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I have a piece of code here that is supposed to return the least common element in a list of elements, ordered by commonality A list uses an internal array to handle its data, and automatically resizes the array when adding more elements to the list than its current capacity, which makes it more easy to use than an array, where you need to know the capacity beforehand. From collections import counter c = counte.
The first way works for a list or a string When items are appended or inserted, the array of references is resized. 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 (). Note that the question was about pandas tolist vs to_list Pandas.dataframe.values returns a numpy array and numpy indeed has only tolist
Indeed, if you read the discussion about the issue linked in the accepted answer, numpy's tolink is the reason why pandas used tolink and why they did not deprecate it after introducing to_list. If it was public and someone cast it to list again, where was the difference 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?
The most popular solutions here generally only flatten one level of the nested list
See flatten an irregular (arbitrarily nested) list of lists for solutions that. 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. In c# if i have a list of type bool
What is the fastest way to determine if the list contains a true value I don’t need to know how many or where the true value is I just need to know if one e. Do you want to simply append, or do you want to merge the two lists in sorted order
What output do you expect for [1,3,6] and [2,4,5]
Can we assume both sublists are already sorted (as in your example)? The implementation uses a contiguous array of references to other objects, and keeps a pointer to this array This makes indexing a list a [i] an operation whose cost is independent of the size of the list or the value of the index
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