Numpy: How to randomly split/select an matrix into n-different matrices

Question

• I have a numpy matrix with shape of (4601, 58).
• I want to split the matrix randomly as per 60%, 20%, 20% split based on number of rows
• This is for Machin

Answer 1

If you want to randomly select rows, you could just use random.sample from the standard Python library:

import random

population = range(4601) # Your number of rows
choice = random.sample(population, k) # k being the number of samples you require

random.sample samples without replacement, so you don't need to worry about repeated rows ending up in choice. Given a numpy array called matrix, you can select the rows by slicing, like this: matrix[choice].

Of, course, k can be equal to the number of total elements in the population, and then choice would contain a random ordering of the indices for your rows. Then you can partition choice as you please, if that's all you need.

Answer 2

Since you need it for machine learning, here is a method I wrote:

import numpy as np

def split_random(matrix, percent_train=70, percent_test=15):
    """
    Splits matrix data into randomly ordered sets 
    grouped by provided percentages.

    Usage:
    rows = 100
    columns = 2
    matrix = np.random.rand(rows, columns)
    training, testing, validation = \
    split_random(matrix, percent_train=80, percent_test=10)

    percent_validation 10
    training (80, 2)
    testing (10, 2)
    validation (10, 2)

    Returns:
    - training_data: percentage_train e.g. 70%
    - testing_data: percent_test e.g. 15%
    - validation_data: reminder from 100% e.g. 15%
    Created by Uki D. Lucas on Feb. 4, 2017
    """

    percent_validation = 100 - percent_train - percent_test

    if percent_validation < 0:
        print("Make sure that the provided sum of " + \
        "training and testing percentages is equal, " + \
        "or less than 100%.")
        percent_validation = 0
    else:
        print("percent_validation", percent_validation)

    #print(matrix)  
    rows = matrix.shape[0]
    np.random.shuffle(matrix)

    end_training = int(rows*percent_train/100)    
    end_testing = end_training + int((rows * percent_test/100))

    training = matrix[:end_training]
    testing = matrix[end_training:end_testing]
    validation = matrix[end_testing:]
    return training, testing, validation

# TEST:
rows = 100
columns = 2
matrix = np.random.rand(rows, columns)
training, testing, validation = split_random(matrix, percent_train=80, percent_test=10) 

print("training",training.shape)
print("testing",testing.shape)
print("validation",validation.shape)

print(split_random.__doc__)

• training (80, 2)
• testing (10, 2)
• validation (10, 2)

Answer 3

A complement to HYRY's answer if you want to shuffle consistently several arrays x, y, z with same first dimension: x.shape[0] == y.shape[0] == z.shape[0] == n_samples.

You can do:

rng = np.random.RandomState(42)  # reproducible results with a fixed seed
indices = np.arange(n_samples)
rng.shuffle(indices)
x_shuffled = x[indices]
y_shuffled = y[indices]
z_shuffled = z[indices]

And then proceed with the split of each shuffled array as in HYRY's answer.

Answer 4

you can use numpy.random.shuffle

import numpy as np

N = 4601
data = np.arange(N*58).reshape(-1, 58)
np.random.shuffle(data)

a = data[:int(N*0.6)]
b = data[int(N*0.6):int(N*0.8)]
c = data[int(N*0.8):]