如何在groovy中找到数组元素的差异？

Question

如何找到groovy中映射的数组元素的差异？

def first = [['fruit':'grapes', 'drink':'tea'], 
['fruit':'apple', 'drink':'milk']]

def second = [['fruit':'grapes', 'drink':'tea'], 
['fruit':'melon', 'drink':'milk'],
['fruit':'coconut', 'drink':'soda']]

def diff = first.minus(second)

echo "diff ${diff}

实际输出为

diff []

预期输出为

[['fruit':'melon'],['fruit':'coconut', 'drink':'soda']]

Answer 1

当您想要执行此操作并查找差异时遇到的一个问题是 - 当您比较元素时您优先考虑哪个数组？鉴于您最初的问题，看来您优先考虑较大的一个，因为所需输出中的 fruit 是 melon （但也可能是 apple ）。

考虑到这一点，这是我为您创建的解决方案，它优先考虑更大的数组（您也可以在这里测试它 http: //tpcg.io/03UE84）：

def first = [['fruit':'grapes', 'drink':'tea'], 
['fruit':'apple', 'drink':'milk']]

def second = [['fruit':'grapes', 'drink':'tea'], 
['fruit':'melon', 'drink':'milk'],
['fruit':'coconut', 'drink':'soda']]

def diff = []

def biggerArray
def smallerArray

if ( first.size() > second.size() ) {
    biggerArray = first;
    smallerArray = second;
}
else if ( first.size() < second.size() ) {
    biggerArray = second;
    smallerArray = first;
}
else {
    //You have to decide how the diff looks in case they are the same size
}

for ( i = 0; i < biggerArray.size(); i++ ) {
    if ( smallerArray[i] && smallerArray[i] != biggerArray[i] ) {
        def pair = [:]
        if ( smallerArray[i].get("fruit") != biggerArray[i].get("fruit") ) {
            pair["fruit"] = biggerArray[i].get("fruit");
        }
        
        if ( smallerArray[i].get("drink") != biggerArray[i].get("drink") ) {
            pair["drink"] = biggerArray[i].get("drink");
        }
        
        if ( !pair.isEmpty ) {
            diff.add(pair);
        }
    }
    else if ( !smallerArray[i] ) {
        diff.add(biggerArray[i]);
    }
}

println(diff);

Answer 2

因此，我在 groovysh 中尝试了您的代码，并得到了接近您正在寻找的答案的内容：

def first = [['fruit':'grapes', 'drink':'tea'], 
            ['fruit':'apple', 'drink':'milk']]
def second = [['fruit':'grapes', 'drink':'tea'], 
             ['fruit':'melon', 'drink':'milk'],
             ['fruit':'coconut', 'drink':'soda']]
println( first.minus(second) )
println( second.minus(first) )

输出：

 [[fruit:apple, drink:milk]]
 [[fruit:melon, drink:milk], [fruit:coconut, drink:soda]]

minus 方法将依赖于顺序。它不会返回第二个列表中不在第一个列表中的项目。它只会返回第一个列表中的项目。因此“melon”、“coconut”不会与 first.minus(second) 一起出现。当然，明显的问题是 drink: milk 出现，因为它包含在地图中，其中 1 个成员不在第一个地图中，因此它包含该地图的所有成员。这就是 minus 方法的工作原理，因此它不能成为解决方案。

因此，这是另一种方法，可以满足您的期望（但它仍然依赖于顺序）：

def first = [['fruit':'grapes', 'drink':'tea'],
             ['fruit':'apple', 'drink':'milk']]
def second = [['fruit':'grapes', 'drink':'tea'],
              ['fruit':'melon', 'drink':'milk'],
              ['fruit':'coconut', 'drink':'soda']]

def diff( srcA, srcB ) {
    srcA.findResults { item ->
       def out = item.findAll { x ->
         !srcB.find { y -> y.containsKey(x.key) && y[x.key] == x.value }
       }
       out ?: null
    }
}

println( "first diff second = ${diff( first, second )}" )
println( "second diff first = ${diff( second, first )}" )

输出：

first diff second = [[fruit:apple]]
second diff first = [[fruit:melon], [fruit:coconut, drink:soda]]

Answer 3

这对我来说有效，

def first = [['fruit':'grapes', 'drink':'tea'], ['fruit':'apple', 'drink':'milk']] 
def second = [['fruit':'grapes','drink':'tea'],
['fruit':'melon', 'drink':'milk'], ['fruit':'coconut', 'drink':'soda']]

def firstDiff = first.findAll { !second.contains(it) }
def secondDiff = second.findAll { !first.contains(it) }

println(firstDiff);
println(secondDiff);

输出：

['fruit':'apple', 'drink':'milk']
[['fruit':'melon', 'drink':'milk'],['fruit':'coconut', 'drink':'soda']]