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REFERENCEnumpy

numpy Documentation

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np.zeros()

AI & DATA SCIENCE // np-zeros

np.zeros() creates a new array of a given shape, filled entirely with zeros of a specified dtype.

Syntax

np.zeros(shape, dtype=float)

Deep Dive Course

np.zeros(shape) allocates an array of the requested shape, an integer for 1D, or a tuple for multi-dimensional arrays, and initializes every element to 0, using float64 by default unless a different dtype is given. It's the standard way to pre-allocate an array of a known size before filling it in with computed values, which is far more efficient than growing an array one element at a time, since NumPy arrays have a fixed size and any append-like operation actually allocates a brand-new array.

1Understanding np.zeros()

np.zeros(shape) allocates an array of the requested shape, an integer for 1D, or a tuple for multi-dimensional arrays, and initializes every element to 0, using float64 by default unless a different dtype is given. It's the standard way to pre-allocate an array of a known size before filling it in with computed values, which is far more efficient than growing an array one element at a time, since NumPy arrays have a fixed size and any append-like operation actually allocates a brand-new array.

💡

Pre-allocate with np.zeros() (or np.empty()) whenever you know the final shape of an array ahead of time and plan to fill it in via indexing, instead of building it up incrementally with concatenation, which reallocates memory on every step.

editor.html
import numpy as np

arr = np.zeros(5)
print(arr)
localhost:3000

2Practical Example

Here is a real-world application of np.zeros() showing how it is used in production NumPy code.

editor.html
import numpy as np

matrix = np.zeros((2, 3), dtype=np.int32)
print(matrix)
localhost:3000

3Best Practices

Follow these guidelines when working with np.zeros():

1. Pre-allocate an output array with np.zeros() before filling it in a loop by index, instead of repeatedly concatenating/appending arrays

2. Specify an explicit dtype (like np.int32) when the default float64 wastes memory for data that's actually meant to be integers

3. Use np.zeros_like(other_array) instead of np.zeros(other_array.shape) when you want a zero-filled array matching another array's shape and dtype exactly

⚠️

Tip: Pre-allocate with np.zeros() (or np.empty()) whenever you know the final shape of an array ahead of time and plan to fill it in via indexing, instead of building it up incrementally with concatenation, which reallocates memory on every step.

editor.html
import numpy as np

arr = np.zeros(5)
print(arr)
localhost:3000

Examples

Example 01Basic Usage
import numpy as np

arr = np.zeros(5)
print(arr)
Example 02Advanced Example
import numpy as np

matrix = np.zeros((2, 3), dtype=np.int32)
print(matrix)

Best Practices

  • Pre-allocate an output array with np.zeros() before filling it in a loop by index, instead of repeatedly concatenating/appending arrays
  • Specify an explicit dtype (like np.int32) when the default float64 wastes memory for data that's actually meant to be integers
  • Use np.zeros_like(other_array) instead of np.zeros(other_array.shape) when you want a zero-filled array matching another array's shape and dtype exactly

Interview Question

Why is pre-allocating an array with np.zeros() usually faster than building it up with repeated concatenation?

Hint: Think about what happens to memory every time you concatenate or append to a NumPy array.

NumPy arrays have a fixed size once created — there's no way to grow one in place. Every concatenation or append-like operation actually allocates a brand-new, larger array and copies all the existing elements into it before adding the new ones, which costs progressively more work as the array grows, similar to the cost of repeated string concatenation. Pre-allocating the full array once with np.zeros() and then filling in values by index avoids all of that repeated copying entirely.

Exercises

MediumPractice using np.zeros() in a real scenario.
View Solution
import numpy as np

arr = np.zeros(5)
print(arr)

Frequently Asked Questions

Why is pre-allocating an array with np.zeros() usually faster than building it up with repeated concatenation?

NumPy arrays have a fixed size once created — there's no way to grow one in place. Every concatenation or append-like operation actually allocates a brand-new, larger array and copies all the existing elements into it before adding the new ones, which costs progressively more work as the array grows, similar to the cost of repeated string concatenation. Pre-allocating the full array once with np.zeros() and then filling in values by index avoids all of that repeated copying entirely.

Related Functions

np-onesnp-emptynp-full