如何快速从二维数组中删除特定元素

Question

我有一个[[String]]喜欢

myArr = [["1_2","1_3","1_4"], ["2_1","2_2","2_3"], ["3_1","3_2","3_3"]]然后

我想删除这样的元素 bypassArr = ["1_2","2_3","3_2"]

如何bypassArr从myArr使用元素中删除。或者我怎样才能得到这种类型的结果。

result = [["1_3","1_4"], ["2_1","2_2"], ["3_1","3_3"]]

Answer 1

如果目的是从bypassArr\xe2\x80\x9cinner\xe2\x80\x9d 数组中删除所有元素，则可以使用和 的myArr组合来实现：map()filter()

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let myArr = [["1_2","1_3","1_4"], ["2_1","2_2","2_3"], ["3_1","3_2","3_3"]]\nlet bypassArr = ["1_2","2_3","3_2"]\n\nlet result = myArr.map { innerArray in\n    innerArray.filter { elem in\n        !bypassArr.contains(elem)\n    }\n}\n\nprint(result)\n// [["1_3", "1_4"], ["2_1", "2_2"], ["3_1", "3_3"]]\n

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内部数组被过滤以删除给定的元素，外部数组被映射到过滤结果。

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或者使用简写参数表示法：

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let result = myArr.map { $0.filter { !bypassArr.contains($0) }}\n

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