我想将现有的python函数转换为lua函数。但我的lua函数并不像python函数那样产生相同的结果。任何帮助表示赞赏。将Python函数转换为Lua函数
Python函数:
import json
test = '{"http://localhost:8080/":{"phone":{"-detail/phone detail.template.html":"5167n,a,7,2","s/motorola-xoom-with-wifi.json":"516a0,5,4,3"},"favicon.ico":"016ad,3,3,2","img/phones/motorola-xoom-with-wi-fi.":{"1.jpg":"*02s,2s,4v,h3|116da,o,l,6","2.jpg":"*02s,2s,4v,kp|116da,j,i,8","3.jpg":"*02s,2s,4v,ob|116da,o,m,8,7,,7,7,7","4.jpg":"*02s,2s,4v,rx|116da,o,m,9,8,,7,7,7","5.jpg":"*02s,2s,4v,vj|116da,p,m,a,8,,7,7,7"}}}'
def tri(param):
t = {}
for key in param:
if key not in param:
continue
if isinstance(param[key], dict) and param[key] is not None:
flat = tri(param[key])
for x in flat:
if x not in flat:
continue
t[key + x] = flat[x]
else:
t[key] = param[key]
return t
print(tri(json.loads(test)))
Lua代码(其不产生相同的结果蟒功能)
local json = require('cjson')
local test = '{"http://localhost:8080/":{"phone":{"-detail/phone-detail.template.html":"5167n,a,7,2","s/motorola-xoom-with-wi-fi.json":"516a0,5,4,3"},"favicon.ico":"016ad,3,3,2","img/phones/motorola-xoom-with-wi-fi.":{"1.jpg":"*02s,2s,4v,h3|116da,o,l,6","2.jpg":"*02s,2s,4v,kp|116da,j,i,8","3.jpg":"*02s,2s,4v,ob|116da,o,m,8,7,,7,7,7","4.jpg":"*02s,2s,4v,rx|116da,o,m,9,8,,7,7,7","5.jpg":"*02s,2s,4v,vj|116da,p,m,a,8,,7,7,7"}}}'
local function tri(param)
t = {}
for key in pairs(param) do
if param[key] == nil then end
if type(param[key]) == "table" then
flat = tri(param[key])
for k in pairs(flat) do
t[key .. k] = flat[k]
end
else
t[key] = param[key]
end
end
return t
end
print(json.encode(tri(json.decode(test))))
如果你告诉人们该函数应该做什么以及它做什么,变量应该是可能的。只要让你自己清楚你的表t发生了什么,因为每当你调用tri()时它就是全局的。 – Piglet