julia - 如何从字典中删除键？

Question

我想从字典中删除一个键值对。

我现在正在创建一个新字典:

julia> dict = Dict(1 => "one", 2 => "two")
Dict{Int64,String} with 2 entries:
  2 => "two"
  1 => "one"

julia> dict = Dict(k => v for (k, v) in dict if k != 2)
Dict{Int64,String} with 1 entry:
  1 => "one"

但我想更新现有的字典。我怎样才能在 Julia 中做到这一点？

Answer 1

delete! 如果键存在，将从字典中删除键值对，如果键不存在则无效。它返回对字典的引用:

julia> dict = Dict(1 => "one", 2 => "two")
Dict{Int64,String} with 2 entries:
  2 => "two"
  1 => "one"

julia> delete!(dict, 1)
Dict{Int64,String} with 1 entry:
  2 => "two"

使用 pop! 如果您需要使用与键关联的值。但是如果key不存在就会报错:

julia> dict = Dict(1 => "one", 2 => "two");

julia> value = pop!(dict, 2)
"two"

julia> dict
Dict{Int64,String} with 1 entry:
  1 => "one"

julia> value = pop!(dict, 2)
ERROR: KeyError: key 2 not found

您可以避免使用 pop! 的三参数版本引发错误。 .第三个参数是在键不存在的情况下返回的默认值:

julia> dict = Dict(1 => "one", 2 => "two");

julia> value_or_default = pop!(dict, 2, nothing)
"two"

julia> dict
Dict{Int64,String} with 1 entry:
  1 => "one"

julia> value_or_default = pop!(dict, 2, nothing)

使用 filter! 根据一些谓词函数批量删除键值对:

julia> dict = Dict(1 => "one", 2 => "two", 3 => "three", 4 => "four");

julia> filter!(p -> iseven(p.first), dict)
Dict{Int64,String} with 2 entries:
  4 => "four"
  2 => "two"