如果hash ['a']不存在,如何分配hash ['a'] ['b'] ='c'? [英] How to assign hash['a']['b']= 'c' if hash['a'] doesn't exist?
问题描述
有没有比
简单的方法 如果hash.key?('a')
hash [' a'] ['b'] ='c'
else
hash ['a'] = {} $ b $ hash ['a'] ['b'] ='c'
end
最简单的方法是用块参数构建你的Hash :
hash = Hash.new {| h,k | h [k] = {}}
hash ['a'] ['b'] = 1
hash ['a'] ['c'] = 1
hash ['b '] ['c'] = 1
puts hash.inspect
#{a=> {b=> 1,c=> 1},b => {c=> 1}}
$ c> new 创建一个新的空Hash作为默认值。您不希望这样:
hash = Hash.new({})
,因为它将为所有默认条目使用完全相同哈希。
$ b $另外,正如Phrogz所指出的那样,你可以使用
default_proc
: hash = Hash.new {| h,k | h [k] = Hash.new(& h.default_proc)}
UPDATE :我想我应该对 Hash.new({})
进行澄清。当你这样说:
h = Hash.new({})
这就好像这样说:
h = Hash.new
h.default = {}
然后,当您访问 h
将 h [:k] [:m] = y
赋值,它的行为就像您这样做:
if(h.has_key?(:k))
h [:k] [:m] = y
else
h.default [:m] = y
end
然后,如果你 h [:k2] [:n] = z
,你最终将分配 h.default [:n] = ž
。请注意, h
仍然表示 h.has_key?(:k)
为false。
然而,当你这样说的时候:
h = Hash.new(0)
一切都会好起来的,因为你永远不会修改 h [k]
在这里,你只能从 h
(它将使用默认值,如果需要的话)读取一个值,或者给赋值一个新值。 h
。
Is there any way simpler than
if hash.key?('a')
hash['a']['b'] = 'c'
else
hash['a'] = {}
hash['a']['b'] = 'c'
end
The easiest way is to construct your Hash with a block argument:
hash = Hash.new { |h, k| h[k] = { } }
hash['a']['b'] = 1
hash['a']['c'] = 1
hash['b']['c'] = 1
puts hash.inspect
# "{"a"=>{"b"=>1, "c"=>1}, "b"=>{"c"=>1}}"
This form for new
creates a new empty Hash as the default value. You don't want this:
hash = Hash.new({ })
as that will use the exact same hash for all default entries.
Also, as Phrogz notes, you can make the auto-vivified hashes auto-vivify using default_proc
:
hash = Hash.new { |h, k| h[k] = Hash.new(&h.default_proc) }
UPDATE: I think I should clarify my warning against Hash.new({ })
. When you say this:
h = Hash.new({ })
That's pretty much like saying this:
h = Hash.new
h.default = { }
And then, when you access h
to assign something as h[:k][:m] = y
, it behaves as though you did this:
if(h.has_key?(:k))
h[:k][:m] = y
else
h.default[:m] = y
end
And then, if you h[:k2][:n] = z
, you'll end up assigning h.default[:n] = z
. Note that h
still says that h.has_key?(:k)
is false.
However, when you say this:
h = Hash.new(0)
Everything will work out okay because you will never modified h[k]
in place here, you'll only read a value from h
(which will use the default if necessary) or assign a new value to h
.
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