Question

我的数据：

datas=structure(list(Id = 1:4, Product1 = structure(c(3L, 2L, 2L, 1L
), .Label = c("1,2,5", "1,3", "5"), class = "factor"), Product2 = structure(c(4L, 
3L, 1L, 2L), .Label = c("A", "A,B,E", "B,D", "D"), class = "factor")), .Names = c("Id", 
"Product1", "Product2"), class = "data.frame", row.names = c(NA, 
-4L))

视觉上给予（人们将购买两种产品和每种产品或几种原因。对于产品1（原因= 1,2,3,4,5）和产品2（原因= A，B，C，D），E）原因可以合并。

Id  Product1    Product2
1   5           D
2   1,3         B,D
3   1,3         A
4   1,2,5       A,B,E

我想重塑它如下

对于每个ID计算：

Id= 1

        A   B   C   D   E
    1   0   0   0   0   0
    2   0   0   0   0   0
    3   0   0   0   0   0
    4   0   0   0   0   0
    5   0   0   0   1   0

Id= 2 : here 0.25 because we have B1,B3,D1,D3 so 1/4 for each one

    A   B       C   D       E
1   0   0,25    0   0,25    0
2   0   0       0   0       0
3   0   0,25    0   0,25    0
4   0   0       0   0       0
5   0   0       0   1       0


on so on Id = 4, we have : a1,a2,a5,b1,b2,b5,e1,e2,e5, so 1/9 for each one.

    A               B           C       D         E
1   0,11+0,5    0,11+0,25       0      0,25      0,11
2   0,11+0      0,11+0          0      0         0,11
3   0,5         0,25            0      0,25      0
4   0           0               0      0         0
5   0,11+0      0,11+0          0      1         0,11

如何轻松地，或者我应该计算每个矩阵并在每次迭代中得到总和？

非常感谢！

Answer 1

试试这个：

library(dplyr)
library(reshape2)

# split strings
L <- lapply(datas[,2:3],function(v) strsplit(as.character(v),','))

# generate all combinations of products
d <- mapply(expand.grid,L$Product1, L$Product2,SIMPLIFY = F,stringsAsFactors=F)

df <- melt(d,id.vars=c('Var1','Var2')) %>% # convert to long format
    group_by(L1) %>%
    mutate(weight=1/n()) %>%               # calculate weights
    group_by(Var1,Var2) %>%                #
    summarize(sm=sum(weight))              # calculate sums

dcast(df,Var1~Var2)
# Note that it ignores column C and row 4 because no data were available for them
#   Var1         A         B    D         E
# 1    1 0.6111111 0.3611111 0.25 0.1111111
# 2    2 0.1111111 0.1111111   NA 0.1111111
# 3    3 0.5000000 0.2500000 0.25        NA
# 4    5 0.1111111 0.1111111 1.00 0.1111111

Answer 2

datas=structure(list(Id = 1:4, Product1 = structure(c(3L, 2L, 2L, 1L), .Label = c("1,2,5", "1,3", "5"), class = "factor"),
                     Product2 = structure(c(4L, 3L, 1L, 2L), .Label = c("A", "A,B,E", "B,D", "D"), class = "factor")),
                .Names = c("Id", "Product1", "Product2"), class = "data.frame", row.names = c(NA, -4L))


f <- function(p1, p2, lvls = 1:5) {
  # p1 <- datas$Product1[2]; p2 <- datas$Product2[2]
  p1 <- strsplit(as.character(p1), ',')[[1]]
  p2 <- strsplit(as.character(p2), ',')[[1]]
  t2 <- factor(rep(p2, each = length(p1)), levels = LETTERS[lvls])
  t1 <- factor(rep(p1, length(p2)), levels = lvls)
  tbl <- table(t1, t2)
  tbl / sum(tbl)
}

对于单个ID

Map(f, datas$Product1, datas$Product2)

# [[1]]
#    t2
# t1  A B C D E
#   1 0 0 0 0 0
#   2 0 0 0 0 0
#   3 0 0 0 0 0
#   4 0 0 0 0 0
#   5 0 0 0 1 0
# 
# [[2]]
#    t2
# t1     A    B    C    D    E
#   1 0.00 0.25 0.00 0.25 0.00
#   2 0.00 0.00 0.00 0.00 0.00
#   3 0.00 0.25 0.00 0.25 0.00
#   4 0.00 0.00 0.00 0.00 0.00
#   5 0.00 0.00 0.00 0.00 0.00
# 
# [[3]]
#    t2
# t1    A   B   C   D   E
#   1 0.5 0.0 0.0 0.0 0.0
#   2 0.0 0.0 0.0 0.0 0.0
#   3 0.5 0.0 0.0 0.0 0.0
#   4 0.0 0.0 0.0 0.0 0.0
#   5 0.0 0.0 0.0 0.0 0.0
# 
# [[4]]
#    t2
# t1          A         B         C         D         E
#   1 0.1111111 0.1111111 0.0000000 0.0000000 0.1111111
#   2 0.1111111 0.1111111 0.0000000 0.0000000 0.1111111
#   3 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
#   4 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
#   5 0.1111111 0.1111111 0.0000000 0.0000000 0.1111111

累积所有这些

Reduce(`+`, Map(f, datas$Product1, datas$Product2), accumulate = TRUE)

# [[1]]
#    t2
# t1  A B C D E
#   1 0 0 0 0 0
#   2 0 0 0 0 0
#   3 0 0 0 0 0
#   4 0 0 0 0 0
#   5 0 0 0 1 0
# 
# [[2]]
#    t2
# t1     A    B    C    D    E
#   1 0.00 0.25 0.00 0.25 0.00
#   2 0.00 0.00 0.00 0.00 0.00
#   3 0.00 0.25 0.00 0.25 0.00
#   4 0.00 0.00 0.00 0.00 0.00
#   5 0.00 0.00 0.00 1.00 0.00
# 
# [[3]]
#    t2
# t1     A    B    C    D    E
#   1 0.50 0.25 0.00 0.25 0.00
#   2 0.00 0.00 0.00 0.00 0.00
#   3 0.50 0.25 0.00 0.25 0.00
#   4 0.00 0.00 0.00 0.00 0.00
#   5 0.00 0.00 0.00 1.00 0.00
# 
# [[4]]
#    t2
# t1          A         B         C         D         E
#   1 0.6111111 0.3611111 0.0000000 0.2500000 0.1111111
#   2 0.1111111 0.1111111 0.0000000 0.0000000 0.1111111
#   3 0.5000000 0.2500000 0.0000000 0.2500000 0.0000000
#   4 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
#   5 0.1111111 0.1111111 0.0000000 1.0000000 0.1111111

通过加权变量重塑矩阵

2 个答案: