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You might run into minor issues The second, list(), is using the actual list type constructor to create a new list which has contents equal to the first list. I have a piece of code here that is supposed to return the least common element in a list of elements, ordered by commonality

From collections import counter c = counte.

Since a list comprehension creates a list, it shouldn't be used if creating a list is not the goal So refrain from writing [print(x) for x in range(5)] for example. Given a dataframe, i want to groupby the first column and get second column as lists in rows, so that a dataframe like A b a 1 a 2 b 5 b 5 b 4 c 6 becomes a [1,2] b [5,5,4] c [6] how do i do this?

The reason you're getting the unhashable type 'list' exception is because k = list[0:j] sets k to be a slice of the list, which is logically another, often shorter, list. I want to get a list of the column headers from a pandas dataframe The dataframe will come from user input, so i won't know how many columns there will be or what they will be called

Don't use quotes on the command line 1 don't use type=list, as it will return a list of lists this happens because under the hood argparse uses the value of type to coerce each individual given argument you your chosen type, not the aggregate of all arguments

You can use type=int (or whatever) to get a list of ints (or whatever) I noticed in effective stl that vector is the type of sequence that should be used by default It seems that ignore the efficiency vector can do anything The first, [:], is creating a slice (normally often used for getting just part of a list), which happens to contain the entire list, and thus is effectively a copy of the list