "39", "40", "41", "42", "43")), .Names = "d"))
> dput(head(get.dist(fnn[3])))
structure(c(0.747353073126963, 0.13138178555406, 0.0295423956931359,
0.0139075818752804, 0.0119560340960072, 0.0151861821444594, 0.0243382682817407,
0.00697622252131, 0.00255720053835801, 0.00161507402422611, 0.00293853746074473,
0.00116644235082997, 0.004419021982952, 0.0018842530282638, 0.000628084342754598,
0.00053835800807537, 0.000448631673396142, 0.000493494840735756,
0.000650515926424406, 0.000403768506056528, 0.000269179004037685,
0.000179452669358457, 0.000269179004037685, 0.000179452669358457,
8.97263346792284e-05, 0.000246747420367878, 4.48631673396142e-05,
4.48631673396142e-05, 4.48631673396142e-05, 2.24315836698071e-05,
2.24315836698071e-05, 4.48631673396142e-05, 2.24315836698071e-05,
2.24315836698071e-05, 2.24315836698071e-05, 2.24315836698071e-05,
2.24315836698071e-05, 2.24315836698071e-05, 2.24315836698071e-05
), class = "table", .Dim = 39L, .Dimnames = structure(list(d = c("0",
"1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12",
"13", "14", "15", "16", "17", "18", "19", "20", "21", "22", "23",
"24", "25", "26", "27", "28", "30", "32", "33", "34", "36", "37",
"38", "43", "54", "67")), .Names = "d"))
> dput(head(get.dist(fnn[4])))
structure(c(0.217743382682817, 0.49416778824585, 0.135150291610588,
0.0331987438313145, 0.0243831314490803, 0.0431135038133692, 0.022790489008524,
0.00912965455361149, 0.00614625392552714, 0.00937640197397936,
0.00244504262000897, 0.000560789591745177, 0.000493494840735756,
0.000448631673396142, 0.000336473755047106, 0.000112157918349035,
0.000201884253028264, 4.48631673396142e-05, 4.48631673396142e-05,
2.24315836698071e-05, 2.24315836698071e-05, 4.48631673396142e-05,
2.24315836698071e-05), class = "table", .Dim = 23L, .Dimnames = structure(list(
d = c("0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "10",
"11", "12", "13", "14", "15", "16", "17", "18", "19", "23",
"25", "45")), .Names = "d"))
答案 0 :(得分：3)
这是一个使用Reduce的选项，似乎在提供数据时起作用：
# make a list including your 3 dput parts
keylist <- list(d1,d2,d3)
result <- Reduce(function(...) merge(..., by="d", all=T), keylist)
result <- transform(result,row.names=d,d=NULL)
result <- t(result)
rownames(result) <- NULL
似乎有效：
> result[,c(1:2,44:45)]
0 1 54 67
[1,] 0.9992149 0.0001345895 NA NA
[2,] 0.7116196 0.1991475998 NA NA
[3,] 0.7473531 0.1313817856 2.243158e-05 2.243158e-05
答案 1 :(得分：1)
我使用的是数据帧，但如果d1和d2是矩阵，那么如果您删除了unlist次调用，这仍然有用：
M <- matrix(0, nrow=2, ncol=12 )
colnames(M) <- as.character(0:11)
M[1 , as.character(d1[1 , 2:7]) ] <- unlist(d1[2, 2:7 ])
M
# 0 1 2 3 4 5 6 7 8 9 10 11
#[1,] 25 30 100 10 10 0 0 0 0 10 0 0
#[2,] 0 0 0 0 0 0 0 0 0 0 0 0
M[2 , as.character(d2[1 , 2:7]) ] <- unlist(d2[2, 2:7 ])
M
#-------------------
0 1 2 3 4 5 6 7 8 9 10 11
[1,] 25 30 100 10 10 0 0 0 0 10 0 0
[2,] 25 30 0 100 0 10 0 10 0 0 0 10
将我的示例转换为矩阵（从矩阵类继承索引）：
d1a <-data.matrix(d1[,-1])
rownames(d1a) <- d1[,1]
d2a <-data.matrix(d2[,-1])
rownames(d2a) <- d2[,1]
M[1 , as.character(d1a[1 , ]) ] <-d1a[2, ]
M[2 , as.character(d2a[1 , ]) ] <-d2a[2, ]
M
#---------
0 1 2 3 4 5 6 7 8 9 10 11
[1,] 25 30 100 10 10 0 0 0 0 10 0 0
[2,] 25 30 0 100 0 10 0 10 0 0 0 10
如果aslatemail认为（虽然我没有）这些是一行表，那么它就更容易了：
M[2 , colnames(d2b) ] <-d2b
M[2 , colnames(d1b) ] <-d1b
M
0 1 2 3 4 5 6 7 8 9 10 11
[1,] 25 30 100 10 10 0 0 0 0 10 0 0
[2,] 25 30 0 100 0 10 0 10 0 0 0 10
请，请，请，不要在这些上使用for循环：
> M[ , !colSums(M==0)==2]
0 1 2 3 4 5 7 9 11
[1,] 25 30 100 10 10 0 0 10 0
[2,] 25 30 0 100 0 10 10 0 10
如果您不创建任何零列，则无需删除任何零列：
你可以用这种方式创建dist.list：
dist.list= lapply(fnn, get.dist)
# 3 element example built from your example
dist.list<-{}
dist.list[[1]] <-
structure(c(0.999214894571557, 0.000134589502018843, 4.48631673396142e-05,
