iterate through 2d array python numpy

How can I loop through 2D NumPy array using x and y coordinates without getting out of bounds error?

stackoverflow.com › questions › 30499857 › how-can-i-loop-through-2d-numpy-array-using-x-and-y-coordinates-without-getting

You get prettier code with:

for iy, ix in np.ndindex(a.shape):
    print(a[iy, ix])

resulting in:

Answer from Markus Dutschke on Stack Overflow

W3Schools

w3schools.com › python › numpy › numpy_array_iterating.asp

NumPy Array Iterating

import numpy as np arr = np.array([1, 2, 3]) for x in np.nditer(arr, flags=['buffered'], op_dtypes=['S']): print(x) Try it Yourself » · We can use filtering and followed by iteration.

Stack Overflow

stackoverflow.com › questions › 30499857 › how-can-i-loop-through-2d-numpy-array-using-x-and-y-coordinates-without-getting

How can I loop through 2D NumPy array using x and y ...

Top answer

1 of 4

120

You get prettier code with:

for iy, ix in np.ndindex(a.shape):
    print(a[iy, ix])

resulting in:

2 of 4

a.shape[0] is the number of rows and the size of the first dimension, while a.shape[1] is the size of the second dimension. You need to write:

for x in range(0, rows):
    for y in range(0, cols):
        print a[x,y]

Note how rows and cols have been swapped in the range() function.

Edit: It has to be that way because an array can be rectangular (i.e. rows != cols). a.shape is the size of each dimension in the order they are indexed. Therefore if shape is (10, 5) when you write:

a[x, y]

the maximum x is 9 and the maximum y is 4. x and y are actually poor names for array indices because they do not represent cartesian coordinates, but a location in memory. You can use i and j instead:

for i in range(0, rows):
    for j in range(0, cols):
        print a[i,j]

The documentation is a bit long but has a good, in-depth description of indices and shapes.

python - Iterate 2D numpy array using for and while loops

How to iterate through each element in a 2D array in python?

python 3.x - Iterations over 2d numpy arrays with while and for statements

Create 2d Array in Python Using For Loop Results

Python Numpy 2-dimensional array iteration

More results from stackoverflow.com

Discussions

A quick question about iterating through 2d arrays with Python.

Sure. Create a new list, loop through the rows of the old list, and append the elements you want to the new list. Basically, do what the comprehension would’ve done, just with more code. More on reddit.com

r/learnprogramming

September 24, 2021

Is there a NumPy'-thonic'/efficient way to iterate over each elem in a 2D array?

This code will slide a window over the entire board and return a 3x3 array for each cell, I would suggest putting a border around the board so each cell is looked at: import numpy as np gameOfLifeBoard = [[0,0,0,0,0,0,0,0,0], [0,1,0,0,0,0,0,0,0], [0,0,1,1,0,0,0,0,0], [0,1,1,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,0], [0,0,0,0,0,0,0,0,0]] result = np.lib.stride_tricks.sliding_window_view(gameOfLifeBoard,(3,3)) print(result) The output: [[[[0 0 0] [0 1 0] [0 0 1]] [[0 0 0] [1 0 0] [0 1 1]] [[0 0 0] [0 0 0] [1 1 0]] [[0 0 0] [0 0 0] [1 0 0]] [[0 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]]] [[[0 1 0] [0 0 1] [0 1 1]] [[1 0 0] [0 1 1] [1 1 0]] [[0 0 0] [1 1 0] [1 0 0]] [[0 0 0] [1 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]]] [[[0 0 1] [0 1 1] [0 0 0]] [[0 1 1] [1 1 0] [0 0 0]] [[1 1 0] [1 0 0] [0 0 0]] [[1 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]]] [[[0 1 1] [0 0 0] [0 0 0]] [[1 1 0] [0 0 0] [0 0 0]] [[1 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]]] [[[0 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]] [[0 0 0] [0 0 0] [0 0 0]]]] You can then reshape the returned data in to rows which makes it easier to work with: result = result.reshape((result.shape[0]*result.shape[1], 9)) print(result) The output: [[0 0 0 0 1 0 0 0 1] [0 0 0 1 0 0 0 1 1] [0 0 0 0 0 0 1 1 0] [0 0 0 0 0 0 1 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 1 0 0 0 1 0 1 1] [1 0 0 0 1 1 1 1 0] [0 0 0 1 1 0 1 0 0] [0 0 0 1 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 1 0 1 1 0 0 0] [0 1 1 1 1 0 0 0 0] [1 1 0 1 0 0 0 0 0] [1 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 1 1 0 0 0 0 0 0] [1 1 0 0 0 0 0 0 0] [1 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0]] You are then able to easily sum and search the rows. Note that each row is a cell and it's neighbors and index 4 is the cell. # sum of neighbors for x in result: s = sum(x) - x[4] More on reddit.com

r/learnpython

March 25, 2022

Why can't I easily loop over numpy arrays

You can loop over numpy arrays: In [31]: a = np.array([1, 2, 3, 4, 5]) ...: for v in a: ...: print(v) ...: 1 2 3 4 5 It's just not recommended because it's slow. It's not slower than iterating through a normal Python list, but it's much slower than it would otherwise be vectorized (which is a big deal if you're dealing with thousands of values, which is pretty common for many numpy users). Also, np.where() returns two values for a 2D array, first the row indices (i.e., y), and then the column indices (i.e., x). Arrays are indexed by (row, column) which is (y, x), not (x, y) which is what you might be used to. This is true of most programming languages, FWIW and has nothing to do with numpy (a pure python 2d list would be indexed the same). I don't understand what your question is at the end; the above may or may not have shed light on it, but please expand a bit more if that didn't help. More on reddit.com

r/learnpython

December 12, 2018

Looping over first two indices in 4-Dimensional Array?

Just for reference, the correct way to write this as for loops would look like this: for i in range(window.shape[0]): for j in range(window.shape[1]): But this being NumPy, the above is not the best way to solve this. You almost never need to manually loop over NumPy arrays. I'm not quite following what you are doing in the loop body, since your condition appears to be essentially a == 1 and a-1 == 1 and a-1 == 1, so simultaneously redundant and contradictory. Also, P11 is a 3x3 array that you only ever modify one value of. But as an example, if you wanted to count how many of the middle values of the 3x3 arrays in window are equal to 1, you could write this vectorized like so: num_11_ones = (window[:, :, 1, 1] == 1).sum() This exploits the fact that True and False convert to 1 and 0 automatically for arithmetic operations, so summing up a boolean array gives you the number of Trues it contains. If there are more conditions than that, you can create one boolean array for each condition, combine them with the & (AND) or | (OR) operators as appropriate, and then sum them up at the end: num_11_ones_or_zeros = ((window[:, :, 1, 1] == 1) | (window[:, :, 1, 1] == 0)).sum() More on reddit.com

r/learnpython

November 3, 2023

Videos