正如标题所说，计算了从节点u开始到节点v结束的所有路径，我必须通过以下类型计算节点u对节点v“I（u，v）”的影响：
# define matlab dir
MDIR = /Applications/MATLAB_R2017b.app
# compiles mex files using g++
#CC = gcc
# compiler flags for g++
#CCFLAGS = -O3 -fpic
# to use the intel compiler instead, uncomment CC and CCFLAGS below:
# compiles mex file using the intel compiler
CC = icc
# compiler flags for intel compiler
CCFLAGS = -O3 -fPIC -D__amd64
# Figure out which platform we're on
UNAME = $(shell uname -s)
# Linux
ifeq ($(findstring Linux,${UNAME}), Linux)
# define which files to be included
CINCLUDE = -I$(MDIR)/extern/include -Ic++ -shared
# define extension
EXT = mexa64
endif
# Mac OS X
ifeq ($(findstring Darwin,${UNAME}), Darwin)
# define which files to be included
CINCLUDE = -L$(MDIR)/bin/maci64 -Ic++ -shared -lmx -lmex -lmat -lmwblas
# define extension
EXT = mexmaci64
# CCFLAGS += -std=c++11
endif
SRC:=$(wildcard *.c)
*.o: $(SRC)
for i in $(SRC) ; do \
$(CC) $(CCFLAGS) -I$(MDIR)/extern/include -c -Ic++ $$i -o $${i%.c}.o; \
done
OBJ0:=bwblkslv.o sdmauxFill.o sdmauxRdot.o
OBJ1:=choltmpsiz.o
all:
$(CC) $(CCFLAGS) $(CINCLUDE) $(OBJ0) -o bwblkslv.$(EXT)
$(CC) $(CCFLAGS) $(CINCLUDE) $(OBJ1) -o choltmpsiz.$(EXT)
# clean up
clean:
rm -f *.o *.$(EXT)
这就是我这样做的方式：
I(u,v) = Sum(for all the paths from u to v)[ Product of all the weights of consecutive edges from u to v ]其中Alpha [u，v]是I（u，v），Flist [i，j]是路径i，直到第j个节点前进到v的概率。
排队
Alpha<-array(0,dim = c(length(V(g)),length(V(g))))
if( u!=v ){
paths<-all_simple_paths(g, from = V(g)[u])#computing all paths from u
UpathV<-paths[unlist(lapply(paths, function(p){p[length(p)] %in% v}))]#extracting those that end to v
if(length(UpathV)!=0){#computing probability matrix from u to v
Flist<-array(NA,dim = c(length(UpathV),max(lengths(UpathV))))
Flist[1:length(UpathV)]<-1
for(i in 1:length(UpathV)){
for(j in 2:length(UpathV[[i]])){
Flist[i,j]<-Flist[i,j-1]*E(g)[get.edge.ids(g,as.numeric(UpathV[[i]][(j-1):j]))]$weight
}
}
}#if length(pspaths)
if(length(UpathV)!=0){
for(i in 1:length(UpathV)){
Alpha[u,v]<-Alpha[u,v]+Flist[i,length(UpathV[[i]])]
}#computing I(u,v)
}
}else{
Alpha[u,v]<-1
}
你可以看到，如果第三条路径是1-> 2-> 3-> 4则它的长度为4，然后Flist [3,4]是边缘w（1-> 2）的权重*瓦特（2→3）* W（3→4）。
我的解决方案正在运行，但是对于大型网络，它需要花费很多时间来计算矩阵。鉴于我必须从每个u到每个vi计算I（u，v）需要知道是否有更有效的方法来做我的计算没有改变行3-4。提前做了很多！
答案 0 :(得分：0)
过了一段时间，我找到了一种方法，通过更改1行代码并添加另一行代码来加快计算速度。
当计算Flist数组而不是从E（g）$ weight获得每个权重值时，我们应该从邻接矩阵中获取它。我发现在R中使用矩阵比使用数据帧更快。所以添加：
AjM<-as_adjacency_matrix(g,attr = "weight")
和：
Flist[i,j]<-Flist[i,j-1]*AjM[as.numeric(UpathV[[i]][j-1]),as.numeric(UpathV[[i]][j])]