Question

我有一个哈希：

{
  "grey" => ["anf_94748_01_prod1", "anf_94748_01_model1", "anf_94748_01_model2"],
  "heather blue" => ["anf_106537_01_prod1", "anf_106537_01_model1", "anf_106537_01_model2"],
  "indigo" => [],
  "dark grey" => ["anf_94747_01_prod1"]
}

如何根据图像ID数组替换其值：

[317, 318, 319, 320, 340, 358, 365]

如果散列数组为空，则跳过它并转到下一个散列键，并为该映像分配id。期望的输出将是：

{
  "grey" => [317, 318, 319],
  "heather blue" => [320, 340, 358],
  "indigo" => [],
  "dark grey" => [365]
}

Answer 1

h = {
  "grey" => ["anf_94748_01_prod1", "anf_94748_01_model1", "anf_94748_01_model2"],
  "heather blue" => ["anf_106537_01_prod1", "anf_106537_01_model1", "anf_106537_01_model2"],
  "indigo" => [],
  "dark grey" => ["anf_94747_01_prod1"]
}
a = [317, 318, 319, 320, 340, 358, 365]

h.each_value{|v| v.map!{a.shift}}
# =>
# {
#   "grey"=>[317, 318, 319],
#   "heather blue"=>[320, 340, 358],
#   "indigo"=>[],
#   "dark grey"=>[365]
# }

Answer 2

＆lt; p＆gt;作为@ sawa的答案的变体，如果你不想改变＆lt; code＆gt; h＆lt; / code＆gt;，OP的示例哈希，你可以这样做：＆lt; / p为H. ＆lt; pre＆gt;＆lt; code＆gt; a = [317,318,319,320,340,358,365] h.merge（h）{| _，v | a.shift（v.size）} ＃=＆GT; {＆＃34; grey＆＃34; =＆gt; [317,318,319]，＆＃34;石南花蓝＆＃34; =＆gt; [320,340,358]，＃＆＃34; indigo＆＃34; =＆gt; []，＆＃34;深灰色＆＃34; =＆gt; [365] ＆LT; /代码＆GT;＆LT; /预＆GT; ＆lt; p＆gt;这使用＆lt; a href =＆＃34; http：//docs.ruby-lang.org/en/2.0.0/Hash.html#method-i-merge"的rel =＆＃34; nofollow的＆＃34;＆GT;哈希＃合并＆LT; / A＆GT;使用块来确定两个哈希中存在的键的值，这里是所有键。＆lt; / p＆gt;

Answer 3

reduce是一个很好的功能，可以在累积某些结果的同时通过数据结构。

imgs = [317, 318, 319, 320, 340, 358, 365]
input = {"grey"=>["anf_94748_01_prod1", "anf_94748_01_model1", "anf_94748_01_model2"], "heather blue"=>["anf_106537_01_prod1", "anf_106537_01_model1", "anf_106537_01_model2"], "indigo"=>[], "dark grey"=>["anf_94747_01_prod1"]}

input.reduce({}) { |acc, (k, xs)| acc[k] = imgs.shift(xs.count); acc}

# => {"grey"=>[317, 318, 319], "heather blue"=>[320, 340, 358], "indigo"=>[], "dark grey"=>[365]}

从其他数组更新哈希值数组

3 个答案: