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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 Asked 16 years, 5 months ago modified 2 years, 10 months ago viewed 286k times From collections import counter c = counte.
The first way works for a list or a string In python, how do i index a list with another list 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 (). 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.
List_of_values doesn't have to be a list It can be set, tuple, dictionary, numpy array, pandas series, generator, range etc And isin() and query() will still work. 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
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.
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.
List is an interface, you cannot instantiate an interface, because interface is a convention, what methods should have your classes
In order to instantiate, you need some realizations (implementations) of that interface. 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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