Pie*_*gen 1 python generator go
我正在学习围棋的语言,并试图重写一些使用golang我的Python代码.我写了一个生成器函数,它逐行读取文本文件并发送(使用yield关键字)只有"有效"行(忽略空白行,重新构造未完成的行).
示例文件(myfile.txt):
#123= FOOBAR(1.,'text');
#126= BARBAZ('poeazpfodsp',
234,56);
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parse.py:
#!/usr/bin/python
def validlines(filename):
with open(filename) as fdin:
buff = ''
for line in fdin.readlines():
line = line.strip()
if line == '':
continue
buff += line
if line[-1] != ';':
continue
yield buff
buff = ''
fdin.close()
for line in validlines('myfile.txt'):
print(line)
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显示:
#123= FOOBAR(1.,'text');
#126= BARBAZ('poeazpfodsp',234,56);
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现在,我尝试使用golang中的闭包以相同的方式执行此操作:
parse.go:
package main
import (
"bufio"
"fmt"
"os"
"strings"
)
func validLines(filename string) (func() (string, bool)) {
file, _ := os.Open(filename)
scanner := bufio.NewScanner(file)
return func() (string, bool) {
buff := ""
for scanner.Scan() {
line := scanner.Text()
line = strings.TrimSpace(line)
if line == "" {
continue
}
buff += line
if line[len(line)-1] != ';' {
continue
}
return buff, true
}
file.Close()
return "", false
}
}
func main() {
vline := validLines("myfile.txt")
for line, ok := vline(); ok; {
fmt.Println(line)
}
}
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显示:
#123= FOOBAR(1.,'text');
#123= FOOBAR(1.,'text');
#123= FOOBAR(1.,'text');
#123= FOOBAR(1.,'text');
#123= FOOBAR(1.,'text');
#123= FOOBAR(1.,'text');
#123= FOOBAR(1.,'text');
...
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在golang中这样做的正确方法是什么?
小智 6
在Go中你可以使用渠道而不是收益,这非常方便.
包主
import (
"bufio"
"fmt"
"os"
"strings"
)
func ValidLines(filename string) (c chan string) {
c = make(chan string)
buff := ""
go func() {
file, err := os.Open(filename)
if err != nil {
close(c)
return
}
reader := bufio.NewReader(file)
for {
line, err := reader.ReadString('\n')
if err != nil {
close(c)
return
}
line = strings.TrimSpace(line)
if line == "" {
continue
}
buff += line
if line[len(line)-1] != ';' {
continue
}
c <- buff
buff = ""
}
}()
return c
}
func main() {
for line := range ValidLines("myfile.txt") {
fmt.Println(line)
}
}
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