Another way to solve this problem is by using function numpy.amax() >>> import numpy as np >>> arr = [0, 0, 1, 0, 0, 1] , [0, 1, 0, 2, 0, 0] , [0, 0, 2, 0, 0, 1] , [0, 1, 0, 3, 0, 0] , [0, 0, 0, 0, 4, 0] >>> np.amax(arr)
>>> L = [[1, 2, 3], [1, 1, 1]] >>> sum(x.count(1) for x in L) 4
playersList.get(0) Java has limited operator polymorphism. So you use the get() method on List objects, not the array index operator ([])
EDIT As of groovy 1.8.6 you can use the collate method on lists def origList = [1, 2, 3, 4, 5, 6, 7, 8, 9] assert [[1, 2, 3, 4], [5, 6, 7, 8], [9]] == origList.collate(4) Another method using inject and metaClasses List.metaClass.partition = { size -> def rslt = delegate.inject( [ [] ] … Read more
I think the reason is simple. lists are ordered, dicts (prior to Python 3.6/3.7) and sets are not. So modifying a lists as you iterate may be not advised as best practise, but it leads to consistent, reproducible, and guaranteed behaviour. You could use this, for example let’s say you wanted to split a list … Read more
In summary: getting an item from a NumPy array requires new Python objects to be created, whereas this is not the case for lists. Also, indexing is more slightly more complicated for NumPy arrays than lists which may add some additional overhead. To recap, the NumPy operations you have listed do the following: b[100][100] returns … Read more
You can create the enumeration of the elements by something like this: mylist = list(xrange(10)) Then you can use the random.choice function to select your items: import random … random.choice(mylist) As Asim Ihsan correctly stated, my answer did not address the full problem of the OP. To remove the values from the list, simply list.remove() … Read more
Using list comprehension: >>> a = [] >>> a.append((1,2,4)) >>> a.append((2,3,4)) >>> [x for xs in a for x in xs] [1, 2, 4, 2, 3, 4] Using itertools.chain.from_iterable: >>> import itertools >>> list(itertools.chain.from_iterable(a)) [1, 2, 4, 2, 3, 4]
Use frozenset, >>> set([frozenset([1,2]), frozenset([2,3])]) set([frozenset([1, 2]), frozenset([2, 3])]) To represent a set of sets, the inner sets must be frozenset objects for the reason that the elements of a set must be hashable (all of Python’s immutable built-in objects are hashable). frozenset is immutable and set is mutable.