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Ruby用可变条件划分数组

ruby arrays

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dawg · 技术社区 · 4 年前

假设我有一个这样的数组:

arr=[["1"], ["1"], ["2", "2"], ["2.1", "0.8"], ["2.2", "0.2"], 
     ["1"], ["2", "3"], ["2", "0.8"], ["2", "0.4"], ["1"], ["2"], 
     ["1", "0.8"], ["2", "0.4"], ["3", "0.3"]]

即长度为1的一系列子列表(1个或多个)后接一系列长度大于1的子列表(一个或多个子列表)。

我想把这份清单分为:

[[["1"], ["1"], ["2", "2"], ["2.1", "0.8"], ["2.2", "0.2"]], 
 [["1"], ["2", "3"], ["2", "0.8"], ["2", "0.4"]], 
 [["1"], ["2"], ["1", "0.8"], ["2", "0.4"], ["3", "0.3"]]]

即长度为1的子列表与长度大于1的以下子列表组合为这些子列表的单个子列表。

这样可以:

data = []

while arr.length > 0  
    tmp = []

    while arr.length > 0 && arr[0].length() == 1
        tmp << arr.shift 
    end

    while arr.length > 0 && arr[0].length() > 1
        tmp << arr.shift 
    end

    data << tmp 
end

p data
# [[["1"], ["1"], ["2", "2"], ["2.1", "0.8"], ["2.2", "0.2"]], [["1"], ["2", "3"], ["2", "0.8"], ["2", "0.4"]], [["1"], ["2"], ["1", "0.8"], ["2", "0.4"], ["3", "0.3"]]]

但这似乎非常笨拙。有没有 .groupby 或者我缺少的触发器或某种形式的Ruby枚举器,以便更容易地做到这一点?

2 回复 | 直到 4 年前

1

3

Chris 4 年前

作为对@dawg答案的改进,如果我们通过的块 :slice_when 检查 b 大于 a :

data = arr.slice_when { |a, b| b.length < a.length }.to_a

结果是:

[[["1"], ["1"], ["2", "2"], ["2.1", "0.8"], ["2.2", "0.2"]], 
 [["1"], ["2", "3"], ["2", "0.8"], ["2", "0.4"]], 
 [["1"], ["2"], ["1", "0.8"], ["2", "0.4"], ["3", "0.3"]]]

这只能正确处理数组有长度的情况 1 或 2 。为了使其更加稳健,我们可以检查我们只在 b 具有长度 1. 。

data = arr.slice_when { |a, b| b.length == 1 && b.length < a.length }.to_a

现在,如果:

arr = [["1"], ["1"], ["2", "2"], ["2.1", "0.8"], ["2.2", "0.2"], ["1", "3", "4"], ["1", "1"], 
       ["1"], ["2", "3"], ["2", "0.8"], ["2", "0.4"], ["1"], ["2"], 
       ["1", "0.8"], ["2", "0.4"], ["3", "0.3"]]

结果是:

[[["1"], ["1"], ["2", "2"], ["2.1", "0.8"], ["2.2", "0.2"], ["1", "3", "4"], ["1", "1"]], 
 [["1"], ["2", "3"], ["2", "0.8"], ["2", "0.4"]], 
 [["1"], ["2"], ["1", "0.8"], ["2", "0.4"], ["3", "0.3"]]]