# Hướng dẫn remove e in python

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.

martineau

asked Mar 18, 2009 at 15:27

2

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'
``````

Contents

## Update

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

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

answered Oct 19, 2015 at 16:41

Aziz AltoAziz Alto

4

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

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

19:05

nmichaelsnmichaels

3

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
``````

answered May 24, 2019 at 15:34

Josh JanjuaJosh Janjua

1

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='-')
``````

answered Sep 25, 2019 at 21:06

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Dennis GolomazovDennis Golomazov

0

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

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

joanis

Saran ZebSaran Zeb

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
``````

10, 2017 at 3:56

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'
``````

answered Jun 26, 2021 at 21:31

qwerty_urlqwerty_url

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
``````

FriskyGrub

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9393 gold

answered Sep 19, 2017 at 10:40

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`

answered Feb 6, 2015 at 8:43

U-571U-571

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'
``````

gsamaras

answered Mar 18, 2009 at 15:34

30.9k17 gold

1

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}`

answered Jul 2, 2018 at 17:16

Jared MarksJared Marks

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.

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Testing:

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

answered May 18, 2021 at 13:52

SurpriseDogSurpriseDog

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]))
``````

answered Apr 13, 2018 at 23:07

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.

answered Jul 22, 2019 at 1:52

Alex SAlex S

0

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

``````from decimal import Decimal

Decimal(2/25500)
#output:0.00007843137254901961000728982664753630160703323781490325927734375
``````