WebApr 4, 2010 · a[4:] creates a list of elements, from (including) index 4. This results in an empty list. [1, 2, 3][999:] == [] It's how the slicing operation works. If a was ['123', '2', 4, 5, 6], it would return [5, 6] .He then adds the element 'sss', creating a new list, which is ['sss'] in this case. The final step [0] just takes the first element, thus returning 'sss', but would … WebAug 4, 2016 · Here is a simple way to get: A list of indexes one or more if there are more than one, or an empty list if there are none: list = [ {'id':'1234','name':'Jason'}, {'id':'2345','name':'Tom'}, {'id':'3456','name':'Art'}] [i for i, d in enumerate (list) if 'Tom' in d.values ()] Output: >>> [1]
algorithm - Complexity of list.index(x) in Python - Stack Overflow
WebJul 18, 2012 · Previous solution. a.index(max(a)) will do the trick. Built-in function max(a) will find the maximum value in your list a, and list function index(v) will find the index of value v in your list. By combining them, you get what you are looking for, in this case the index value 3.. Note that .index() will find the index of the first item in the list that … WebMar 14, 2024 · Using operator.itemgetter () Accessing multiple items in a given list This technique is the most pythonic and elegant method to perform this particular task. This function zips the elements of the original list with the index required from the other, hence the fasted method to achieve this task. Python3 from operator import itemgetter southland christian church kids
python - Find index of element in a list using recursion - Stack Overflow
WebDec 2, 2024 · When working with Python lists, you may need to find out the index at which a particular item occurs. You can do this by: Looping through the list and checking if the item at the current index is equal to the particular value Using the built-in list method index() You’ll learn both of the above in this tutorial. Let’s get started.👩🏽💻 Python Lists, … WebIf you want to find the list that has an item, the simplest way to do it is: i = 4 index = b_list [0].index ( filter (lambda 1D_list: i in index , b_list [0]) ) Or if you know there are more than one matches for the item, then you can do: WebJul 6, 2024 · [docs] [issue37487] PyList_GetItem() document regarding index Terry J. Reedy 6 Jul 2024 6 Jul '19 southland christian church podcast