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我有这样的循环，试图在这里实现解决方案，使用虚拟变量这样

aaa <- DFM %*% t(DFM)   #DFM is Quanteda dfm-sparse-matrix
for(i in 1:nrow(aaa)) aaa[i,] <- aaa[i,][order(aaa[i,], decreasing = TRUE)]

但现在

for(i in 1:nrow(mmm)) mmm[i,] <- aaa[i,][order(aaa[i,], decreasing = TRUE)]

wheremmm尚不存在，目标是做与mmm <- t(apply(a, 1, sort, decreasing = TRUE)). 但现在在 for 循环之前我需要初始化mmmelse Error: object 'mmm' not found。aaa和mmm的类型dgCMatrix由两个 Quanteda DFM 矩阵的矩阵乘法给出。

结构

aaaFunc由矩阵乘法给出，DFM %*% t(DFM)其中 DFM 是 Quanteda Sparse dfm-matrix。结构是这样的

> str(aaaFunc)
Formal class 'dgCMatrix' [package "Matrix"] with 6 slots
  ..@ i       : int [1:39052309] 0 2 1 0 2 2616 2880 3 4 5 ...
  ..@ p       : int [1:38162] 0 2 3 7 8 10 13 15 16 96 ...
  ..@ Dim     : int [1:2] 38161 38161
  ..@ Dimnames:List of 2
  .. ..$ : chr [1:38161] "90120000" "90120000" "90120000" "86140000" ...
  .. ..$ : chr [1:38161] "90120000" "90120000" "90120000" "86140000" ...
  ..@ x       : num [1:39052309] 1 1 1 1 2 1 1 1 2 1 ...
  ..@ factors : list()

使用此处提到的方法在 DFM 上出现错误，关于复制没有其内容但其结构/等的 R 对象的一般问题。

A.错误aaaFunc.mt[]<- NA

> aaaFunc.mt <- aaaFunc[0,]; aaaFunc.mt[] <- NA; aaaFunc.mt[1,]
Error in intI(i, n = x@Dim[1], dn[[1]], give.dn = FALSE) : index larger than maximal 0

B.错误mySparseMatrix.mt[nrow(mySparseMatrix),]<-

> aaaFunc.mt <- aaaFunc[0,]; aaaFunc.mt[nrow(aaaFunc),] <- NA
Error in intI(i, n = di[margin], dn = dn[[margin]], give.dn = FALSE) : 
  index larger than maximal 0

C.错误replace(...,NA)

Browse[2]> mmmFunc <- replace(aaaFunc,NA);
Error in replace(aaaFunc, NA) : 
  argument "values" is missing, with no default
Browse[2]> mmmFunc <- replace(aaaFunc,,NA);
Error in `[<-`(`*tmp*`, list, value = NA) : 
  argument "list" is missing, with no default
Browse[2]> mmmFunc <- replace(aaaFunc,c(),NA);
Error in .local(x, i, j, ..., value) : 
  not-yet-implemented 'Matrix[<-' method

您如何初始化由两个 Quanteda DFM 矩阵的矩阵乘法给出的空 dgCMatrix？

4

1 回答 1

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以下将初始化一个空的稀疏矩阵或重置现有的稀疏矩阵，同时保留维度和dimnames

library(Matrix)

i <- c(1,3:8)
j <- c(2,9,6:10)
x <- 7 * (1:7)
A <- sparseMatrix(i, j, x = x)
rownames(A) <- letters[seq_len(nrow(A))]

A2 <- sparseMatrix(i = integer(0), j = integer(0), dims = A@Dim, dimnames = A@Dimnames)

A@i <- integer(0)
A@p[] <- 0L
A@x <- numeric(0)

setequal(A, A2)
[1] TRUE
于 2017-01-12T13:29:07.360 回答