Here’s my code:

x = 1.0
y = 100000.0    
print x/y

My quotient displays as 1.00000e-05.

Is there any way to suppress scientific notation and make it display as
0.00001? I’m going to use the result as a string.

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'

'%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

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

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

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

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='-')

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

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

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'

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

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'

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}

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))

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

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

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

from decimal import Decimal

Decimal(2/25500)
#output:0.00007843137254901961000728982664753630160703323781490325927734375

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]))

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.