how do I simplify searching for a list item by an index?

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I have a simple program:

pos = [1,2]
searched = [
    [1,3,4,6],
    [2,6,7,8],
    [0,1,2,8],
    [5,6,9,2]
]
print(searched[pos[0]][pos[1]])
7

Now what I want is some way to get rid off of searched[pos[0]][pos[1]] and just type something like searched[[pos]].

Is there a way to this, or do I have to write out this every time

I have gotten a lot of suggestions, but what I am searching for is a way to do this in one neat line, simplifying everything. That means that things like using a function, converting to a specific dictionary or even enumerate don't work for me, so for anyone looking at this post later: . I suggest using np.array(variable) while defining said variable so you can use variable[pos]

You can convert this to a numpy array and get the search your looking for:

import numpy as np
pos = [1,2] 
searched = np.array([ 
  [1,3,4,6], 
  [2,6,7,8], 
  [0,1,2,8], 
  [5,6,9,2] 
  ]) 
print(searched[1,2])  
# 7

How to simplify your codebase with map(), reduce(), and filter() in , It surprised me how simplified the codebase became. but it's so compact that it was hard not to put it in the list. The filter() method creates a new array with all elements that pass the Indexing an array of objects (lookup table) object where the user's id represents a key (with constant searching time). There are times when you may observe different behavior when searching for an item or file in list/library by ID. Scenario 1: Recently we noticed that when we try to search for an item/document using the ID it no longer returns any results, or the results do not include the item/document with that ID as part of the result set.

Don't know if this is acceptable to you, but a function will do:

def search_for_list_item_by_index(a_list, row, col):
    return a_list[row][col]

print(search_for_list_item_by_index(searched, 1, 2))

This prints 7, as expected.

How to use INDEX and MATCH, In a nutshell, INDEX retrieves values at a given location in a list or table. For example, let's We need a way to locate the position of things we're looking for. Enter the MATCH MATCH returns 3, since "Peach" is the third item in the range . Search an Item in a List. The BinarySearch method uses the binary search algorithm to find an item in the sorted List. The following code snippet finds an item in a List.

You can do this by converting your array to numpy as suggested by @oppressionslayer. One other way to do that is to create a dictionary and use that as follows:

pos = [1,2]
searched = [
    [1,3,4,6],
    [2,6,7,8],
    [0,1,2,8],
    [5,6,9,2]
]
m=4 # width of the searched array
n=4 # hight of the searched array

searched = {(i,j):searched[i][j] for j in range(m) for i in range(n)}

print(searched[1,2]) # prints 7
print(searched[tuple(pos)]) # prints 7

Hope this helps!!

NumPy Searching Arrays, Which means that the value 4 is present at index 3, 5, and 6. Example. Find the indexes where the values are even: import numpy as np arr =� Index of element "is" in the list is: 2 Python: Get last index of item in list. To get the last index of an item in list, we can just reverse the contents of list and then use index() function, to get the index position. But that will give the index of element from last. Where we want index of last occurrence of element from start.

You can use the following function:

def func(idx, lst):
    for i in idx:
        lst = lst[i]
    return lst

func(pos, searched)
# 7

Python List Tutorial: Lists, Loops, and More! – Dataquest, To quickly find the index of a list element, identify its position number in the list, Looking back at the list row_1 from the code example above, we can retrieve the above is a much more simplified and abstracted version of the code below:� 2 how do I simplify searching for a list item by an index? Dec 11 '19. 1 Removing all variables in a list except strings Oct 23 '19. 0 How to manipulate value in

I'd use the dictionary approach like HNMN3 but what is fun is you can generalize to non rectangular lists by using enumerate:

searched = {(row,col):item
              for row,elems in enumerate(searched)
                for col,item in enumerate(elems)}

or use flatten_data from my answer here which does the same thing but works for totally arbitrary nesting including dictionaries. Probably not the direction you want but worth mentioning.

data = flatten_data(searched)
pos = [1,2]
print(data[1,2])
print(data[tuple(pos)])

Python List index(), The list index() method can take a maximum of three arguments: element - the element to be searched; start (optional) - start searching from this index; end (� The IndexOf(T, Int32) method overload is used to search the list beginning with index location 3 and continuing to the end of the list, and finds the second occurrence of the string. Finally, the IndexOf(T, Int32, Int32) method overload is used to search a range of two entries, beginning at index location two; it returns -1 because there are no

8.5. bisect — Array bisection algorithm — Python 2.7.18 , For long lists of items with expensive comparison operations, this can it is better to search a list of precomputed keys to find the index of the� To find index of first occurrence of an item in a list, you can use index() method of List class with the item passed as argument. index=mylist.index(item). You can also provide start and end, where the elements between the positions start and end in the list are considered. list.index(x, [start[, end]])

5. Data Structures — Python 3.8.5 documentation, If no index is specified, a.pop() removes and returns the last item in the list. notation and are used to limit the search to a particular subsequence of the list. Lists are central constructs in the Wolfram Language, used to represent collections, arrays, sets, and sequences of all kinds. Lists can have any structure and size and can routinely involve even millions of elements.

Search engine indexing, Search engine optimisation indexing collects, parses, and stores data to facilitate fast and Inverted index: Stores a list of occurrences of each atomic search criterion, The following is a simplified illustration of an inverted index: of characters which represent words and other elements, such as punctuation, which are� Still doesn't do what OP wanted. You set n to zero which would be a valid index. So when no Person aged twenty is found your code would suggest the first entry on the list would fit. At least set n to an invalid number for an index, like -1 for instance.

Comments
  • No, Python list objects don't support that. You can always write a function, though
  • There are many options. You can write helper function e.g, def get_val(matrix, pos): that returns this position and then call it get_val(searched, pos)
  • Or use numpy as this will support your desired indexing syntax