2.24315836698071e-05, 6.72947510094213e-05, 8.97263346792284e-05,
2.24315836698071e-05, 4.48631673396142e-05, 4.48631673396142e-05,
2.24315836698071e-05, 2.24315836698071e-05, 6.72947510094213e-05,
2.24315836698071e-05, 2.24315836698071e-05, 4.48631673396142e-05,
2.24315836698071e-05, 6.72947510094213e-05, 2.24315836698071e-05
), class = "table", .Dim = 18L, .Dimnames = structure(list(d = c("0",
"1", "2", "3", "4", "5", "8", "9", "11", "12", "15", "16", "17",
"18", "20", "22", "24", "31")), .Names = "d"))
dist.list[[2]] <-
structure(c(0.71161956034096, 0.199147599820547, 0.0644010767160162,
0.0147599820547331, 0.00327501121579183, 0.000807537012113055,
6.72947510094213e-05, 0.000785105428443248, 0.000179452669358457,
0.000134589502018843, 0.000112157918349035, 4.48631673396142e-05,
6.72947510094213e-05, 0.00307312696276357, 0.00107671601615074,
0.000336473755047106, 6.72947510094213e-05, 2.24315836698071e-05,
2.24315836698071e-05), class = "table", .Dim = 19L, .Dimnames = structure(list(
d = c("0", "1", "2", "3", "4", "5", "6", "9", "10", "11",
"35", "36", "37", "38", "39", "40", "41", "42", "43")), .Names = "d"))
dist.list[[3]] <-
structure(c(0.747353073126963, 0.13138178555406, 0.0295423956931359,
0.0139075818752804, 0.0119560340960072, 0.0151861821444594, 0.0243382682817407,
0.00697622252131, 0.00255720053835801, 0.00161507402422611, 0.00293853746074473,
0.00116644235082997, 0.004419021982952, 0.0018842530282638, 0.000628084342754598,
0.00053835800807537, 0.000448631673396142, 0.000493494840735756,
0.000650515926424406, 0.000403768506056528, 0.000269179004037685,
0.000179452669358457, 0.000269179004037685, 0.000179452669358457,
8.97263346792284e-05, 0.000246747420367878, 4.48631673396142e-05,
4.48631673396142e-05, 4.48631673396142e-05, 2.24315836698071e-05,
2.24315836698071e-05, 4.48631673396142e-05, 2.24315836698071e-05,
2.24315836698071e-05, 2.24315836698071e-05, 2.24315836698071e-05,
2.24315836698071e-05, 2.24315836698071e-05, 2.24315836698071e-05
), class = "table", .Dim = 39L, .Dimnames = structure(list(d = c("0",
"1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12",
"13", "14", "15", "16", "17", "18", "19", "20", "21", "22", "23",
"24", "25", "26", "27", "28", "30", "32", "33", "34", "36", "37",
"38", "43", "54", "67")), .Names = "d"))
all.names <- lapply(dist.list, names)
uniq.names <- unique(unlist(all.names))
M <- matrix(0, nrow=length(dist.list), ncol=length(uniq.names) )
colnames(M) <- uniq.names
for (i in seq_along(dist.list) ) {
M[i, all.names[[i]] ] <- dist.list[[i]] }
M
前20列
0 1 2 3 4
[1,] 0.9992149 0.0001345895 4.486317e-05 2.243158e-05 6.729475e-05
[2,] 0.7116196 0.1991475998 6.440108e-02 1.475998e-02 3.275011e-03
[3,] 0.7473531 0.1313817856 2.954240e-02 1.390758e-02 1.195603e-02
5 8 9 11 12
[1,] 8.972633e-05 2.243158e-05 4.486317e-05 4.486317e-05 2.243158e-05
[2,] 8.075370e-04 0.000000e+00 7.851054e-04 1.345895e-04 0.000000e+00
[3,] 1.518618e-02 2.557201e-03 1.615074e-03 1.166442e-03 4.419022e-03
15 16 17 18 20
[1,] 2.243158e-05 6.729475e-05 2.243158e-05 2.243158e-05 4.486317e-05
[2,] 0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00
[3,] 5.383580e-04 4.486317e-04 4.934948e-04 6.505159e-04 2.691790e-04
# remainder excluded
答案 2 :(得分：0)
如果您将d1和d2转换为data.table，则可以通过公用密钥轻松合并它们：
library(data.table)
> d1 <- data.table(value = c(0, 1, 2, 3, 4, 9), freq = c(25, 30, 100, 10, 10, 10))
> d2 <- data.table(value = c(0, 1, 3, 5, 7, 11), freq = c(25, 30, 100, 10, 10, 10))
> setkey(d1, value)
> setkey(d2, value)
> merge(d1, d2, all = TRUE)
value freq.x freq.y
1: 0 25 25
2: 1 30 30
3: 2 100 NA
4: 3 10 100
5: 4 10 NA
6: 5 NA 10
7: 7 NA 10
8: 9 10 NA
9: 11 NA 10
然后，您可以将生成的data.table转换为矩阵，将NA替换为0 s等。