CODEWITHSUNDEEP
pythonnumpyprecision
KOB
KOB

Reputation: 4545

Range of floats with a given precision

I want to create an array of all floating point numbers in the range [0.000, 1.000], all with 3 decimal places / precision of 4.

e.g.

>>> np.arange(start=0.000, stop=1.001, decimals=3)
[0.000, 0.001, ..., 0.100, 0.101, ..., 0.900, ..., 0.999, 0.000]

Can something along the lines of this be done?

Upvotes: 5

Views: 6909

Answers (1)

MSeifert
MSeifert

Reputation: 152637

You could use np.linspace:

>>> import numpy as np
>>> np.linspace(0, 1, 1001)
array([ 0.   ,  0.001,  0.002, ...,  0.998,  0.999,  1.   ])

or np.arange using integers and then dividing:

>>> np.arange(0, 1001) / 1000
array([ 0.   ,  0.001,  0.002, ...,  0.998,  0.999,  1.   ])

However, that's not really 3 decimals because all of these values are floats and floats are inexact. That means some of those numbers may look like they have 3 decimals but they haven't!

>>> '{:.40f}'.format((np.arange(0, 1001) / 1000)[1])  # show 40 decimals of second element
'0.0010000000000000000208166817117216851329'

NumPy doesn't support fixed decimals so in order to get a "perfect result" you need to use Python. For example a list with fractions.Fraction:

>>> from fractions import Fraction
>>> [Fraction(i, 1000) for i in range(0, 1001)]
[Fraction(0, 1), Fraction(1, 1000), ..., Fraction(999, 1000), Fraction(1, 1)]

or decimal.Decimal:

>>> from decimal import Decimal
>>> [Decimal(i) / 1000 for i in range(1, 1001)]
[Decimal('0.001'), Decimal('0.002'), ..., Decimal('0.999'), Decimal('1')]

Upvotes: 5

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