替代python替换

我想知道下面的代码是否有替代方案，因为，1：有一个错误，2：我不想经常这样做：

restring=string.replace("a","01").replace("b","02")...

要替换的代码：

chars=["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","t","s","u","v","w","x","y","z"]
numbs=["01","02","03","04","05","06","07","08","09","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26"]
string="test";
restring=string.replace(chars,numbs);
print restring;

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在这里，由于您只有1个字符的替换“键”，因此我将使用dict和

连接：

>>> chars=["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","t","s","u","v","w","x","y","z"]
>>> numbs=["01","02","03","04","05","06","07","08","09","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26"]
>>> d = dict(zip(chars,numbs))
>>> my_sentence = "foobar"
>>> ''.join(d.get(c,c) for c in my_sentence)
'061515020118'

这不会扩展到多个字符替换键，不过…
至少使用一个循环。第一位可以压缩为
{v:'{:02d}。enumerate（string.ascii_lowercase，1）}
中的k，v格式（k）或避免硬编码长列表。@DSM+1，把它放在答案中@DSM——是的，我已经考虑过了。但我不想把事情和OP最初的想法扯得太远