Get ALL items in a python list?

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it's very easy to get the number of items in a list, len(list), but say I had a matrix like: [[1,2,3],[1,2,3]] Is there a pythonic way to return 6? Or do I have to iterate.

You can use chain

from itertools import chain
l = [[1,2,3],[1,2,3]]
len(list(chain(*l))) # give you 6

the expression list(chain(*l)) give you flat list: [1, 2, 3, 1, 2, 3]

Python, Accessing an element from its index is easier task in python, just using the Iterating over the index list to get the corresponding elements from list into new list  Varun September 16, 2019 Python: How to find all indexes of an item in a List? In this article we will discuss different ways to find indexes of all occurrences of an item in the list. Also a way to find indexes of items in a list that satisfy a certain condition. To find indexes of an item in a list, we are going to use a function i.e.

Make the matrix numpy array like this

mat = np.array([[1,2,3],[1,2,3]])

Make the array 1D like this

arr = mat.ravel()

Print length

print(len(arr))

Python Lists and List Manipulation, Python - Lists - The most basic data structure in Python is the sequence. Each element of a sequence is assigned a number - its position or index. The first index is in this tutorial. There are certain things you can do with all sequence types. To get items or elements from the Python list, you can use list index number. Remember that Python lists index always starts from 0. So, the first item of list is indexed at 0.

l = [[1,2,3],[1,2,3]]    
len([item for innerlist in l for item in innerlist])

gives you 6

Python - Lists, Python's list data type provides this method to find the first index of a given element in list or a sub list i.e.. You can also get a sub-list or a range of items from the list by providing the range of index. In the following example, we will create a Python list and then access a range of items from 2nd item until 5th item, three items in total excluding the last index.

Python: How to find all indexes of an item in a List? – thispointer.com, And Python never creates a new list if you assign a list to a variable. seq = L[​start:stop:step] seq = L[::2] # get every other item, starting with the first seq = L[1::2​]  The list defined above has items that are all of the same type (int), but all the items of a list do not need to be of the same type as you can see below. # Define a list heterogenousElements = [3, True, 'Michael', 2.0] The list contains an int, a bool, a string, and a float. Access Values in a List. Each item in a list has an assigned index value.

An Introduction to Python Lists, In Python programming, a list is created by placing all the items (elements) inside square When we run the above program, we will get the following output: e p Everything in Python is an object, including lists. All objects have a header of some sort in the C implementation. Lists and other similar builtin objects with a "size" in Python, in particular, have an attribute called ob_size, where the number of elements in the object is cached. So checking the number of objects in a list is very fast.

Python List (With Examples), How To Get The Last Element Of A List In Your List. The answer to this question is an addition to  The items() method returns a view object. The view object contains the key-value pairs of the dictionary, as tuples in a list. The view object will reflect any changes done to the dictionary, see example below.

Comments
  • use: l = [[1,2,3],[1,2,3]] sum(map(len, l)) to get the 6, but for you an item is a number, but an item for a list is just an item that can be another list, number, floating, object, etc.
  • The operation you're looking for is "flatten" the list. Then take the length of the flattened list
  • Probably iteration is the most readable solution IMO
  • stackoverflow.com/questions/2158395/…
  • Whenever you're doing anything fancy with list, the answer is always "just use numpy".
  • Would this work if the Matrix was 3 dimensional as well ex [[[1,2,3]]]
  • @eyllanesc thanks for the edit. I am now from a mobile device. I was unable to format the code from here.