执行矩阵的成对比较 [英] Perform pairwise comparison of matrix
问题描述
Transportation.services Recreational.goods.and.vehicles Recreation.services Other.services
2.958003 -0.25983789 5.526694 2.8912009
2.857370 -0.03425164 5.312857 2.9698044
2.352275 0.30536569 4.596742 2.9190123
2.093233 0.65920773 4.192716 3.2567390
1.991406 0.92246531 3.963058 3.6298314
2.065791 1.06120930 3.692287 3.4422340
我试着在下面运行for循环，但是我知道R循环很慢。 p>
Difference.Matrix< - function（data）{
n <-2
new.cols =New列
list = list（）
for（i in 1：ncol（data））{$ b $ （数据））{
名称< - paste（diff，i，j，data [，i]，data [，j] ，
new< - data [，i] -data [，j]
list [[new.cols]]< -c（name）
data< （数据，新）
}
n = n + 1
}
结果< -list（data = data）
return（results）
}
正如我之前所说，代码运行速度很慢，甚至还没有完成一个单一的运行。另外，我对初学者的编码表示歉意。另外我知道这个代码将原始数据留在矩阵上，但是我可以稍后删除它。
是否可以使用apply函数或foreach数据？
您可以找到 combn code> apply $ / code>来创建结果：
$ $ p $ apply $（combn（ncol（d） ，2），2，function（x）d [，x [1]] - d [，x [2]]）
## [，1] [，2] [，3] [，4] [，5] [，6]
## [1，] 3.217841 -2.568691 0.0668021 -5.786532 -3.151039 2.6354931
## [2，] 2.891622 -2.455487 -0.1124344 -5.347109 -3.004056 2.3430526
## [3] 2.046909 -2.244467 -0.5667373 -4.291376 -2.613647 1.6777297
## [4，] 1.434025 -2.099483 -1.1635060 -3.533508 -2.597531 0.9359770
## [5，] 1.068941 -1.971652 -1.6384254 -3.040593 -2.707366 0.3332266
## [6，] 1.004582 -1.626496 -1.3764430 -2.631078 -2.381025 0.2500530
您可以使用a添加适当的名称nother apply 。这里的列名非常长，这会损害格式，但是标签会告诉我们每一列有什么不同：
x< ; - apply（combn（ncol（d），2），2，function（x）d [，x [1]] - d [，x [2]]）
colnames（x）< apply （combn（ncol（d），2），2，function（x）paste（names（d）[x]，collapse =' - '））
> x
Transportation.services - Recreational.goods.and.vehicles Transportation.services - Recreation.services
[1，] 3.217841 -2.568691
[2，] 2.891622 -2.455487
[ 3，] 2.046909 -2.244467
[4，] 1.434025 -2.099483
[5，] 1.068941 -1.971652
[6，] 1.004582 -1.626496
Transportation.services - Other.services Recreational.goods.and.vehicles - Recreation.services
[1，] 0.0668021 -5.786532
[2，] -0.1124344 -5.347109
[3，] -0.5667373 -4.291376
[4，] -1.1635060 -3.533508
[5，] -1.6384254 -3.040593
[6，] -1.3764430 -2.631078
Recreational.goods.and.vehicles - Other.services Recreation.services - Other.services
[1，] -3.151039 2.6354931
[ 2，] -3.004056 2.3430526
[3，] -2.613647 1.6777297
[4，] -2.597531 0.9359770
[5，] -2.707366 0.3332266
[6，] -2.381025 0.2500530
I have a matrix of n variables and I want to make an new matrix that is a pairwise difference of each vector, but not of itself. Here is an example of the data.
Transportation.services Recreational.goods.and.vehicles Recreation.services Other.services
2.958003 -0.25983789 5.526694 2.8912009
2.857370 -0.03425164 5.312857 2.9698044
2.352275 0.30536569 4.596742 2.9190123
2.093233 0.65920773 4.192716 3.2567390
1.991406 0.92246531 3.963058 3.6298314
2.065791 1.06120930 3.692287 3.4422340
I tried running a for loop below, but I'm aware that R is very slow with loops.
Difference.Matrix<- function(data){
n<-2
new.cols="New Columns"
list = list()
for (i in 1:ncol(data)){
for (j in n:ncol(data)){
name <- paste("diff",i,j,data[,i],data[,j],sep=".")
new<- data[,i]-data[,j]
list[[new.cols]]<-c(name)
data<-merge(data,new)
}
n= n+1
}
results<-list(data=data)
return(results)
}
As I said before the code is running very slow and has not even finished a single run through yet. Also I apologize for the beginner level coding. Also I am aware this code leaves the original data on the matrix, but I can delete it later.
Is it possible for me to use an apply function or foreach on this data?
You can find the pairs with combn and use apply to create the result:
apply(combn(ncol(d), 2), 2, function(x) d[,x[1]] - d[,x[2]])
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 3.217841 -2.568691 0.0668021 -5.786532 -3.151039 2.6354931
## [2,] 2.891622 -2.455487 -0.1124344 -5.347109 -3.004056 2.3430526
## [3,] 2.046909 -2.244467 -0.5667373 -4.291376 -2.613647 1.6777297
## [4,] 1.434025 -2.099483 -1.1635060 -3.533508 -2.597531 0.9359770
## [5,] 1.068941 -1.971652 -1.6384254 -3.040593 -2.707366 0.3332266
## [6,] 1.004582 -1.626496 -1.3764430 -2.631078 -2.381025 0.2500530
You can add appropriate names with another apply. Here the column names are very long, which impairs the formatting, but the labels tell what differences are in each column:
x <- apply(combn(ncol(d), 2), 2, function(x) d[,x[1]] - d[,x[2]])
colnames(x) <- apply(combn(ncol(d), 2), 2, function(x) paste(names(d)[x], collapse=' - '))
> x
Transportation.services - Recreational.goods.and.vehicles Transportation.services - Recreation.services
[1,] 3.217841 -2.568691
[2,] 2.891622 -2.455487
[3,] 2.046909 -2.244467
[4,] 1.434025 -2.099483
[5,] 1.068941 -1.971652
[6,] 1.004582 -1.626496
Transportation.services - Other.services Recreational.goods.and.vehicles - Recreation.services
[1,] 0.0668021 -5.786532
[2,] -0.1124344 -5.347109
[3,] -0.5667373 -4.291376
[4,] -1.1635060 -3.533508
[5,] -1.6384254 -3.040593
[6,] -1.3764430 -2.631078
Recreational.goods.and.vehicles - Other.services Recreation.services - Other.services
[1,] -3.151039 2.6354931
[2,] -3.004056 2.3430526
[3,] -2.613647 1.6777297
[4,] -2.597531 0.9359770
[5,] -2.707366 0.3332266
[6,] -2.381025 0.2500530
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