我有一个由三列组成的组件列表：产品，组件和使用的组件数量：
a <- structure(list(prodName = c("prod1", "prod1", "prod2", "prod3",
"prod3", "int1", "int1", "int2", "int2"), component = c("a",
"int1", "b", "b", "int2", "a", "b", "int1", "d"), qty = c(1L,
2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L)), row.names = c(NA, -9L), class = c("data.table",
"data.frame"))
prodName component qty
1 prod1 a 1
2 prod1 int1 2
3 prod2 b 3
4 prod3 b 4
5 prod3 int2 5
6 int1 a 6
7 int1 b 7
8 int2 int1 8
9 int2 d 9
以prod开头的产品是最终产品，以int开头的产品是中间产品，带有字母的产品是原材料。
我需要完整的最终产品组件列表，其中只包含原材料。也就是说，我想将任何int转换为原材料。
在此示例中，我的预期结果是（我明确说明了结果数的计算）：
prodName |component |qty
prod1 |a |1+2*6 = 13
prod1 |b |0+2*7 = 14
prod2 |b |3
prod3 |b |4+5*8*7 = 284
prod3 |a |0+5*8*6 = 240
prod3 |d |0+5*9 = 45
我通过创建一个非常繁琐的merge连接序列解决了这个问题。虽然这种方法适用于玩具数据，但我不太可能将其应用于真实数据。
#load data.table
library(data.table)
# split the tables between products and different levels of intermediate
a1 <- a[prodName %like% "prod",]
b1 <- a[prodName %like% "int1",]
c1 <- a[prodName %like% "int2",]
# convert int2 to raw materials
d1 <- merge(c1,
b1,
by.x = "component",
by.y = "prodName",
all.x = TRUE)[
is.na(component.y),
component.y := component][
is.na(qty.y),
qty.y := 1][,
.(prodName, qty = qty.x*qty.y),
by = .(component = component.y)]
# Since int1 is already exploded into raw materials, rbind both tables:
d1 <- rbind(d1, b1)
# convert all final products into raw materials, except that the raw mats that go directly into the product won't appear:
e1 <- merge(a1,
d1,
by.x = "component",
by.y = "prodName",
all.x = TRUE)
# rbind the last calculated raw mats (those coming from intermediate products) with those coming _directly_ into the final product:
result <- rbind(e1[!is.na(qty.y),
.(prodName, qty = qty.x * qty.y),
by = .(component = component.y)],
e1[is.na(qty.y),
.(prodName, component, qty = qty.x)])[,
.(qty = sum(qty)),
keyby = .(prodName, component)]
我知道我可以将数据拆分为表并执行连接，直到每个中间产品都表示为仅由原材料组成，但是如上所述，由于数据的大小和级别，这将是最后的选择中间产品的递归。
是否有更简便/更好的方法来进行这种递归联接？
答案 0 :(得分：4)
从本质上来说，您的数据代表了有向图中的加权边列表。以下代码使用igraph库直接计算从原始组件->最终产品到每个简单路径的（产品）距离之和：
library(igraph)
## transform edgelist into graph
graph <- graph_from_edgelist(as.matrix(a[, c(2, 1)])) %>%
set_edge_attr("weight", value = unlist(a[, 3]))
## combinations raw components -> final products
out <- expand.grid(prodname = c("prod1", "prod2", "prod3"), component = c("a", "b", "d"), stringsAsFactors = FALSE)
## calculate quantities
out$qty <- mapply(function(component, prodname) {
## all simple paths from component -> prodname
all_paths <- all_simple_paths(graph, from = component, to = prodname)
## if simple paths exist, sum over product of weights for each path
ifelse(length(all_paths) > 0,
