How do I get the number of keys in a hash table in Lua?

问题

myTable = {}
myTable["foo"] = 12
myTable["bar"] = "blah"
print(#myTable) -- this prints 0

Do I actually have to iterate through the items in the table to get the number of keys?

numItems = 0
for k,v in pairs(myTable) do
    numItems = numItems + 1
end
print(numItems) -- this prints 2

回答1:

I experimented with both the # operator and table.getn(). I thought table.getn() would do what you wanted but as it turns out it's returning the same value as #, namely 0. It appears that dictionaries insert nil placeholders as necessary.

Looping over the keys and counting them seems like the only way to get the dictionary size.

回答2:

The length operator:

The length of a table t is defined to be any integer index n such that t[n] is not nil and t[n+1] is nil; moreover, if t[1] is nil, n can be zero. For a regular array, with non-nil values from 1 to a given n, its length is exactly that n, the index of its last value. If the array has "holes" (that is, nil values between other non-nil values), then #t can be any of the indices that directly precedes a nil value (that is, it may consider any such nil value as the end of the array).

so only way to get length is iterate over it.

回答3:

Aside from iterating through the keys manually, it is simple to automatically keep track of it via metamethods. Considering you probably don't want to keep track of every table you make, you can just write a function that will allow you to convert any table into a key-countable object. The following isn't perfect, but I think it would illustrate the point:

function CountedTable(x)
  assert(type(x) == 'table', 'bad parameter #1: must be table')

  local mt = {}
  -- `keys`  will represent the number of non integral indexes
  -- `indxs` will represent the number of integral indexes
  -- `all`   will represent the number of both 
  local keys, indxs, all = 0, 0, 0

  -- Do an initial count of current assets in table. 
  for k, v in pairs(x) do
    if (type(k) == 'number') and (k == math.floor(k)) then indxs = indxs + 1
    else keys = keys + 1 end

    all = all + 1
  end

  -- By using `__nexindex`, any time a new key is added, it will automatically be
  -- tracked.
  mt.__newindex = function(t, k, v)
    if (type(k) == 'number') and (k == math.floor(k)) then indxs = indxs + 1
    else keys = keys + 1 end

    all = all + 1
    t[k] = v
  end

  -- This allows us to have fields to access these datacounts, but won't count as
  -- actual keys or indexes.
  mt.__index = function(t, k)
    if k == 'keyCount' then return keys 
    elseif k == 'indexCount' then return indxs 
    elseif k == 'totalCount' then return all end
  end

  return setmetatable(x, mt)
end

Examples of using this would include:

-- Note `36.35433` would NOT be counted as an integral index.
local foo = CountedTable { 1, 2, 3, 4, [36.35433] = 36.35433, [54] = 54 }
local bar = CountedTable { x = 23, y = 43, z = 334, [true] = true }
local foobar = CountedTable { 1, 2, 3, x = 'x', [true] = true, [64] = 64 }

print(foo.indexCount)    --> 5
print(bar.keyCount)      --> 4
print(foobar.totalCount) --> 6

Live Working Example

Hope this helped! :)

回答4:

Lua stores table as two separates parts : a hash part and an array part, the len operator only deal with the array part, meaning value indexed by a number value, plus using rules mentioned below, so you don't have any choice for counting "hash" value you need to iterate over the table with pairs() function.

来源：https://stackoverflow.com/questions/652957/how-do-i-get-the-number-of-keys-in-a-hash-table-in-lua

标签

lua

hashtable