An array and nested list version:

In [163]: A=np.arange(12).reshape(3,4)
In [164]: Al = A.tolist()

For sliced indexing, a list comprehension (or mapping equivalent) works fine:

In [165]: A[:,1:3]
Out[165]: 
array([[ 1,  2],
       [ 5,  6],
       [ 9, 10]])
In [166]: [l[1:3] for l in Al]
Out[166]: [[1, 2], [5, 6], [9, 10]]

For advanced indexing, the list requires a further level of iteration:

In [167]: A[:,[0,2,3]]
Out[167]: 
array([[ 0,  2,  3],
       [ 4,  6,  7],
       [ 8, 10, 11]])

In [169]: [[l[i] for i in [0,2,3]] for l in Al]
Out[169]: [[0, 2, 3], [4, 6, 7], [8, 10, 11]]

Again there are various mapping alternatives.

In [171]: [operator.itemgetter(0,2,3)(l) for l in Al]
Out[171]: [(0, 2, 3), (4, 6, 7), (8, 10, 11)]

itemgetter uses tuple(obj[i] for i in items) to generate those tuples.

Curiously, itemgetter returns tuples for the list index, and lists for slices:

In [176]: [operator.itemgetter(slice(1,3))(l) for l in Al]
Out[176]: [[1, 2], [5, 6], [9, 10]]
Answer from hpaulj on Stack Overflow
🌐
Stack Overflow
stackoverflow.com › questions › 43329235 › indexing-of-python-2d-list
Indexing of python 2D list - Stack Overflow
Top answer
1 of 2
3

An array and nested list version:

In [163]: A=np.arange(12).reshape(3,4)
In [164]: Al = A.tolist()

For sliced indexing, a list comprehension (or mapping equivalent) works fine:

In [165]: A[:,1:3]
Out[165]: 
array([[ 1,  2],
       [ 5,  6],
       [ 9, 10]])
In [166]: [l[1:3] for l in Al]
Out[166]: [[1, 2], [5, 6], [9, 10]]

For advanced indexing, the list requires a further level of iteration:

In [167]: A[:,[0,2,3]]
Out[167]: 
array([[ 0,  2,  3],
       [ 4,  6,  7],
       [ 8, 10, 11]])

In [169]: [[l[i] for i in [0,2,3]] for l in Al]
Out[169]: [[0, 2, 3], [4, 6, 7], [8, 10, 11]]

Again there are various mapping alternatives.

In [171]: [operator.itemgetter(0,2,3)(l) for l in Al]
Out[171]: [(0, 2, 3), (4, 6, 7), (8, 10, 11)]

itemgetter uses tuple(obj[i] for i in items) to generate those tuples.

Curiously, itemgetter returns tuples for the list index, and lists for slices:

In [176]: [operator.itemgetter(slice(1,3))(l) for l in Al]
Out[176]: [[1, 2], [5, 6], [9, 10]]
2 of 2
0

Wasteful but should work:

list(zip(*(list(zip(*A))[0:9])))

Slightly more economical using itertools.isclice:

list(zip(*(itertools.islice(zip(*A), 0, 9))))

Or one could use map and operator.itemgetter:

list(map(operator.itemgetter(slice(0,9)), A))
Indexing values in a 2D array in python
Python: How do I assign the indexes of an item in a 2D list to two integer variables?
How to return the index of an element in a 2d array in python?
Iterating on Python 2d array indexes and values
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Beauty and Joy of Computing
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2D Lists in Python
A list within another list is called 2-Dimensional (2D). And just like in a 2D cartesian graph, retrieving an element requires two index values (essentially "the x and y position").
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However, I find that this syntax is quite clunky and doesn't sit well in a big block of code. It would also become worse with a higher dimensional list. My question: is there an easier way to do this index? ... Since you're working with a 2D-list, it might be a good idea to use numpy. More on stackoverflow.com
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[Python] 2D list indexing and slicing
I don't think you can just do it with slicing. I recommend creating a function for this. You can create a result list and append elements to it, use a list comprehension, or even use map with a lambda expression. def get_column(matrix, col): # Your code here More on reddit.com
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Stack Overflow
stackoverflow.com › questions › 5775352 › python-return-2-ints-for-index-in-2d-lists-given-item
Python: Return 2 ints for index in 2D lists given item - Stack Overflow
Top answer
1 of 8
26

Try this:

def index_2d(myList, v):
    for i, x in enumerate(myList):
        if v in x:
            return (i, x.index(v))

Usage:

>>> index_2d(myList, 3)
(1, 0)
2 of 8
5

If you are doing many lookups you could create a mapping.

>>> myList = [[1,2],[3,4],[5,6]]
>>> d = dict( (j,(x, y)) for x, i in enumerate(myList) for y, j in enumerate(i) )
>>> d
{1: (0, 0), 2: (0, 1), 3: (1, 0), 4: (1, 1), 5: (2, 0), 6: (2, 1)}
>>> d[3]
(1, 0)
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Statistics Globe
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geeksforgeeks.org › python-using-2d-arrays-lists-the-right-way
Python | Using 2D arrays/lists the right way - GeeksforGeeks
11:48
This peculiar functioning is because Python uses shallow lists which we will try to understand. In method 1a, Python doesn't create 5 integer objects but creates only one integer object, and all the indices of the array arr point to the same int object as shown. If we assign the 0th index to another integer say 1, then a new integer object is created with the value of 1 and then the 0th index now points to this new int object as shown below · Similarly, when we create a 2d array as "arr = [[0]*cols]*rows" we are essentially extending the above analogy.
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geeksforgeeks.org › python › indexing-lists-of-lists-in-python
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Stack Overflow
stackoverflow.com › questions › 46856519 › indexing-a-2d-list-with-another-list
python - Indexing a 2D List with another List - Stack Overflow
Top answer
1 of 3
2

Since you're working with a 2D-list, it might be a good idea to use numpy. You'll then simply need to define index as a tuple. Index 3 would be out of range, though:

>>> import numpy as np
>>> a = np.array([[1,2,3], [4,5,6], [7,8,9]])
>>> index = (1, 2)
>>> a[index]
6

The method you're looking for is called Array#dig in Ruby:

[[1,2,3], [4,5,6], [7,8,9]].dig(1, 2)
# 6

but I couldn't find any plain Python equivalent.

2 of 3
2

You could just create a simple function that iterates over the index. For every element in index just fetch item from object and assign that as a new object. Once you have iterated over the whole index return current object. As @EricDuminil noted it works with dicts and all other objects that support __getitem__:

def index(obj, idx):
    for i in idx:
        obj = obj[i]

    return obj

LST = [[1,2,3], [4,[5],6], [{'foo': {'bar': 'foobar'}},8,9]]
INDEXES = [[2, 2], [1, 1, 0], [2, 0, 'foo', 'bar']]

for i in INDEXES:
    print('{0} -> {1}'.format(i, index(LST, i)))

Output:

[2, 2] -> 9
[1, 1, 0] -> 5
[2, 0, 'foo', 'bar'] -> foobar
🌐
Reddit
reddit.com › r/learnprogramming › [python] 2d list indexing and slicing
r/learnprogramming on Reddit: [Python] 2D list indexing and slicing
February 20, 2020 -

Hi all

how to print every first element of each row?

tried a few but failed 

Matrix[0:][0:1]

Matrix[0::,0:1:]

your help will be highly appreciated!

