将行与列匹配并计算相同的出现次数 R

Matching rows to columns and counting same occurences R

我有一个具有以下形式的数据集:-

a <- data.frame(X1=c("A", "B", "C", "A", "B", "C"),
                X2=c("B", "C", "C", "A", "A", "B"),
                X3=c("B", "E", "A", "A", "A", "B"),
                X4=c("E", "C", "A", "A", "A", "C"),
                X5=c("A", "C", "C", "A", "B", "B")
               )

我还有另一组如下表格:-

b <- data.frame(col_1=c("ASD", "ASD", "BSD", "BSD"),
                col_2=c(1, 1, 1, 1),
                col_3=c(12, 12, 31, 21),
                col_4=("A", "B", "B", "A")
               )

我想要做的是从集合 b 中取出列 col_4 并匹配集合 a 中的行,以便它告诉我哪一行有多少元素来自 col_4 的新专栏。新列的名称无关紧要。

例如:- 集合 a 中的第一行和第五行包含集合 b.

col_4 的所有元素

此外,不应找到重复项。对于前。集合 a 中的第六行有 3 "B"。但是由于集合 b 中的 col_4 只有两个 "B",它应该告诉我 2 而不是 3。

预期输出的形式为:-

c <- data.frame(X1=c("A", "B", "C", "A", "B", "C"),
                X2=c("B", "C", "C", "A", "A", "B"),
                X3=c("B", "E", "A", "A", "A", "B"),
                X4=c("E", "C", "A", "A", "A", "C"),
                X5=c("A", "C", "C", "A", "B", "B"),
                found=c(4, 1, 2, 2, 4, 2)
               )

使用data.table的选项:

library(data.table)

#convert a into a long format
m <- melt(setDT(a)[, rn:=.I], id.vars="rn", value.name="col_4")

#order by row number and create an index for identical occurrences in col_4
setorder(m, rn, col_4)[, vidx := rowid(col_4), rn]

#create a similar index for b
setDT(b, key="col_4")[, vidx := rowid(col_4)]

#count occurrences and lookup this count into original data
a[b[m, on=.(col_4, vidx), nomatch=0L][, .N, rn], on=.(rn), found := N]

输出:

   X1 X2 X3 X4 X5 rn found
1:  A  B  B  E  A  1     4
2:  B  C  E  C  C  2     1
3:  C  C  A  A  C  3     2
4:  A  A  A  A  A  4     2
5:  B  A  A  A  B  5     4
6:  C  B  B  C  B  6     2

我们可以使用 vecsets::vintersect 来处理重复项。

按行使用apply,我们可以计算b$col4a中每一行之间有多少共同值。

apply(a, 1, function(x) length(vecsets::vintersect(b$col_4, x)))
#[1] 4 1 2 2 4 2

另一个有效地操作集合的想法是 count 并比较每行 b$col_4 的元素 occurences a:

b1 = c(table(b$col_4))
#b1
#A B 
#2 2

a1 = table(factor(as.matrix(a), names(b1)), row(a))
#a1
#   
#    1 2 3 4 5 6
#  A 2 0 2 5 3 0
#  B 2 1 0 0 2 3

最后,确定每个元素(对于每一行)出现的最少次数并求和:

colSums(pmin(a1, b1))
#1 2 3 4 5 6 
#4 1 2 2 4 2

如果尺寸更大 a "data.frame" 和更多元素,Matrix::sparseMatrix 提供了一个合适的替代方案:

library(Matrix)

a.fac = factor(as.matrix(a), names(b1))
.i = as.integer(a.fac)
.j = c(row(a))

noNA = !is.na(.i)  ## need to remove NAs manually
.i = .i[noNA]
.j = .j[noNA]

a1 = sparseMatrix(i = .i, j = .j, x = 1L, dimnames = list(names(b1), 1:nrow(a)))

a1
#2 x 6 sparse Matrix of class "dgCMatrix"
#  1 2 3 4 5 6
#A 2 . 2 5 3 .
#B 2 1 . . 2 3

colSums(pmin(a1, b1))
#1 2 3 4 5 6 
#4 1 2 2 4 2