R：两个矩阵的列之间的成对欧几里德距离
以下循环运行时间太长（2 分钟/迭代） tumor_signals 的大小为 950000x422 Normal_signals 的大小为 950000x772 关于如何加快速度有什么想法吗？
for(i in 1:ncol(tumor_signals)){
x <- as.vector(tumor_signals[,i])
print("Assigned x")
y <- t((t(normal_signals) - x)^2)
print("assigned y")
y <- t(sqrt(colSums(y)))
print("done")
#all_distance <- cbind(all_distance,matrix(distance))
print(i)
}
The following loop takes too lonng to run (2mins/iteration)
The tumor_signals is size 950000x422
The normal_signals is size 950000x772
Any ideas for how to speed it up?
for(i in 1:ncol(tumor_signals)){
x <- as.vector(tumor_signals[,i])
print("Assigned x")
y <- t((t(normal_signals) - x)^2)
print("assigned y")
y <- t(sqrt(colSums(y)))
print("done")
#all_distance <- cbind(all_distance,matrix(distance))
print(i)
}
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评论（1）
您的代码中有一个错误 - 您不需要对
normal_signals进行转置。据我了解，您正在尝试计算所有i = 1,2,...422和j=1,2,...,772，tumor_signals[,i]和normal_signals[,j]之间的欧几里德距离。您可能希望结果采用 422 x 772 矩阵形式。fields包中有一个函数rdist()可以为您执行此操作：顺便说一句，Google 搜索
[R Euclidean distance]会很容易找到这个包。There's a bug in your code -- you don't need to take the transpose of
normal_signals. As I understand it, you are trying to compute, for alli = 1,2,...422, andj=1,2,...,772, the Euclidean distance betweentumor_signals[,i]andnormal_signals[,j]. You would probably want the results in a 422 x 772 matrix. There's a functionrdist()in the packagefieldsthat will do this for you:Incidentally, a Google search for
[R Euclidean distance]would have easily found this package.