在给定条件下如何从MutableSet中删除项目? [英] How to remove an item from a MutableSet given a certain condition?

问题描述

var chart_values: MutableSet<MutableMap.MutableEntry<String, Any>>? = mutableSetOf()

打印图表值:

[ground={}, 
ground_level={0=115, 1=4, 2=0, 3=37, 4=63, 5=44, 6=40, 7=9}, 
ground_over={0=3, 1=3, 2=3, 3=3, 4=3, 5=3, 6=3}
date_of_birth=1988-07-18T00:00Z]

我想从 chart_values

推荐答案

鉴于chartValues类型为MutableSet<MutableMap.MutableEntry<String, Any>>?，您可以执行以下操作以删除任何以空映射作为值的条目:

Given that chartValues is of type MutableSet<MutableMap.MutableEntry<String, Any>>? you can do the following to remove any entry with an empty map as a value:

chartValues?.removeAll { (_, value) ->
    (value as? Map<*,  *>)?.isEmpty() == true
}

as?被称为安全强制转换运算符，它将返回强制转换的对象；如果强制转换失败，则返回null.

as? is called safe cast operator and will return the casted object or null if the cast did not succeed.

注意:

• 使用MutableMap<String, Any>
可能会更好
• 使用val而不是var，因为您要更改集合而不是对其的引用

• You might be better off using a MutableMap<String, Any>
• use val instead of var, since you want to mutate the collection an not the reference to it

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