Question

我的原始列表是：

[[string: firstString, id: 2], 
[string: secondString, id: 1], 
[string: secondString, id: 3], 
[string: firstString, id: 1], 
[string: firstString, id: 3], 
[string: secondString, id: 2]]

我希望按字符串对其进行分组，然后按ID对其进行排序：

[[string: firstString, id: 1], 
[string: firstString, id: 2], 
[string: firstString, id: 3], 
[string: secondString, id: 1], 
[string: secondString, id: 2], 
[string: secondString, id: 3]]

Answer 1

def firstString = 'first'

def secondString = 'second'

def source =
[[string: firstString, id: 2], 
[string: secondString, id: 1], 
[string: secondString, id: 3], 
[string: firstString, id: 1], 
[string: firstString, id: 3], 
[string: secondString, id: 2]]

您声明的预期结果：

def expected =
[[string: firstString, id: 1], 
[string: firstString, id: 2], 
[string: firstString, id: 3], 
[string: secondString, id: 1], 
[string: secondString, id: 2], 
[string: secondString, id: 3]]

def result = source.sort { [it.string, it.id] }
assert result == expected

ID按顺序排序：

def expected =
[[string: firstString, id: 3], 
[string: firstString, id: 2], 
[string: firstString, id: 1], 
[string: secondString, id: 3], 
[string: secondString, id: 2], 
[string: secondString, id: 1]]

def result = source.sort { [it.string, -it.id] }
assert result == expected

真正分组，不仅按字符串排序：

def expected =
[(firstString):
  [[string: firstString, id: 1], 
  [string: firstString, id: 2], 
  [string: firstString, id: 3]], 
(secondString):
  [[string: secondString, id: 1], 
  [string: secondString, id: 2], 
  [string: secondString, id: 3]]]

def result = source.groupBy { it.string }
result.each { it.value.sort { it.id } }
assert result == expected

真正分组，不仅按字符串排序，ID反向排序：

def expected =
[(firstString):
  [[string: firstString, id: 3], 
  [string: firstString, id: 2], 
  [string: firstString, id: 1]], 
(secondString):
  [[string: secondString, id: 3], 
  [string: secondString, id: 2], 
  [string: secondString, id: 1]]]

def result = source.groupBy { it.string }
result.each { it.value = it.value.sort { it.id }.reverse() }
assert result == expected

真正分组并且ID已经变平：

def expected = 
[(firstString): [1, 2, 3], 
(secondString): [1, 2, 3]]

def result = source.groupBy { it.string }
result.each { it.value = it.value.collect { it.id }.sort() }
assert result == expected

真正分组，ID变平，ID反向排序：

def expected = 
[(firstString): [3, 2, 1], 
(secondString): [3, 2, 1]]

def result = source.groupBy { it.string }
result.each { it.value = it.value.collect { it.id }.sort().reverse() }
assert result == expected

Groovy - 组和排序列表

1 个答案: