Question

我有一个这样的字符串列表：-

listOf("abc", "a", "bb", "aa", "aaa", "bb", "a")

我想要这样的输出：-

listOf("a", "a", "aa", "bb", "bb", "aaa", "abc")

首先，我想按长度对列表进行排序，然后再次按字母对长度组进行排序。

到目前为止，我尝试使用以下代码

fun main() {
    val result = listOf("abc", "a", "bb", "aa", "aaa", "bb", "a").groupBy { it.length }
    val valueList = ArrayList(result.values).flatMap { it.toList() }
    println(valueList)
}

但是我得到的结果如下

[abc, aaa, a, a, bb, aa, bb]

@Sergey Lagutin的重复评论后我也尝试过

val sortedList = a.sortedWith(compareBy { it.length })

没有返回期望的结果

Answer 1

val a = listOf("abc", "a", "bb", "aa", "aaa", "bb", "a")
a.sortedWith(compareBy({ it.length }, { it })) // [a, a, aa, bb, bb, aaa, abc]

Answer 2

您可以尝试这种方式

val yourList = listOf("abc", "a", "bb", "aa", "aaa", "bb", "a")
val yourSortedList = yourList.sorted().sortedBy { it.length }

已排序将根据您的自然排序顺序对您进行排序。在这种情况下，它将是字母顺序。
使用 sortyBy ，您可以精确地确定排序顺序就是字符串的长度。

结果[a, a, aa, bb, bb, aaa, abc]

Answer 3

在集合上使用class SaveEveryEpoch(Callback): def __init__(self, model_name, *args, **kwargs): self.model_checkpoints_with_loss = [] self.model_name = model_name super().__init__(*args, **kwargs) def on_epoch_end(self, epoch, logs): # I suppose here it's a Functional model print(logs['acc']) path_to_checkpoint = ( str(datetime.datetime.now()).split(' ')[0] + f'_{self.model_name}' + f'_{epoch:02d}.hdf5' ) self.model.save(path_to_checkpoint) self.model_checkpoints_with_loss.append((logs['loss'], path_to_checkpoint))函数

sortedWith

在某些条件下更改列表元素的位置

3 个答案: