sprintf的Python等价物

有人知道如何将PHP函数移植到python吗

/**
 * converts id (media id) to the corresponding folder in the data-storage
 * eg: default mp3 file with id 120105 is stored in
 * /(storage root)/12/105/default.mp3
 * if absolute paths are needed give path for $base
 */

public static function id_to_location($id, $base = FALSE)
{
    $idl = sprintf("%012s",$id);
    return $base . (int)substr ($idl,0,4) . '/'. (int)substr($idl,4,4) . '/' . (int)substr ($idl,8,4);
}

您希望对Python 3中的字符串使用format（）方法：

或者查看Python 2.X的字符串插值文档

---

对于python 2.x，您有以下选项：

[最佳选项]更新的和完整的，例如

较旧的语法，例如

"I like %s" % "berries"
"I like %(food)s" % {"food": "cheese"}

，例如

---

好的-找到了一个方法-我认为不是很好，但工作

def id_to_location(id):
    l = "%012d" % id
    return '/%d/%d/%d/' % (int(l[0:4]), int(l[4:8]), int(l[8:12]))

---

可以使用默认参数引入基准。也许你想要这样：

def id_to_location(id,base=""):
   l = "%012d" % id
   return '%s/%d/%d/%d/' % (base,int(l[0:4]), int(l[4:8]), int(l[8:12]))

在一行中，（Python 2.x）：

然后：

def id_to_location(id):
    l = "%012d" % id
    return '/%d/%d/%d/' % (int(l[0:4]), int(l[4:8]), int(l[8:12]))

def id_to_location(id,base=""):
   l = "%012d" % id
   return '%s/%d/%d/%d/' % (base,int(l[0:4]), int(l[4:8]), int(l[8:12]))

id_to_location = lambda i: '/%d/%d/%d/' % (int(i)/1e8, int(i)%1e8/1e4, int(i)%1e4)

print id_to_location('001200230004')
'/12/23/4/'