我有两个数组:teams = [1,2,3]和drivers = [4,5,6]。通过排列，我成功地显示了这两个数组的所有组合，但也成功地定义了我希望从每个数组中使用的值的数量。所以我在"车队"中使用了1个值，在"车手"中使用了2个值。我只想显示总和小于或等于10的组合，并删除任何重复项。
teams = [1,2,3]
drivers = [4,5,6]
team = teams.permutation(1).to_a
driver = drivers.permutation(2).to_a
array = team.product(driver)
target = 11
这成功地输出了两个数组的所有组合，其中1个来自车队，2个来自车手，如下所示:
[[1], [4, 5]], [[1], [4, 6]], [[1], [5, 4]], [[1], [5, 6]], [[1], [6, 4]], [[1], [6, 5]], [[2], [4, 5]],etc…
仅显示小于或等于10的值，我的预期结果将是:[[1], [4, 5]], [[1], [5, 4]],
，然后没有重复，我就只剩下:[[1], [4, 5]]
我已经尝试添加下面的代码行，但我得到一个未定义的方法' <='
错误:@array = array[0].product(*array[1..-1]).select { |a| a.reduce(:+) <= target }
我也试过这个，没有运气:
result = array.combination(1).select{|combi| combi.sum <= target}
@array = result
我猜这和排列有关?
teams = [1,2,3]
drivers = [2,5,4,5,6,4,5,7]
max_driver_sum = 10
我假设drivers可以包含重复的元素(就像我的例子中一样)，但我将在最后解释如果没有重复元素，计算将如何简化。
作为第一步，让我们将drivers划分为重复值和非重复值。
counts = drivers.tally
#=> {2=>1, 5=>3, 4=>2, 6=>1, 7=>1}
dup_drivers, uniq_drivers = counts.partition { |_d,n| n > 1 }
.map { |arr| arr.map(&:first) }
#=> [[5, 4], [2, 6, 7]]
因此,
dup_drivers
#=> [5, 4]
uniq_drivers
#=> [2, 6, 7]
参见Enumerable#tally和Enumerable#partition。
,
counts.partition { |_d,n| n > 1 }
#=> [[[5, 3], [4, 2]], [[2, 1], [6, 1], [7, 1]]]
首先计算两个车手相等的唯一组合:
dup_combos = teams.each_with_object([]) do |t,arr|
max_driver = (max_driver_sum - t)/2
dup_drivers.each do |d|
arr << [[t],[d,d]] if d <= max_driver
end
end
#=> [[[1], [4, 4]], [[2], [4, 4]]]
接下来，计算两个驱动程序不相等的唯一组合:
all_uniq = uniq_drivers + dup_drivers
#=> [2, 6, 7, 5, 4]
all_uniq_combos = all_uniq.combination(2).to_a
#=> [[2, 6], [2, 7], [2, 5], [2, 4], [6, 7], [6, 5],
# [6, 4], [7, 5], [7, 4], [5, 4]]
uniq_combos = teams.each_with_object([]) do |t,arr|
adj_driver_sum = max_driver_sum - t
all_uniq_combos.each do |combo|
arr << [[t],combo] if combo.sum <= adj_driver_sum
end
end
#=> [[[1], [2, 6]], [[1], [2, 7]], [[1], [2, 5]], [[1], [2, 4]],
# [[1], [5, 4]], [[2], [2, 6]], [[2], [2, 5]], [[2], [2, 4]],
# [[3], [2, 5]], [[3], [2, 4]]]
看到数组#组合。
最后一步是合并两组组合:
a1 = dup_combos + uniq_combos
#=> [[[1], [4, 4]], [[2], [4, 4]], [[1], [2, 6]], [[1], [2, 7]],
# [[1], [2, 5]], [[1], [2, 4]], [[1], [5, 4]], [[2], [2, 6]],
# [[2], [2, 5]], [[2], [2, 4]], [[3], [2, 5]], [[3], [2, 4]]]
排序后，结果如下:
a1.sort
#=> [[[1], [2, 4]], [[1], [2, 5]], [[1], [2, 6]], [[1], [2, 7]],
# [[1], [4, 4]], [[1], [5, 4]],
# [[2], [2, 4]], [[2], [2, 5]], [[2], [2, 6]], [[2], [4, 4]],
# [[3], [2, 4]], [[3], [2, 5]]]
注意，前面没有使用array# uniq。如果需要，当然可以替换掉上面的一些变量。
如果drivers不包含重复项，则所需数组由uniq_combos给出，其中all_uniq在计算all_uniq_combos时被drivers取代。例如，
teams = [1,2,3]
drivers = [2,5,4,6,7]
max_driver_sum = 10
然后
all_uniq_combos = drivers.combination(2).to_a
#=> [[2, 5], [2, 4], [2, 6], [2, 7], [5, 4], [5, 6],
# [5, 7], [4, 6], [4, 7], [6, 7]]
combos = teams.each_with_object([]) do |t,arr|
adj_driver_sum = max_driver_sum - t
all_uniq_combos.each do |combo|
arr << [[t],combo] if combo.sum <= adj_driver_sum
end
end
#=> [[[1], [2, 5]], [[1], [2, 4]], [[1], [2, 6]], [[1], [2, 7]],
# [[1], [5, 4]], [[2], [2, 5]], [[2], [2, 4]], [[2], [2, 6]],
# [[3], [2, 5]], [[3], [2, 4]]]
combos.sort
#=> [[[1], [2, 4]], [[1], [2, 5]], [[1], [2, 6]], [[1], [2, 7]],
# [[1], [5, 4]],
# [[2], [2, 4]], [[2], [2, 5]], [[2], [2, 6]],
# [[3], [2, 4]], [[3], [2, 5]]]
这是一个方法
teams = [1, 2, 3]
drivers = [2, 6, 5, 4]
team = teams.permutation(1).to_a
driver = drivers.permutation(2).to_a
array = team.product(driver)
target = 10
res = array.select {|i| i.map(&:sum).sum <= target}.compact
==> [[[1], [2, 6]], [[1], [2, 5]], [[1], [2, 4]], [[1], [6, 2]],
[[1], [5, 2]], [[1], [5, 4]], [[1], [4, 2]], [[1], [4, 5]],
[[2], [2, 6]], [[2], [2, 5]], [[2], [2, 4]], [[2], [6, 2]],
[[2], [5, 2]], [[2], [4, 2]], [[3], [2, 5]], [[3], [2, 4]],
[[3], [5, 2]], [[3], [4, 2]]]
获取唯一项(修改后也适用于teams>驱动程序)
t1 = res.map {|i| i[0]}
d2 = res.map {|i| i[1].flatten.sort}
t1.zip(d2).uniq
==> [[[1], [2, 6]], [[1], [2, 5]], [[1], [2, 4]], [[1], [4, 5]],
[[2], [2, 6]], [[2], [2, 5]], [[2], [2, 4]], [[3], [2, 5]],
[[3], [2, 4]]]