unwind ['医保疾病名称','症状','体征'] as lab
match (n) where any(label in labels(n) WHERE label=lab) with lab,keys(n) as kk
unwind kk as k
return lab,collect(distinct k)

相关文章:

  • 2021-07-27
  • 2022-12-23
  • 2021-11-15
  • 2021-09-30
  • 2022-12-23
  • 2022-12-23
  • 2021-08-10
  • 2021-10-16
猜你喜欢
  • 2021-10-30
  • 2021-12-18
  • 2021-12-27
  • 2021-10-14
  • 2022-12-23
  • 2022-12-23
相关资源
相似解决方案