How to suppress scientific notation when printing float values?

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

Using the newer version ''.format (also remember to specify how many digit after the . you wish to display, this depends on how small is the floating number). See this example:

>>> a = -7.1855143557448603e-17
>>> '{:f}'.format(a)
'-0.000000'

as shown above, default is 6 digits! This is not helpful for our case example, so instead we could use something like this:

>>> '{:.20f}'.format(a)
'-0.00000000000000007186'

Update

Starting in Python 3.6, this can be simplified with the new formatted string literal, as follows:

>>> f'{a:.20f}'
'-0.00000000000000007186'

Method 2

'%f' % (x/y)

but you need to manage precision yourself. e.g.,

'%f' % (1/10**8)

will display zeros only.
details are in the docs

Or for Python 3 the equivalent old formatting or the newer style formatting

Method 3

With newer versions of Python (2.6 and later), you can use ''.format() to accomplish what @SilentGhost suggested:

'{0:f}'.format(x/y)

Method 4

Another option, if you are using pandas and would like to suppress scientific notation for all floats, is to adjust the pandas options.

import pandas as pd
pd.options.display.float_format = '{:.2f}'.format

Method 5

Most of the answers above require you to specify precision. But what if you want to display floats like this, with no unnecessary zeros:

1
0.1
0.01
0.001
0.0001
0.00001
0.000001
0.000000000001

numpy has an answer: np.format_float_positional

import numpy as np

def format_float(num):
    return np.format_float_positional(num, trim='-')

Method 6

This will work for any exponent:

def getExpandedScientificNotation(flt):
    str_vals = str(flt).split('e')
    coef = float(str_vals[0])
    exp = int(str_vals[1])
    return_val = ''
    if int(exp) > 0:
        return_val += str(coef).replace('.', '')
        return_val += ''.join(['0' for _ in range(0, abs(exp - len(str(coef).split('.')[1])))])
    elif int(exp) < 0:
        return_val += '0.'
        return_val += ''.join(['0' for _ in range(0, abs(exp) - 1)])
        return_val += str(coef).replace('.', '')
    return return_val

Method 7

This is using Captain Cucumber’s answer, but with 2 additions.

1) allowing the function to get non scientific notation numbers and just return them as is (so you can throw a lot of input that some of the numbers are 0.00003123 vs 3.123e-05 and still have function work.

2) added support for negative numbers. (in original function, a negative number would end up like 0.0000-108904 from -1.08904e-05)

def getExpandedScientificNotation(flt):
    was_neg = False
    if not ("e" in flt):
        return flt
    if flt.startswith('-'):
        flt = flt[1:]
        was_neg = True 
    str_vals = str(flt).split('e')
    coef = float(str_vals[0])
    exp = int(str_vals[1])
    return_val = ''
    if int(exp) > 0:
        return_val += str(coef).replace('.', '')
        return_val += ''.join(['0' for _ in range(0, abs(exp - len(str(coef).split('.')[1])))])
    elif int(exp) < 0:
        return_val += '0.'
        return_val += ''.join(['0' for _ in range(0, abs(exp) - 1)])
        return_val += str(coef).replace('.', '')
    if was_neg:
        return_val='-'+return_val
    return return_val

Method 8

You can use the built-in format function.

>>> a = -3.42142141234123e-15
>>> format(a, 'f')
'-0.000000'
>>> format(a, '.50f') # Or you can specify precision
'-0.00000000000000342142141234122994048466990874926279'

Method 9

In case of numpy arrays you can suppress with suppress command as

import numpy as np
np.set_printoptions(suppress=True)

Method 10

If it is a string then use the built in float on it to do the conversion for instance:
print( "%.5f" % float("1.43572e-03"))
answer:0.00143572

Method 11

In addition to SG’s answer, you can also use the Decimal module:

from decimal import Decimal
x = str(Decimal(1) / Decimal(10000))

# x is a string '0.0001'

Method 12

Since this is the top result on Google, I will post here after failing to find a solution for my problem. If you are looking to format the display value of a float object and have it remain a float – not a string, you can use this solution:

Create a new class that modifies the way that float values are displayed.

from builtins import float
class FormattedFloat(float):

    def __str__(self):
        return "{:.10f}".format(self).rstrip('0')

You can modify the precision yourself by changing the integer values in {:f}

Method 13

A simpler solution to display a float to an arbitrary number of significant digits. No numpy or list comprehensions required here:

def sig(num, digits=3):
    "Return number formatted for significant digits"
    if num == 0:
        return 0
    negative = '-' if num < 0 else ''
    num = abs(float(num))
    power = math.log(num, 10)
    if num < 1:
        step = int(10**(-int(power) + digits) * num)
        return negative + '0.' + '0' * -int(power) + str(int(step)).rstrip('0')
    elif power < digits - 1:
        return negative + ('{0:.' + str(digits) + 'g}').format(num)
    else:
        return negative + str(int(num))

I’m stripping trailing 0s and displaying full integers in the example: sig(31415.9) = 31415 instead of 31400. Feel free to modify the code if that’s not something you’re into.

Testing:

for power in range(-8,8):
    num = math.pi * 10**power
    print(str(num).ljust(25), sig(num))

Method 14

Using 3.6.4, I was having a similar problem that randomly, a number in the output file would be formatted with scientific notation when using this:

fout.write('someFloats: {0:0.8},{1:0.8},{2:0.8}'.format(someFloat[0], someFloat[1], someFloat[2]))

All that I had to do to fix it was to add ‘f’:

fout.write('someFloats: {0:0.8f},{1:0.8f},{2:0.8f}'.format(someFloat[0], someFloat[1], someFloat[2]))

Method 15

As of 3.6 (probably works with slightly older 3.x as well), this is my solution:

import locale
locale.setlocale(locale.LC_ALL, '')

def number_format(n, dec_precision=4):
    precision = len(str(round(n))) + dec_precision
    return format(float(n), f'.{precision}n')

The purpose of the precision calculation is to ensure we have enough precision to keep out of scientific notation (default precision is still 6).

The dec_precision argument adds additional precision to use for decimal points. Since this makes use of the n format, no insignificant zeros will be added (unlike f formats). n also will take care of rendering already-round integers without a decimal.

n does require float input, thus the cast.

Method 16

I was having a similar problem that randomly, using my solution:

from decimal import Decimal

Decimal(2/25500)
#output:0.00007843137254901961000728982664753630160703323781490325927734375

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

0 0 votes

Article Rating