sprin的Python等价物

Question

有人知道如何将tis 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);
}

Answer 1

在一行中（Python 2.x）：

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

然后：

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

Answer 2

您想对Python3中的字符串使用format（）方法：

http://docs.python.org/library/string.html#formatstrings

或者查看Python 2.X的字符串插值文档

http://docs.python.org/library/stdtypes.html

Answer 3

对于Python2.x，有以下选项：

[最佳选择]较新的str.format和完整的format specification，例如

"I like {food}".format(food="chocolate")

较旧的interpolation formatting语法，例如

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

string.Template，例如

string.Template('I like $food').substitute(food="spinach")