1)没有包裹这可以通过暴力来完成。使用df
从问题作为输入确保price
是数字(它是df
问题的)并计算最大的数字mx
对于每个变量。然后创建网格g
变量计数并计算total
每个及相关的价格objective
giving gg
。现在排序gg
按照目标的降序排列,并采取满足约束条件的解决方案。head
将显示前几个解决方案。
price <- as.numeric(as.character(df$price))
mx <- ceiling(20/price)
g <- expand.grid(ana = 0:mx[1], ban = 0:mx[2], cook = 0:mx[3])
gg <- transform(g, total = as.matrix(g) %*% price, objective = sqrt(ana * ban * cook))
best <- subset(gg[order(-gg$objective), ], total <= 20)
giving:
> head(best) # 1st row is best soln, 2nd row is next best, etc.
ana ban cook total objective
1643 3 9 5 19.96 11.61895
1929 3 7 6 19.80 11.22497
1346 3 10 4 19.37 10.95445
1611 4 6 5 19.88 10.95445
1632 3 8 5 19.21 10.95445
1961 2 10 6 19.88 10.95445
2) dplyr这也可以使用 dplyr 包很好地表达。使用g
and price
从上面:
library(dplyr)
g %>%
mutate(total = c(as.matrix(g) %*% price), objective = sqrt(ana * ban * cook)) %>%
filter(total <= 20) %>%
arrange(desc(objective)) %>%
top_n(6)
giving:
Selecting by objective
ana ban cook total objective
1 3 9 5 19.96 11.61895
2 3 7 6 19.80 11.22497
3 3 10 4 19.37 10.95445
4 4 6 5 19.88 10.95445
5 3 8 5 19.21 10.95445
6 2 10 6 19.88 10.95445