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34. While the answer from Niklas B. is pretty comprehensive, when we want to find an item in a list it is sometimes useful to get its index: next ( (i for i, x in enumerate (lst) if [condition on x]), [default value]) answered Dec 4, 2015 at 6:53. Vincent Cantin. 17.1k23658. You want to simplify it a bit.
2 de feb. de 2014 · 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. When items are appended or inserted, the array of references is resized. Some algorithm is applied to improve the performance of appending items repeatedly; when the array must be grown, some extra space is allocated so the next few times don’t require an actual resize i ...
20 de sept. de 2010 · The question does not make clear what exactly you want to achieve. List has the append method, which appends its argument to the list:
11 de nov. de 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?
I was wondering what the simplest way is to convert a string representation of a list like the following to a list: x = '[ "A","B","C" , " D"]' Even in cases
If playerOne and playerTwo are set/list/tuple of values, then compute their union and test if it's a subset of board: if {*playerOne,*playerTwo}.issubset(board): Also if the question is
joined_list = [item for list_ in [list_one, list_two] for item in list_] It has all the advantages of the newest approach of using Additional Unpacking Generalizations - i.e. you can concatenate an arbitrary number of different iterables (for example, lists, tuples, ranges, and generators) that way - and it's not limited to Python 3.5 or later.
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 ...
Try to understand this. Let's say that my_list is in the heap memory at location X, i.e., my_list is pointing to the X. Now by assigning new_list = my_list you're letting new_list point to the X. This is known as a shallow copy. Now if you assign new_list = my_list[:], you're simply copying each object of my_list to new_list. This is known as a ...
3 de dic. de 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]
