求Lua 实现单词计数的代码
答案:1 悬赏:40 手机版
解决时间 2021-04-03 02:19
- 提问者网友：你给我的爱
- 2021-04-02 19:07
求Lua 实现单词计数的代码
最佳答案
- 五星知识达人网友：逃夭
- 2021-04-02 19:36
给你个全功能的wc （linux命令wc 即word count）
local BUFSIZE = 2^13 -- 8K
local f = io.input(arg[1]) -- open input file
local cc, lc, wc = 0, 0, 0 -- char, line, and word counts
for lines, rest in io.lines(arg[1], BUFSIZE, "*L") do
if rest then lines = lines .. rest end
cc = cc + #lines
-- count words in the chunk
local _, t = string.gsub(lines, "%S+", "")
wc = wc + t
-- count newlines in the chunk
_,t = string.gsub(lines, "\n", "\n")
lc = lc + t
end
print(lc, wc, cc) --打印行数，单词数，字符数
示例来自 《Programming in Lua third edition》
local BUFSIZE = 2^13 -- 8K
local f = io.input(arg[1]) -- open input file
local cc, lc, wc = 0, 0, 0 -- char, line, and word counts
for lines, rest in io.lines(arg[1], BUFSIZE, "*L") do
if rest then lines = lines .. rest end
cc = cc + #lines
-- count words in the chunk
local _, t = string.gsub(lines, "%S+", "")
wc = wc + t
-- count newlines in the chunk
_,t = string.gsub(lines, "\n", "\n")
lc = lc + t
end
print(lc, wc, cc) --打印行数，单词数，字符数
示例来自 《Programming in Lua third edition》
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