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

Question

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]])

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]

No, Python `list` objects don't support that. You can always write a function, though — juanpa.arrivillaga, Dec 11 '19 at 17:20
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)` — Andrej Kesely, Dec 11 '19 at 17:21
Or use `numpy` as this will support your desired indexing syntax — modesitt, Dec 11 '19 at 17:22

score 2 · Answer 1 · answered Dec 11 '19 at 17:23

2

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.

answered Dec 11 '19 at 17:23

asymmetryFan

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score 2 · Answer 2 · answered Dec 11 '19 at 17:38

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!!

score 1 · Accepted Answer · answered Dec 11 '19 at 17:28

1

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

answered Dec 11 '19 at 17:28

oppressionslayer

6,942
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score 1 · Answer 4 · answered Dec 11 '19 at 18:07

1

You can use the following function:

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

func(pos, searched)
# 7

answered Dec 11 '19 at 18:07

Mykola Zotko

15,583
3
71
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Tadhg McDonald-Jensen · Answer 5 · 2019-12-11T20:00:05.283

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)])

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

5 Answers5