Am 22.08.2014 um 02:05 schrieb mattn:
It need to do like below with MapSort()
echo MapSort([{"foo":{"bar": "3"}}, {"foo":{"bar": "3 0"}}], '0 +
v:val["foo"]["bar"]')
If using my patch, it is possible easy to avoid this.
:echo sort(["3 0", "3 0 1"], "a:lhs < a:rhs")
I suppose your examples should read as follows (?):
:echo MapSort([{"foo":{"bar": "3"}}, {"foo":{"bar": "3 0"}}],
'v:val.foo.bar')
:echo sort([{"foo":{"bar": "3"}}, {"foo":{"bar": "3 0"}}], 'v:lhs.foo.bar
< v:rhs.foo.bar')
With MapSort(), you have to derive strings from you list items, as plain sort()
can only compare strings.
It's easy to turn positive integers into strings for comparison.
(It's hard to turn floats into strings for comparison.)
And MapSort() can't look local variable in the expression. For example, this
doesn't work.
---------------------------------------------------
function! s:foo()
let rank = {"queen": 12, "king": 13}
return MapSort(["queen", "king"], 'rank[v:val]')
endfunction
echo s:foo()
It can use local variables, with a little change to the func sig:
MapSort(list, selexpr, ...)
now you can place a:1, a:2, ... in your map expression:
function! s:foo()
let rank = {"queen": 9, "king": 10}
return MapSort(["queen", "king"], 'NumStr(a:1[v:val])', rank)
endfunction
echo s:foo()
(where NumStr() turns 9 into 'A9' and 10 into 'B10')
--
Andy
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