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groovy - Groovy 中的分组依据

转载 作者:行者123 更新时间:2023-12-01 09:17:22 25 4
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我有一个对象列表如下

[["GX 470","Model"],["Lexus","Make"],["Jeep","Make"],["Red","Color"],["blue","Color"]]

我想把它转换成
{"Model":["GX 470"],"Make":["Lexus","Jeep"],"Color":["Red", "blue"]}

我试过了
list.groupBy{ it[1] }

但这让我回归
{"Model":[["GX 470","Model"]],"Make":[["Lexus","Make"],["Jeep","Make"]],"Color":[["Red","Color"],["blue","Color"]]}

最佳答案

您可以将中间结果进一步转换为:

list.groupBy{ it[1] }.collectEntries{ k, v -> [(k):v*.get(0)] }

关于groovy - Groovy 中的分组依据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40981393/

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