Is there an elegant way to check if index can be requested in a numpy array?

Question

I am looking for an elegant way to check if a given index is inside a numpy array (for example for BFS algorithms on a grid).

The following code does what I want:

import numpy as np

def isValid(np_shape: tuple, index: tuple):
    if min(index) < 0:
        return False
    for ind,sh in zip(index,np_shape):
        if ind >= sh:
            return False
    return True

arr = np.zeros((3,5))
print(isValid(arr.shape,(0,0))) # True
print(isValid(arr.shape,(2,4))) # True
print(isValid(arr.shape,(4,4))) # False

But I’d prefer something build-in or more elegant than writing my own function including python for-loops (yikes)

Asked By: schajan

||

Source

Answer 1

It seems to me that simple

(np_shape > index) and all(ind >= 0 for ind in index)

is equivalent to your isValid function.

Answered By: Dmitri Chubarov

Answer 2

You can try:

def isValid(np_shape: tuple, index: tuple):
    index = np.array(index)
    return (index >= 0).all() and (index < arr.shape).all()

arr = np.zeros((3,5))
print(isValid(arr.shape,(0,0))) # True
print(isValid(arr.shape,(2,4))) # True
print(isValid(arr.shape,(4,4))) # False

Answered By: mozway

Answer 3

Thank you for your answers, I have benchmarked them and it turns out the fastest way is indeed @Dmitri Chubarov ‘s answed:

def isValid(np_shape: tuple, index: tuple):
    return (np_shape > index) and all(ind >= 0 for ind in index)

Method 1 by @Dmitri Chubarov, Method 2 by @mozway. Times measured for index inside array and outside array (example: For a 4×4 array, (2,3) is inside, (6,3) is not inside), times averaged over 10’000 inputs. Number of array dimensions is the shape size; for a normal mxn matrix it would be 2.

import numpy as np
import time
import random
import matplotlib.pyplot as plt

def isValid_1(np_shape: tuple,index: tuple):
    return np.all(np_shape > index) and all(ind >= 0 for ind in index)

def isValid_2(np_shape: tuple, index: tuple):
    index = np.array(index)
    return (index >= 0).all() and (index < np_shape).all()

nExp = 10000
max_dimensions = 100
max_shape_content = 100

dims = []
times_1_valid = []
times_1_invalid = []
times_2_valid = []
times_2_invalid = []

for counter,dimensions in enumerate(range(1,max_dimensions+1)):
    
    print(f"Progress {(1000*counter//max_dimensions)/10}%",end="r")
    dims.append(dimensions)
    times_1_valid.append(0)
    times_1_invalid.append(0)
    times_2_valid.append(0)
    times_2_invalid.append(0)
    
    for i in range(nExp):
        
        np_shape = tuple(random.randint(1,max_shape_content) for i in range(dimensions))
        valid_index = tuple(random.randint(0,n_i-1) for n_i in np_shape)
        invalid_index = list(valid_index); 
        invalid_index[random.randint(0,dimensions-1)] = max_shape_content + 1
        invalid_index = tuple(invalid_index)
        
        t0 = time.process_time()
        isValid_1(np_shape,valid_index)
        t1 = time.process_time()
        isValid_1(np_shape,invalid_index)
        t2 = time.process_time()
        isValid_2(np_shape,valid_index)
        t3 = time.process_time()
        isValid_2(np_shape,invalid_index)
        t4 = time.process_time()
        
        times_1_valid[-1] += (t1-t0)
        times_1_invalid[-1] += (t2-t1)
        times_2_valid[-1] += (t3-t2)
        times_2_invalid[-1] += (t4-t3)

times_1_valid = np.array(times_1_valid) / nExp
times_1_invalid = np.array(times_1_invalid) / nExp
times_2_valid = np.array(times_2_valid) / nExp
times_2_invalid = np.array(times_2_invalid) / nExp

plt.plot(dims,times_1_valid,label="Method 1; index inside array")
plt.plot(dims,times_1_invalid,label="Method 1; index outside array")
plt.plot(dims,times_2_valid,label="Method 2; index inside array")
plt.plot(dims,times_2_invalid,label="Method 2; index outside array")
plt.legend()
plt.xlabel("Number of array dimensions")
plt.ylabel("Average Time [s]")
plt.show()

Answered By: schajan

Is there an elegant way to check if index can be requested in a numpy array?

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