Convert scientific notation to decimals

Question

I have numbers in a file (so, as strings) in scientific notation, like:

8.99284722486562e-02 

but I want to convert them to:

0.08992847

Is there any built-in function or any other way to do it?

Answer 1

I'm pretty sure you can do this with:

float("8.99284722486562e-02") # and now with 'rounding' "{:.8f}".format(float("8.99284722486562e-02")) 

Answer 2

I'm making this answer since the top voted one has misinformation and so i can explain my improvements.

TL;DR: Use ("%.17f" % n).rstrip('0').rstrip('.')

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By default Python formats to scientific notation if there's 5 or more zeroes at the beginning.
0.00001 / 1e-05 formats to "1e-05".
0.0001 / 1e-04 formats to "0.0001".

So of course 8.99284722486562e-02 will format to "0.0899284722486562" already.
A better example would've been 8.99284722486562e-05. (0.00008992847224866)

We can easily format to raw decimal places with "%f" which is same as "%.6f" by default.
"%f" % 8.99284722486562e-05 produces '0.000090'.
"%f" % 0.01 produces '0.010000'.

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By default floats display upto 17 decimal places.
0.1234567898765432123 - (19 dp input)
0.12345678987654321 - (17 dp output)

So if we did "%.17f" % 8.99284722486562e-02 we'd get '0.08992847224865620'. (note the extra 0)
But if we did "%.17f" % 0.0001 we surely wouldn't want '0.00010000000000000'.

So to remove the trailing zeroes we can do: ("%.17f" % n).rstrip('0').rstrip('.')
(Notice we also strip the decimal point incase the number has no fraction left)

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Also there's counterparts to %f:
%f shows standard notation
%e shows scientific notation
%g shows default (scientific if 5 or more zeroes)

Answer 3

The scientific notation can be converted to a floating point number with float.

   In [1]:  float("8.99284722486562e-02")
Out [1]:   0.0899284722486562

The float can be rounded with format and then float can be used on the string to return the final rounded float.

   In [2]:  float("{:.8f}".format(float("8.99284722486562e-02")))
Out [2]:   0.08992847

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2022 edit for No Sound's comment:

I learned this solution from here (archived)

The following solutions work for larger numbers.

Solution 1)

import numpy as np

print(np.format_float_positional(1.32456e-12, trim='-'))
print(np.format_float_positional(1.32456e-24, trim='-'))
print(np.format_float_positional(1.32456e12, trim='-'))
print(np.format_float_positional(1.32456e24, trim='-'))

# Output: 0.00000000000132456
#         0.00000000000000000000000132456
#         1324560000000
#         1324560000000000000000000

Solution 2)

same as above accept this time using a lambda function

import numpy as np

pretty_print = lambda x: np.format_float_positional(x, trim="-")

print(pretty_print(1.32456e-12))
print(pretty_print(1.32456e-24))
print(pretty_print(1.32456e12))
print(pretty_print(1.32456e24))

# Output: 0.00000000000132456
#         0.00000000000000000000000132456
#         1324560000000
#         1324560000000000000000000

Answer 4

As you may know floating point numbers have precision problems. For example, evaluate:

>>> (0.1 + 0.1 + 0.1) == 0.3
False

Instead you may want to use the Decimal class. At the python interpreter:

>>> import decimal
>>> tmp = decimal.Decimal('8.99284722486562e-02')
Decimal('0.0899284722486562')
>>> decimal.getcontext().prec = 7
>>> decimal.getcontext().create_decimal(tmp)
Decimal('0.08992847')