thanks a ton

Top answer
1 of 2
1
I don't think you can just do it with slicing. I recommend creating a function for this. You can create a result list and append elements to it, use a list comprehension, or even use map with a lambda expression. def get_column(matrix, col): # Your code here
2 of 2
1
[row[0] for row in matrix] is another way. I’d recommend using numpys indexing if you’re not limited to the standard library though
🌐
Guru99
guru99.com › home › python › python 2d arrays: two-dimensional list examples
Python 2D Arrays: Two-Dimensional List Examples
July 10, 2026 - The examples below show how to create, access, insert, update, delete, and size 2D arrays. 🧱 Structure: A 2D array is an array of arrays in rows and columns. 🔢 Indexing: Access an element with array[row][column]; indexes start at 0.
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Brainly
brainly.com › computers and technology › high school › indexes of 2d lists are listed _rows_ first and then _columns_.
[FREE] Indexes of 2D lists are listed _rows_ first and then _columns_. - brainly.com
April 11, 2024 - In Python, the indexing of 2D lists follows a common convention where the first index corresponds to the row number and the second index corresponds to the column number.
🌐
Stack Overflow
stackoverflow.com › questions › 46536640 › 2-d-list-and-indexing
python - 2-D List and Indexing - Stack Overflow
The matrix or 2-D list m shows bag weights in pounds for three passengers. The first passenger has 4 bags, the second passenger has 3 bags and the third passenger has 4 bags. Your function will use a row and column index as discussed during the lecture to display passenger information as shown in Example, using format() string method
🌐
Reddit
reddit.com › r/learnpython › help finding the index of a given value within a 2d array
r/learnpython on Reddit: Help finding the index of a given value within a 2D array
November 13, 2022 -

Been messing around with numpy, trying to familiarize myself with it and seeing how I'd be able to utilize its arrays to store (very simple) map data for this text adventure game I've been working on. So far it seems like it'd be pretty darn useful, but I seem to have run into something of an issue.

My code is as follows (note: this is purposefully made with out a main class, because I am just trying to get the basic functionality down in my head before I incorporate it into my main program, and this is just easier for me): 

# import numpy module as np
import numpy as np

# Establish the game map, a 3x3 grid of 0's
mainarray = np.array([[0, 0, 0],
                      [0, 0, 0],
                      [0, 0, 0]])

# Give feedback to make console easier to read
print(mainarray)
print("")
print("Placing player in center...")

# Place player on map (represented by a value of 1)
mainarray[1, 1] = 1

print(mainarray)

print("")
print("Locating player...")

# Attempt to find what the current index is of the value 1
print("")
print("Player is at index: ", np.where(mainarray == 1)[0][0])

In my head, I would like to eventually use this np.where() function (if I can) in one of the functions that moves my character. What I want to do is grab the current index of the "player" (represented by the number 1) and attempt to change that value to a 0, and change the value at an adjacent index to 1 (the tile that the player is moving to). Basically, I would like to use each index of the array as a sort of coordinate that I can take and use to move this "1" around the array, setting each index back to "0" after moving the "1", all based off user input.

Anyways, I am not getting any errors, however, the result that gets printed to the console is: 

Player is at index:  1

Why is it just one number? It remains the same, even when I remove that second [0] in the last line. I don't fully understand how this function works, as this is the first time I've tried to use it, but since there are no errors, I have no clue what is going on.

Shouldn't I be receiving an index with two values, one for the column and one for the row? If not, how can I grab that as a result and use it in the way I described above? Is it even possible, or am I barking up the wrong tree with this function?

Top answer
1 of 3
3
Numpy is a great and powerful tool, but this is not one of the things it's good at. It would be much better for you to store the player's actual position, rather than searching for the player in a map. But if you really really want to: the function you want is np.argwhere.
2 of 3
2
Shouldn’t I be receiving an index with two values, one for the column and one for the row? Columns don't exist, only rows do, in a row-major 3x3 array. That's what makes them row-major. You could have a column-major array instead, but then only the columns would exist and the rows wouldn't, so it doesn't really help. You can't use index for this - you have to iterate over your cells, row by row and column by column, keeping track of your coordinate and then returning it when you find what you're looking for.
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Quora
quora.com › How-do-you-access-a-two-dimensional-list-in-Python
How to access a two-dimensional list in Python - Quora
Answer (1 of 2): You use []s for indexing twice: That said, if you really want a matrix, i.e., a 2-D array, you should use NumPy: Lists are containers, whereas NumPy arrays are mathematical objects. If you’re doing math, use NumPy, but if you’re just holding stuff, use lists.