my_list = [1,2,3,4,5]
len(my_list)
# 5
The same works for tuples:
my_tuple = (1,2,3,4,5)
len(my_tuple)
# 5
And strings, which are really just arrays of characters:
my_string = 'hello world'
len(my_string)
# 11
It was intentionally done this way so that lists, tuples and other container types or iterables didn't all need to explicitly implement a public .length() method, instead you can just check the len() of anything that implements the 'magic' __len__() method.
Sure, this may seem redundant, but length checking implementations can vary considerably, even within the same language. It's not uncommon to see one collection type use a .length() method while another type uses a .length property, while yet another uses .count(). Having a language-level keyword unifies the entry point for all these types. So even objects you may not consider to be lists of elements could still be length-checked. This includes strings, queues, trees, etc.
The functional nature of len() also lends itself well to functional styles of programming.
lengths = map(len, list_of_containers)
Is arr.__len__() the preferred way to get the length of an array in Python? - Stack Overflow
Why is getting the length of an array O(1) and not O(n)?
How do I find the length (or dimensions, size) of a numpy matrix in python? - Stack Overflow
Array length
my_list = [1,2,3,4,5]
len(my_list)
# 5
The same works for tuples:
my_tuple = (1,2,3,4,5)
len(my_tuple)
# 5
And strings, which are really just arrays of characters:
my_string = 'hello world'
len(my_string)
# 11
It was intentionally done this way so that lists, tuples and other container types or iterables didn't all need to explicitly implement a public .length() method, instead you can just check the len() of anything that implements the 'magic' __len__() method.
Sure, this may seem redundant, but length checking implementations can vary considerably, even within the same language. It's not uncommon to see one collection type use a .length() method while another type uses a .length property, while yet another uses .count(). Having a language-level keyword unifies the entry point for all these types. So even objects you may not consider to be lists of elements could still be length-checked. This includes strings, queues, trees, etc.
The functional nature of len() also lends itself well to functional styles of programming.
lengths = map(len, list_of_containers)
The way you take a length of anything for which that makes sense (a list, dictionary, tuple, string, ...) is to call len on it.
l = [1,2,3,4]
s = 'abcde'
len(l) #returns 4
len(s) #returns 5
The reason for the "strange" syntax is that internally python translates len(object) into object.__len__(). This applies to any object. So, if you are defining some class and it makes sense for it to have a length, just define a __len__() method on it and then one can call len on those instances.
I have read that this operation works in constant time. But how is that possible?
Given an array of length n, I would have thought that computing n would require accessing every element. In pseudocode, something like the following:
n = 0
for each item in array:
n = n + 1
return nBut that would clearly be O(n), suggesting that’s not how it actually works.
Can anyone shed light on this?
shape is a property of both numpy ndarray's and matrices.
A.shape
will return a tuple (m, n), where m is the number of rows, and n is the number of columns.
In fact, the numpy matrix object is built on top of the ndarray object, one of numpy's two fundamental objects (along with a universal function object), so it inherits from ndarray
matrix.size according to the numpy docs returns the Number of elements in the array. Hope that helps.
A non-empty array A consisting of N integers is given.
Array A represents a linked list. A list is constructed from this array as follows:
the first node (the head) is located at index 0;
the value of a node located at index K is A[K];
if the value of a node is −1 then it is the last node of the list;
otherwise, the successor of a node located at index K is located at index A[K] (you can assume that A[K] is a valid index, that is 0 ≤ A[K] < N).
For example, for array A such that:
A[0] = 1 A[1] = 4 A[2] = -1 A[3] = 3 A[4] = 2
📷
the following list is constructed:
the first node (the head) is located at index 0 and has a value of 1;
the second node is located at index 1 and has a value of 4;
the third node is located at index 4 and has a value of 2;
the fourth node is located at index 2 and has a value of −1.
Write a function:
def solution(A)
that, given a non-empty array A consisting of N integers, returns the length of the list constructed from A in the above manner.
For example, given array A such that:
A[0] = 1 A[1] = 4 A[2] = -1 A[3] = 3 A[4] = 2
the function should return 4, as explained in the example above.
Assume that:
N is an integer within the range [1..200,000];
each element of array A is an integer within the range [−1..N-1];
it will always be possible to construct the list and its length will be finite.
I read this question and still don't understand why the order of list in that way. Would you please explain? Could not post a photo here to make it easier to understand. Thank you so much!