即使os.path.walk完全没有时间，您在 pf 中的方法也会非常缓慢。在所有现有路径中进行包含 3 个无界闭包的正则表达式匹配会在此处杀死您。这是我引用的Kernighan 和 Pike的代码，这是该任务的正确算法：
/* spname: return correctly spelled filename */
/*
* spname(oldname, newname) char *oldname, *newname;
* returns -1 if no reasonable match to oldname,
* 0 if exact match,
* 1 if corrected.
* stores corrected name in newname.
*/
#include <sys/types.h>
#include <sys/dir.h>
spname(oldname, newname)
char *oldname, *newname;
{
char *p, guess[DIRSIZ+1], best[DIRSIZ+1];
char *new = newname, *old = oldname;
for (;;) {
while (*old == '/') /* skip slashes */
*new++ = *old++;
*new = '\0';
if (*old == '\0') /* exact or corrected */
return strcmp(oldname,newname) != 0;
p = guess; /* copy next component into guess */
for ( ; *old != '/' && *old != '\0'; old++)
if (p < guess+DIRSIZ)
*p++ = *old;
*p = '\0';
if (mindist(newname, guess, best) >= 3)
return -1; /* hopeless */
for (p = best; *new = *p++; ) /* add to end */
new++; /* of newname */
}
}
mindist(dir, guess, best) /* search dir for guess */
char *dir, *guess, *best;
{
/* set best, return distance 0..3 */
int d, nd, fd;
struct {
ino_t ino;
char name[DIRSIZ+1]; /* 1 more than in dir.h */
} nbuf;
nbuf.name[DIRSIZ] = '\0'; /* +1 for terminal '\0' */
if (dir[0] == '\0') /* current directory */
dir = ".";
d = 3; /* minimum distance */
if ((fd=open(dir, 0)) == -1)
return d;
while (read(fd,(char *) &nbuf,sizeof(struct direct)) > 0)
if (nbuf.ino) {
nd = spdist(nbuf.name, guess);
if (nd <= d && nd != 3) {
strcpy(best, nbuf.name);
d = nd;
if (d == 0) /* exact match */
break;
}
}
close(fd);
return d;
}
/* spdist: return distance between two names */
/*
* very rough spelling metric:
* 0 if the strings are identical
* 1 if two chars are transposed
* 2 if one char wrong, added or deleted
* 3 otherwise
*/
#define EQ(s,t) (strcmp(s,t) == 0)
spdist(s, t)
char *s, *t;
{
while (*s++ == *t)
if (*t++ == '\0')
return 0; /* exact match */
if (*--s) {
if (*t) {
if (s[1] && t[1] && *s == t[1]
&& *t == s[1] && EQ(s+2, t+2))
return 1; /* transposition */
if (EQ(s+1, t+1))
return 2; /* 1 char mismatch */
}
if (EQ(s+1, t))
return 2; /* extra character */
}
if (*t && EQ(s, t+1))
return 2; /* missing character */
return 3;
}
注意：此代码是在 ANSI C、ISO C 或 POSIX 之前编写的，当人们读取原始目录文件时，甚至可以想象任何事情。代码的方法比所有的指针吊起来有用得多。