From collections import counter c = counte. The most popular solutions here generally only flatten one level of the nested list List[a:b:c], a is the starting index, b is the ending index and c is the optional step size
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This will give you a list starting at index a (inclusive) and ending at index b (exclusive) picking elements at a step of c
It gets all the elements from the list (or characters from a string) but the last element
Colon in the list index, you are asking for a slice, which is always another list In python you can assign values to both an individual item in a list, and to a slice of the list. Reorder list items on your computer, go to google keep Point to the item you want to move
At the left, click and hold move Drag the item where you want. 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 The second, list(), is using the actual list type constructor to create a new list which has contents equal to the first list.
The first way works for a list or a string
The second way only works for a list, because slice assignment isn't allowed for strings Other than that i think the only difference is speed It looks like it's a little faster the first way Try it yourself with timeit.timeit () or preferably timeit.repeat ().
To_list () is a method of pandas dataframes that returns a list representation of the dataframe Although both methods return the same output, their differences lie in their origins and compatibility If your list of lists comes from a nested list comprehension, the problem can be solved more simply/directly by fixing the comprehension Please see how can i get a flat result from a list comprehension instead of a nested list?
