如何在groovy中组合数组？

Question

下面的java代码存在，但我试图将其转换为groovy。我应该保持它与System.arraycopy一样吗？还是groovy有更好的方法来组合这样的数组？

byte[] combineArrays(foo, bar, start) { 
    def tmp = new byte[foo.length + bar.length] 
    System.arraycopy(foo, 0, tmp, 0, start) 
    System.arraycopy(bar, 0, tmp, start, bar.length) 
    System.arraycopy(foo, start, tmp, bar.length + start, foo.length - start) 
    tmp 
    } 

谢谢

Answer 1

如果你想使用一个数组：

def abc = [1,2,3,4] as Integer[] //Array 
def abcList = abc as List 
def xyz = [5,6,7,8] as Integer[] //Array 
def xyzList = xyz as List 

def combined = (abcList << xyzList).flatten() 

使用列表：

def abc = [1,2,3,4] 
def xyz = [5,6,7,8] 
def combined = (abc << xyz).flatten() 

Answer 2

这是可以做到这样的：

def newCombine(foo,bar,start) { 
    ([].add + foo[0..<start]+bar+foo[start..<foo.size()]).flatten() 
} 

它适用于各种阵列（字节[]）或列表

Answer 3

我与

byte[] combineArrays(foo, bar, int start) { 
    [*foo[0..<start], *bar, *foo[start..<foo.size()]] 
} 

Answer 4

def a = [1, 2, 3] 
def b = [4, 5, 6] 
a.addAll(b) 
println a 

>> [1, 2, 3, 4, 5, 6]

Answer 5

def a = [1, 2, 3] 
def b = [4, 5, 6] 

assert a.plus(b) == [1, 2, 3, 4, 5, 6] 
assert a + b  == [1, 2, 3, 4, 5, 6] 

Answer 6

所有的解决方案如果上面的失败去阵列是未定义：

def a = [1,2] 
def b 
assert a+b == [1, 2, null] 

这可能不是你想要的。

def a = [1,2,3,4] 
def b // null array 
def c = [0,4,null,6] 
def abc = [] 
[a,b,c].each{ if (it) abc += it } 
assert abc == [1, 2, 3, 4, 0, 4, null, 6] 

，或者添加所有然后过滤输出：

(a+b+c).findAll{ it != null } 

（在此假设null不在有效的值，如果添加之前存在

要么测试阵列原数组，这意味着第一个解决方案是好了很多，即使它可能不看的Groovy不够。）

Answer 7

诀窍是扁平化（）方法，即结合d嵌套数组合并为一个：

def a = [1, 2, 3] 
def b = [4, 5, 6] 
def combined = [a, b].flatten() 

assert combined == [1, 2, 3, 4, 5, 6] 

println(combined) 

要删除空值，你可以使用的findAll（）这样的：

def a = null 
def b = [4, 5, 6] 
def combined = [a, b].flatten().findAll{it} 

assert combined == [4, 5, 6] 

println(combined)