numpy boolean array with 1 bit entries

Question

Is there a way in numpy to create an array of booleans that uses just 1 bit for each entry?

The standard np.bool type is 1 byte, but this way I use 8 times the required memory.

On Google I found that C++ has std::vector<bool>.

Asked By: Andrea Zonca

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Source

Answer 1

You want a bitarray:

efficient arrays of booleans — C extension

This module provides an object type which efficiently represents an array of booleans. Bitarrays are sequence types and behave very much like usual lists. Eight bits are represented by one byte in a contiguous block of memory. The user can select between two representations; little-endian and big-endian. All of the functionality is implemented in C. Methods for accessing the machine representation are provided. This can be useful when bit level access to binary files is required, such as portable bitmap image files (.pbm). Also, when dealing with compressed data which uses variable bit length encoding, you may find this module useful…

Answered By: Chris Eberle

Answer 2

You might like to take a look at bitstring (documentation here).

If you create a ConstBitArray or ConstBitStream from a file then it will use mmap and not load it into memory. In this case it won’t be mutable so if you want to make changes it will have to be loaded in memory.

For example to create without loading into memory:

>>> a = bitstring.ConstBitArray(filename='your_file')

or

>>> b = bitstring.ConstBitStream(a_file_object)

Answered By: Scott Griffiths

Answer 3

To do this you can use numpy’s packbits and unpackbits:

import numpy as np
# original boolean array
A1 = np.array([
    [0, 1, 1, 0, 1],
    [0, 0, 1, 1, 1],
    [1, 1, 1, 1, 1],
], dtype=bool)

# packed data
A2 = np.packbits(A1, axis=None)

# checking the size
print(len(A1.tostring())) # 15 bytes
print(len(A2.tostring())) #  2 bytes (ceil(15/8))

# reconstructing from packed data. You need to resize and reshape
A3 = np.unpackbits(A2, count=A1.size).reshape(A1.shape).view(bool)

# and the arrays are equal
print(np.array_equal(A1, A3)) # True

Prior to numpy 1.17.0, the first function is straight-forward to use, but reconstruction required additional manipulations. Here is an example:

import numpy as np
# original boolean array
A1 = np.array([
    [0, 1, 1, 0, 1],
    [0, 0, 1, 1, 1],
    [1, 1, 1, 1, 1],
], dtype=np.bool)

# packed data
A2 = np.packbits(A1, axis=None)

# checking the size
print(len(A1.tostring())) # 15 bytes
print(len(A2.tostring())) #  2 bytes (ceil(15/8))

# reconstructing from packed data. You need to resize and reshape
A3 = np.unpackbits(A2, axis=None)[:A1.size].reshape(A1.shape).astype(np.bool)

# and the arrays are equal
print(np.array_equal(A1, A3)) # True

Answered By: Salvador Dali

numpy boolean array with 1 bit entries

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