Question

如何比较两个列表中的项目并创建一个与Groovy不同的新列表？

Answer 1

我只是使用算术运算符，我认为发生的事情更为明显：

def a = ["foo", "bar", "baz", "baz"]
def b = ["foo", "qux"]

assert ["bar", "baz", "baz", "qux"] == ((a - b) + (b - a))

Answer 2

集合相交可能对你有所帮助，即使反转它有点棘手。也许是这样的：

def collection1 = ["test", "a"]
def collection2 = ["test", "b"]
def commons = collection1.intersect(collection2)
def difference = collection1.plus(collection2)
difference.removeAll(commons)
assert ["a", "b"] == difference

Answer 3

我认为OP要求两个列表之间的exclusive disjunction？

（注意：以上的解决方案都没有处理重复项！）

如果您想在Groovy中自己编写代码，请执行以下操作：

def a = ['a','b','c','c','c'] // diff is [b, c, c]
def b = ['a','d','c']         // diff is [d]

// for quick comparison
assert (a.sort() == b.sort()) == false

// to get the differences, remove the intersection from both
a.intersect(b).each{a.remove(it);b.remove(it)}
assert a == ['b','c','c']
assert b == ['d']
assert (a + b) == ['b','c','c','d']    // all diffs

一个问题是，正在使用整数列表/数组。您（可能）有问题，因为多态方法remove（int）vs remove（Object）。 See here for a (untested) solution

而不是重新发明轮子，您应该只使用现有的库（例如commons-collections）：

@Grab('commons-collections:commons-collections:3.2.1')

import static org.apache.commons.collections.CollectionUtils.*

def a = ['a','b','c','c','c'] // diff is [b, c, c]
def b = ['a','d','c']         // diff is [d]

assert disjunction(a, b) == ['b', 'c', 'c', 'd']

Answer 4

如果它是数字列表，则可以执行以下操作：

def before = [0, 0, 1, 0]
def after = [0, 1, 1, 0]
def difference =[]
for (def i=0; i<4; i++){
    difference<<after[i]-before[i]
}
println difference //[0, 1, 0, 0]

如何比较Groovy中的两个列表

4 个答案: