如何将矢量化事务转换为二进制矩阵事务[重复]

Question

我有一个名为transactions 的data.frame，只有一个名为items 的字段，因此第i 行包含一个包含第i 个事务项的向量，它看起来像这样：

> head(transactions)
                                              items
1                                       Cake, Fudge
2                                       Coffee, Tea
3                                Coffee, Choco, Tea
4                                            Coffee
5                                Bread, Muffin, Jam
6                                            Coffee

我想把它转换成一个二进制矩阵，这样每个元素都表示如果给定的对象是为给定的交易购买的，它应该是这样的：

   Cake  Fudge  Coffee  Tea  Choco  Bread  Muffin  Jam
1     1      1       0    0      0      0       0    0
2     0      0       1    1      0      0       0    0
3     0      0       1    1      1      0       0    0
4     0      0       1    0      0      0       0    0
5     0      0       0    0      0      1       1    1
6     0      0       1    0      0      0       0    0

如果没有阴暗的嵌套 for 循环，我找不到一种方法。这就是从arules 包中应用apriori 的全部内容，如果你们中的任何人能帮我解决这个问题，将不胜感激。

谢谢！

Answer 1

我们可以创建新的列来对每一行进行分组 (row) 以及我们想要在存在值为 1 的情况下表示的值 (spread_value)。我们使用separate_rows 将每个逗号分隔值拆分为单独的行。然后我们 spread 将值从长到宽，如果没有值，我们将 fill 它设为 0。

library(tidyverse)

df %>%
  mutate(row = row_number(), spread_value = 1) %>%
  separate_rows(items, sep = ",") %>%
  mutate(items = trimws(items)) %>%
  spread(items, spread_value, fill = 0) %>%
  select(-row)


#  Bread Cake Choco Coffee Fudge Jam Muffin Tea
#1     0    1     0      0     1   0      0   0
#2     0    0     0      1     0   0      0   1
#3     0    0     1      1     0   0      0   1
#4     0    0     0      1     0   0      0   0
#5     1    0     0      0     0   1      1   0
#6     0    0     0      1     0   0      0   0

Answer 2

有来自splitstackshape 的cSplit_e 函数。

df1 <- splitstackshape::cSplit_e(
  data = df,
  split.col = "items",
  sep = ", ",
  mode = "binary",
  fixed = TRUE,
  type = "character",
  fill = 0L,
  drop = TRUE
)

names(df1) <- sub("^items_", "", names(df1))
df1
#  Bread Cake Choco Coffee Fudge Jam Muffin Tea
#1     0    1     0      0     1   0      0   0
#2     0    0     0      1     0   0      0   1
#3     0    0     1      1     0   0      0   1
#4     0    0     0      1     0   0      0   0
#5     1    0     0      0     0   1      1   0
#6     0    0     0      1     0   0      0   0

数据

df <- structure(list(items = c("Cake, Fudge", "Coffee, Tea", "Coffee, Choco, Tea", 
"Coffee", "Bread, Muffin, Jam", "Coffee")), .Names = "items", class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6"))

Answer 3

非 dplyr 替代方案：

library(magrittr)
library(stringr)

uniq_words <- df[["items"]] %>% 
  strsplit(", ") %>% 
  unlist() %>%
  unique()

sol <- outer(df[["items"]], uniq_words, str_detect) * 1L
colnames(sol) <- uniq_words

sol
     Cake Fudge Coffee Tea Choco Bread Muffin Jam
[1,]    1     1      0   0     0     0      0   0
[2,]    0     0      1   1     0     0      0   0
[3,]    0     0      1   1     1     0      0   0
[4,]    0     0      1   0     0     0      0   0
[5,]    0     0      0   0     0     1      1   1
[6,]    0     0      1   0     0     0      0   0

数据

df <- data.frame(
  items = c(
    "Cake, Fudge", "Coffee, Tea", "Coffee, Choco, Tea", 
    "Coffee", "Bread, Muffin, Jam", "Coffee"
  ),
  stringsAsFactors = FALSE
)