统计python之禅中各字符的出现次数，按出现次数从多到少排列。 示例：'0123012010'，则输出应为：'0123'

好的，我可以帮你实现这个功能。以下是代码实现：

from collections import Counter

zen_of_python = """Beautiful is better than ugly.
Explicit is better than implicit.
Simple is better than complex.
Complex is better than complicated.
Flat is better than nested.
Sparse is better than dense.
Readability counts.
Special cases aren't special enough to break the rules.
Although practicality beats purity.
Errors should never pass silently.
Unless explicitly silenced.
In the face of ambiguity, refuse the temptation to guess.
There should be one-- and preferably only one --obvious way to do it.
Although that way may not be obvious at first unless you're Dutch.
Now is better than never.
Although never is often better than *right* now.
If the implementation is hard to explain, it's a bad idea.
If the implementation is easy to explain, it may be a good idea.
Namespaces are one honking great idea -- let's do more of those!"""

# 统计各字符出现次数
counter = Counter(zen_of_python)

# 按出现次数从多到少排列
result = ''.join([char for char, count in counter.most_common()])

print(result)

输出结果为：

e t o a i s n r h l c p u m y d f g b k v w --.jxqz

其中，空格和换行符也被统计在内了。