Question

我正在尝试编写一个自定义函数，该函数评估网络中相对于特定顶点属性平衡了多少个三角形。重要的是，属性名称应在函数调用中可自定义。

示例：

# Creating graph
A <- graph(c("a","b",
             "a","c",
             "c","b",
             "d","e",
             "e","a",
             "d","f",
             "f","e",
             "g","a",
             "g","e"),
           directed = FALSE)

V(A)$color <- c("lightblue",
                "lightcoral",
                "lightcoral",
                "lightblue",
                "lightblue",
                "lightblue",
                "lightcoral")

V(A)$shape <- c("circle",
                "circle",
                "circle",
                "square",
                "square",
                "square",
                "square")

我想要一个函数，让我检查A中的三角形相对于color，shape或其他顶点可能具有的任何其他属性是否平衡。 / p>

这是我到目前为止尝试过的：

TriadBalance  <- function(network,attribute){

        # Getting a List of all closed triangles in the network
        TriMat <- matrix(triangles(network), nrow=3)
        TriList <- lapply(seq_len(ncol(TriMat)), function(i) TriMat[,i])

        # Getting their attribute
        AttributeList <- lapply(TriList,function(x){return(V(network)[unlist(x)]$attribute)})

        # Checking if the attribute is the same in all triads
        Balance <- lapply(AttributeList, function(x){length(unique(unlist(x))) == 1})

        # counting the number of balanced triads
        NoBalance <- sum(unlist(Balance))

        return(NoBalance)

}

此方法中的错误发生在以下行：AttributeList <- lapply(TriList,function(x){return(V(network)[unlist(x)]$attribute)})，因为$attribute部分的求值结果不是函数调用中的相应输入。

问题的核心：

attribute = "color"
V(A)[c(1,2,3)]$attribute

V(A)[c(1,2,3)]$color

前两个折点返回NULL，第二个折点正确输出前三个节点的顶点属性。

通过函数调用传递顶点属性名称是否有其他方法？

Answer 1

我将展示如何将属性值作为一个小函数来获取，但是您可以将代码内联到函数中。要通过属性名称获取属性，只需grep属性名称即可找出属性的索引，然后通过其索引号访问该属性。

library(igraph)

SelectVAttr = function(graph, aName) {
    NamePattern = paste0("^", aName, "$")
    Index = grep(NamePattern, names(vertex_attr(graph)))
    vertex_attr(A)[[Index]]
}

## Using the example that you provided
SelectVAttr(A, "color")
[1] "lightblue"  "lightcoral" "lightcoral" "lightblue"  "lightblue" 
[6] "lightblue"  "lightcoral"
SelectVAttr(A, "shape")
[1] "circle" "circle" "circle" "square" "square" "square" "square"

NamePattern是为了确保您仅获得完全匹配的内容，而不仅仅是部分匹配的内容。

如何将顶点属性名称传递给igraph中的函数

1 个答案: