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Another alternative: you can check if an item is in a list with if item in list:, but this is order O(n). If you are dealing with big lists of items and all you need to know is whether something is a member of your list, you can convert the list to a set first and take advantage of constant time set lookup:
Sep 20, 2010 · List slicing is quite flexible as it allows to replace a range of entries in a list with a ...
Oct 26, 2019 · Guest Dan Lawson. If you look at Peyton's blouse you can see a bit of a starter belly. Feel free to post more. She’s something else. Her gain was also sort of apparent during Mad Men, startet out the show looking like a twig and then actually seemed like a woman by season 4.
73. The best way is probably to use the list method .index. For the objects in the list, you can do something like: def __eq__(self, other): return self.Value == other.Value. with any special processing you need. You can also use a for/in statement with enumerate (arr) Example of finding the index of an item that has value > 100. for index ...
Dec 3, 2016 · A list of lists named xss can be flattened using a nested list comprehension: flat_list = [ x for xs in xss for x in xs ] The above is equivalent to: flat_list = [] for xs in xss: for x in xs: flat_list.append(x) Here is the corresponding function: def flatten(xss): return [x for xs in xss for x in xs]
Nov 11, 2009 · item_count = 0 for item in list: item_count += 1 return item_count count([1,2,3,4,5]) (The list object must be iterable, implied by the for..in stanza.) The lesson here for new programmers is: You can’t get the number of items in a list without counting them at some point. The question becomes: when is a good time to count them?
@loved.by.Jesus: Yeah, they added optimizations for Python level method calls in 3.7 that were extended to C extension method calls in 3.8 by PEP 590 that remove the overhead of creating a bound method each time you call a method, so the cost to call alist.copy() is now a dict lookup on the list type, then a relatively cheap no-arg function call that ultimately invokes the same thing as slicing.
Generally speaking: all and any are functions that take some iterable and return True, if. in the case of all, no values in the iterable are falsy;
Sep 17, 2012 · @Wouter, it will not. On the one hand, only lists of strings can be joined; so list.join would be inappropriate for an arbitrary list. On the other, the argument of str.join can be any "iterable" sequence of strings, not just a list.
l_x0 = [item for items in l_x for item in items.split(',') if not '[' in items] # Convert the nested lists to lists. l_x1 = [. i[1:-1].split(',') for i in l_x if '[' in i. ] # Add the two lists. l_x = l_x0 + l_x1. This last solution will work on any list stored as a string, nested or not.
