2

嗨，我是 R 新手，我必须用它来制作维恩图。我已经用谷歌搜索了一段时间，我能找到的所有示例都处理二进制变量。但是我有 2 个列表（实际上是 2 个 csv 文件）。列表中的项目只是字符串，例如 PSF113_xxxx。我必须对它们进行比较，看看它们各自的独特之处和共享的东西。我将如何在 R 中制作维恩图？

此外，这些文件中的东西数量也不相同，一个比另一个稍多，这意味着 cbind 函数返回错误。

到目前为止，我已经想出了这个，但这只是给我一个名为 group 1 的圆圈的图像，里面有一个 1，外面有一个 0。

matLis <- list(matrix(A), matrix(B))

n <- max(sapply(matLis, nrow))
do.call(cbind, lapply(matLis, function (x)
     rbind(x, matrix(, n-nrow(x), ncol(x))))) 

x = vennCounts(n)
vennDiagram(x)

这是数据的一个例子

2 PSF113_0018
3 PSF113_0079
4 PSF113_0079a
5 PSF113_0079b

左边的编号不是我做过的任何事情，它补充说，当我从 excel 将文件导入 R 时

head(A)
> head(A)
            V1
1 PSF113_0016a
2  PSF113_0018
3  PSF113_0079
4 PSF113_0079a
5 PSF113_0079b
6 PSF113_0079c

> head(b,10)
             V1
1  PSF113_0016a
2   PSF113_0021
3   PSF113_0048
4   PSF113_0079
5  PSF113_0079a
6  PSF113_0079b
7  PSF113_0079c
8   PSF113_0295
9  PSF113_0324a
10 PSF113_0324b
4

1 回答 1

2

您的代码仍然不能完全重现，因为您还没有定义 A 或 B。这是包中维恩图的指南，venneuler因为我发现它更灵活。

List1 <- c("apple", "apple", "orange", "kiwi", "cherry", "peach")
List2 <- c("apple", "orange", "cherry", "tomatoe", "pear", "plum", "plum")
Lists <- list(List1, List2)  #put the word vectors into a list to supply lapply
items <- sort(unique(unlist(Lists)))   #put in alphabetical order
MAT <- matrix(rep(0, length(items)*length(Lists)), ncol=2)  #make a matrix of 0's
colnames(MAT) <- paste0("List", 1:2)
rownames(MAT) <- items
lapply(seq_along(Lists), function(i) {   #fill the matrix
    MAT[items %in% Lists[[i]], i] <<- table(Lists[[i]])
})

MAT   #look at the results
library(venneuler)
v <- venneuler(MAT)
plot(v)

编辑：头部非常有帮助，因为它给了我们一些可以使用的东西。试试这个方法：

#For reproducibility (skip this and read in the csv files)
A <- structure(list(V1 = structure(1:6, .Label = c("PSF113_0016a", 
    "PSF113_0018", "PSF113_0079", "PSF113_0079a", "PSF113_0079b", 
    "PSF113_0079c"), class = "factor")), .Names = "V1", 
    class = "data.frame", row.names = c("1", 
    "2", "3", "4", "5", "6"))

B <- structure(list(V1 = structure(1:10, .Label = c("PSF113_0016a", 
    "PSF113_0021", "PSF113_0048", "PSF113_0079", "PSF113_0079a", 
    "PSF113_0079b", "PSF113_0079c", "PSF113_0295", "PSF113_0324a", 
    "PSF113_0324b"), class = "factor")), .Names = "V1", 
    class = "data.frame", row.names = c("1", 
    "2", "3", "4", "5", "6", "7", "8", "9", "10"))

从这里运行代码：

#after reading in the csv files start here
Lists <- list(A, B)  #put the word vectors into a list to supply lapply
Lists <- lapply(Lists, function(x) as.character(unlist(x)))
items <- sort(unique(unlist(Lists)))   #put in alphabetical order
MAT <- matrix(rep(0, length(items)*length(Lists)), ncol=2)  #make a matrix of 0's
colnames(MAT) <- paste0("List", 1:2)
rownames(MAT) <- items
lapply(seq_along(Lists), function(i) {   #fill the matrix
    MAT[items %in% Lists[[i]], i] <<- table(Lists[[i]])
})

MAT   #look at the results
library(venneuler)
v <- venneuler(MAT)
plot(v)

这种方法的不同之处在于我没有列出两个数据帧（如果它们是数据帧），然后将它们转换为字符向量。我认为这应该有效。

于 2012-07-30T13:52:53.960 回答