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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 For example i have a list as follows and i want to iterate over a,b and c. From collections import counter c = counte.
The first way works for a list or a string Is the a short syntax for joining a list of lists into a single list ( or iterator) in python 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. 1538 first declare your list properly, separated by commas
You can get the unique values by converting the list to a set. 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.
The second action taken was to revert the accepted answer to its state before it was partway modified to address determine if all elements in one list are in a second list.
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