NumPy Aliasing Explained: How a = b Broke a 52-Card Deck and How .copy() and Vectorization Fix It

Learn why aliasing in NumPy (a = b) corrupts a 52-card deck build, and how .copy(), vectorization with np.tile/np.repeat, avoiding np.concatenate fix issues.

Building a 52-card deck with NumPy looks straightforward: values 1–13 for each suit, and suits encoded as 0 to 3. But a small line like a = b can quietly derail the whole result. Here’s what goes wrong and how to fix it cleanly.

The setup and the unexpected output

The goal is a 2D array where each row is a card: [value, suit]. The value runs from 1 to 13, the suit from 0 to 3. Instead, the first 13 rows end up with suit 1, followed by 26 rows of suit 1, then suits 2 and 3 — effectively producing 0 spades, 26 hearts, 13 clubs, and 13 diamonds.

Problematic code (renamed, same logic)

The following snippet reproduces the issue. Variable names are different, but the behavior matches the original.

import numpy as np
num_suits = 4
deck_chunk = np.empty((13, 2))
deck_chunk[:, 0] = np.arange(1, deck_chunk.shape[0] + 1)
alias = deck_chunk
print(deck_chunk)
for suit_code in range(1, num_suits):
    alias[:, 1] = suit_code
    deck_chunk = np.concatenate([deck_chunk, alias])
print(deck_chunk)

What actually happens

The crux is aliasing the same NumPy array. When alias = deck_chunk is used, both names refer to the exact same block of memory. Assigning alias[:, 1] = suit_code also updates deck_chunk at the same time. As a result, before the first concatenation, the original 13 rows get their suit column overwritten with 1. Subsequent concatenations keep duplicating that mutated block, which is why the first 26 rows end up with suit 1.

The fix

Use a real copy when preparing the chunk for concatenation. That way, modifying the working array does not touch the original block.

import numpy as np
num_suits = 4
deck_chunk = np.zeros((13, 2))
deck_chunk[:, 0] = np.arange(1, deck_chunk.shape[0] + 1)
clone = deck_chunk.copy()
print(deck_chunk)
for suit_code in range(1, num_suits):
    clone[:, 1] = suit_code
    deck_chunk = np.concatenate([deck_chunk, clone])
print(deck_chunk)

The only essential change is using .copy() instead of creating a second reference. With a proper copy, the first 13 rows keep suit 0, and the loop appends suits 1, 2, and 3 as intended. The use of np.zeros ensures the initial suit column starts at 0 as designed.

An even more compact build

You can also generate the whole deck without concatenation by leveraging vectorized construction. This keeps the value column in repeating blocks of 1–13 and fills the suit column in grouped runs of 0–3.

import numpy as np
deck = np.zeros((52, 2))
deck[:, 0] = np.tile(np.arange(13) + 1, 4)
deck[:, 1] = np.repeat(np.arange(4), 13)

Why this matters

NumPy arrays are mutable and frequently shared by reference. Accidentally aliasing an array and then mutating it can cascade into subtle data corruption, especially when intermediate results are reused or concatenated. Understanding when you are holding a view, a reference, or a true copy is critical for predictable numerical code.

Takeaways

If you need an independent working array, create it with .copy() instead of assigning a second name to the same object. Avoid unnecessary concatenations when a direct vectorized construction is available. Most importantly, be mindful that writing through any alias changes the original data too — sometimes far away from the line where the alias was created.