Given this simplified example to sort:
l = [10, '0foo', 2.5, 'foo', 'bar']
I want to sort l so that numeric is always before strings. In this case, I'd like to get [2.5, 10, '0foo', 'foo', 'bar']. Is it possible make numeric and string temporarily comparable (with strings always larger than numeric)?
Note it is not easy to provide a key function to sorted if you are thinking about it. For example, converting numeric to string won't work because "10" < "2.5".
A way that you might do this does involve passing a key to sorted. it looks like this:
sorted(l, key=lambda x:(isinstance(x str), x))
This works because the key returns a tuple with the type of x and its value. Because of the way tuples are compared. Items at index 0 are compared first and if it they are the same, then next two items are compared and so on if they also are the same. This allows for the values to be sorted by type (string or not a string), then value if they are a similar type.
A more robust solution that also can handle further types might use a dictionary in the key function like this:
sorted(l,key=lambda x:({int:0, float:0, str:1, list:2, set:3}[type(x)], x))
further types can be added as necessary.
1 would be ordered larger than 2.5 with your solution, because <class int> is ordered after <class float>.
pandas.DataFrame(...).sort_values(...)? there is nokeyargument to this method.