Groovy迭代时删除集合项目 [英] Groovy remove Collection item while iterating

问题描述

在迭代时是否有Groovy方法去除集合的项目？在Java中，这是通过使用 Iterator.remove（）完成的：

Is there a Groovy way of removing a Collection's item while iterating? In Java this is accomplished using Iterator.remove():

Collection collection = ...
for (Iterator it=collection.iterator(); it.hasNext(); ) {
    Object obj = it.next();
    if (should remove) {
        it.remove();
    }
}

Groovy是否在其语言中提供了迭代式语法，还是我有使用 Iterator.remove（）？

Does Groovy provide remove-while-iterating in its language syntax, or do I have do use Iterator.remove()?

推荐答案

使用 removeAll（） 。

Use removeAll().

> c = [1, 2, 3, 4, 5]
> c.removeAll { it % 2 == 0 }
> println c
[1, 3, 5]

您具体询问while iterating ，你是否试图用/每个对象做某件事？ removeAll 仍然有效，只要封闭的最后一条语句仍然是truthy / falsey（和以前一样）：

You ask specifically about "while iterating", are you trying to do something with/too each object? removeAll still works as long as the closure's last statement is still truthy/falsey (as before):

> c.removeAll { 
*     tmp = it * 10
*     println "ohai ${it}*10=${tmp}"
*     tmp >= 40
* }
ohai 1*10=10
ohai 2*20=20
ohai 3*30=30
ohai 4*40=40
ohai 5*50=50
> println c
[1, 2, 3]

闭包的返回值最后一条语句或明确的 return 值）是truthy / falsey，它将用于确定应该删除的内容。它不需要明确引用每个对象。

The closure's return value (value of the last statement, or an explicit return value) is truthy/falsey, it will be used to determine what should be removed. It doesn't need to refer explicitly to each object.

这篇关于Groovy迭代时删除集合项目的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！