sum(sapply(all_paths, function(path) prod(E(graph, path = path)$weight))), 0)
}, out$component, out$prodname)
out
#> prodname component qty
#> 1 prod1 a 13
#> 2 prod2 a 0
#> 3 prod3 a 240
#> 4 prod1 b 14
#> 5 prod2 b 3
#> 6 prod3 b 284
#> 7 prod1 d 0
#> 8 prod2 d 0
#> 9 prod3 d 45
答案 1 :(得分：3)
这是我尝试使用您的数据集的尝试。
它使用while循环检查来查看components字段中是否还有prodName。循环始终需要具有相同的字段，因此，不需要为递归乘法器添加一列（即末尾为5 * 8 * 7），而是对迭代乘法器进行集成。也就是说，5 * 8 * 7最终变为5 * 56。
library(data.table)
a[, qty_multiplier := 1]
b <- copy(a)
while (b[component %in% prodName, .N] > 0) {
b <- b[a
, on = .(prodName = component)
, .(prodName = i.prodName
, component = ifelse(is.na(x.component), i.component, x.component)
, qty = i.qty
, qty_multiplier = ifelse(is.na(x.qty), 1, x.qty * qty_multiplier)
)
]
}
b[prodName %like% 'prod', .(qty = sum(qty * qty_multiplier)), by = .(prodName, component)]
prodName component qty
1: prod1 a 13
2: prod1 b 14
3: prod2 b 3
4: prod3 b 284
5: prod3 a 240
6: prod3 d 45
答案 2 :(得分：1)
我认为您最好用一组邻接矩阵来表示信息，这些矩阵可以告诉您 “其中有多少是由它构成的”。您需要4个矩阵，对应于所有可能的矩阵 关系。 例如，您将最终产品和中间体之间的关系放在具有3行的矩阵中 和2列是这样的：
QPI <- matrix(0,3,2)
row.names(QPI) <- c("p1","p2","p3")
colnames(QPI) <- c("i1","i2")
QPI["p1","i1"] <- 2
QPI["p3","i2"] <- 5
i1 i2
p1 2 0
p2 0 0
p3 0 5
这告诉您，需要2单位中间产品i1才能制成1单位最终产品 p1。
类似地，您定义其他矩阵：
QPR <- matrix(0,3,3)
row.names(QPR) <- c("p1","p2","p3")
colnames(QPR) <- c("a","b","d")
QPR["p1","a"] <- 1
QPR["p2","b"] <- 3
QPR["p3","b"] <- 4
QIR <- matrix(0,2,3)
row.names(QIR) <- c("i1","i2")
colnames(QIR) <- c("a","b","d")
QIR["i1","a"] <- 6
QIR["i1","b"] <- 7
QIR["i2","d"] <- 9
QII <- matrix(0,2,2)
row.names(QII) <- colnames(QII) <- c("i1","i2")
例如，从QIR来看，我们需要6单位原材料a才能制造一单位中间产品i1。 一旦有了这种方式，就可以总结从原材料到最终原材料的所有可能方式 矩阵乘法的乘积。
您有3个条件：您可以直接从原始[QPR]到最终[QPR] QPR，也可以从原始到中间
到最终[QPI%*%QIR]还是从原始到中间再到其他中间再到最终[QPI%*%QII%*%QIR]
结果最终由矩阵表示
result <- QPI%*%QIR + QPI%*%QII%*%QIR + QPR
我将所有代码放在下面。如果运行它，您将看到结果如下：
a b d
p1 13 14 0
p2 0 3 0
p3 240 284 45
说的完全相同
prodName |component |qty
prod1 |a |1+2*6 = 13
prod1 |b |0+2*7 = 14
prod2 |b |3
prod3 |b |4+5*8*7 = 284
prod3 |a |0+5*8*6 = 240
prod3 |d |0+5*9 = 45
希望这会有所帮助
QPI <- matrix(0,3,2)
row.names(QPI) <- c("p1","p2","p3")
colnames(QPI) <- c("i1","i2")
QPI["p1","i1"] <- 2
QPI["p3","i2"] <- 5
QPR <- matrix(0,3,3)
row.names(QPR) <- c("p1","p2","p3")
colnames(QPR) <- c("a","b","d")
QPR["p1","a"] <- 1
QPR["p2","b"] <- 3
QPR["p3","b"] <- 4
QIR <- matrix(0,2,3)
row.names(QIR) <- c("i1","i2")
colnames(QIR) <- c("a","b","d")
QIR["i1","a"] <- 6
QIR["i1","b"] <- 7
QIR["i2","d"] <- 9
QII <- matrix(0,2,2)
row.names(QII) <- colnames(QII) <- c("i1","i2")
QII["i2","i1"] <- 8
result <- QPI%*%QIR + QPI%*%QII%*%QIR + QPR
print(result)