Groovy .each vs .collect

Question

作为初学者Groovy研讨会的一部分,我们一直在迭代以下列表(fromJson.secrets):

[[floors:10, street:emaseS, url:http://plywoodpeople.com/wp-content/uploads/2012/03/kermit_the_frog.jpg], [floors:2, street:emaseS, url:http://36.media.tumblr.com/tumblr_lp9bg9Lh2x1r0h9bqo1_500.jpg], [floors:2, street:yawdaorB, url:https://montclairdispatch.com/wp-content/uploads/2013/07/broadway1.jpg], [floors:5, street:emaseS, url:AAA], [floors:2, street:yawdaorB, url:AAA], [floors:6, street:albmaR aL, url:AAA], [floors:1, street:teertS llaW, url:AAA], [floors:6, street:daoR yebbA, url:AAA], [floors:3, street:teertS llaW, url:AAA], [floors:4, street:dlican someone help me to understand the difference between the 2 methods in rehstoR, url:AAA]]

最初的计划是使用.collect,但它看起来像使用.each产生了相同的结果(迭代在列表上......).

问题是,有人可以帮助我理解有关我的用例和一般方法之间的区别

每:

reversed_streets = fromJson.secrets.each {
    it.street = it.street.reverse()
    it
}

搜集:

reversed_streets = fromJson.secrets.collect {
    it.street = it.street.reverse()
    it
}

Answer 1

each返回输入each.你的代码在那里操作it.street.所以你回到原来的列表,每个都street被颠倒了.随着collect您创建一个包含操纵项目的新列表.所以显而易见的结果是相同的,但区别在于您创建了一个新容器,但您的原始容器仍然被篡改.一个简单的经验法则:each用于副作用(这是你的例子).虽然collect用于创造新的东西(例如map)