Clojure中实现合并两个map向量的函数：同车型取最低价格

Alright, let's solve this problem where we need to merge two vectors of Clojure maps, keeping only the entry with the lowest :price for each :car value. Here's how to approach it step by step:

First, let's recap the requirements clearly:

• We have two vectors of maps, each containing :id, :car, and :price keys
• For entries with the same :car value, we keep the one with the smallest :price
• All unique :car entries (no duplicates across inputs) should be retained as-is

Step-by-Step Approach #

1. Combine both input vectors: Merge all entries into a single sequence so we can process them uniformly.
2. Group entries by :car: Cluster all maps that share the same car type together.
3. Select the minimum price entry per group: For each car group, pick the map with the lowest :price value.
4. Optional: Sort by :car: If you want the output ordered alphabetically by car type (matching your expected output), add a sort step.

Clojure Code Implementation #

First, define your input vectors:

(def input1 [{:id 1 :car "A" :price 10} {:id 2 :car "B" :price 20} {:id 3 :car "C" :price 30}])
(def input2 [{:id 4 :car "A" :price 5} {:id 5 :car "B" :price 30} {:id 6 :car "D" :price 40}])

Then the merging function:

(defn merge-car-maps [first-maps second-maps]
  (->> (concat first-maps second-maps)
       (group-by :car)
       (map (fn [[_ car-group]]
              (apply min-key :price car-group)))
       (sort-by :car)))

Testing the Function #

When you call the function with your inputs:

(merge-car-maps input1 input2)

You'll get exactly the expected output:

[{:id 4, :car "A", :price 5} {:id 2, :car "B", :price 20} {:id 3, :car "C", :price 30} {:id 6, :car "D", :price 40}]

How It Works #

• concat: Merges the two input vectors into one continuous sequence of maps.
• group-by :car: Creates a map where each key is a car type (like "A"), and the value is a vector of all maps with that car type.
• min-key :price: This function takes a key (:price) and a collection, returning the element with the smallest value for that key. Using apply here passes all elements of the car group to min-key.
• sort-by :car: Sorts the final collection alphabetically by the :car key to match the order in your expected output. If you don't care about the order, you can omit this step.

内容的提问来源于stack exchange，提问作者Kami.Foo