How i can display the value of suffix e+number in python?

Question

i'm trying to display the same original value (inverse of log )after taking the log based 10 as shown in the example, but i'm not able to display the original value, i have follow the answer there but its still not display it, i know its large value . is there away to display it? i'm using python 3.6

import math
a=2172533123329917620172861889109867406049975387320169322608751844460406222231626416365984808960
x=math.log10(a)
print(x) # result there is 93.33696640660276
y=10**x
print(y) #  result there is 2.1725331233299482e+93
print(y*10**93) # result there is 2.1725331233299483e+186

in python you don't have to use `;` at the end of line. But you have to use `#` instead of `//` for comments. — furas, Sep 30 '19 at 07:10
i just use the // for show the answer there, the code is work good sir, its just the problem i need to display the original value of a after taking the inverse of log — hani, Sep 30 '19 at 07:14
In Python, the result of a logarithm and exponentiation is always a float, even if your original value was an integer. There's no real way around this when using standard types. If you need to do exact math, you can try [sympy](https://www.sympy.org/en/index.html). — L3viathan, Sep 30 '19 at 07:16

L3viathan · Accepted Answer · 2019-09-30T07:43:43.620

When using sympy, you can use arbitrary precision math:

>>> import sympy
>>> a = 2172533123329917620172861889109867406049975387320169322608751844460406222231626416365984808960
>>> x = sympy.log(a, 10)
>>> x
1 + log(217253312332991762017286188910986740604997538732016932260875184446040622223162641636598480896)/log(10)
>>> y = 10**x
>>> y
10**(1 + log(217253312332991762017286188910986740604997538732016932260875184446040622223162641636598480896)/log(10))
>>> y*10**93
1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000*10**(1 + log(217253312332991762017286188910986740604997538732016932260875184446040622223162641636598480896)/log(10))
>>> result = y*10**93
>>> result
1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000*10**(1 + log(217253312332991762017286188910986740604997538732016932260875184446040622223162641636598480896)/log(10))
>>> result.simplify()
2172533123329917620172861889109867406049975387320169322608751844460406222231626416365984808960000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

thank you sir, good explaining , working great – hani Sep 30 '19 at 07:37 — hani, Sep 30 '19 at 07:37

score 1 · Answer 2 · answered Sep 30 '19 at 07:18

You can use string formating - '{:f}'

print('{:f}'.format(y))
print('{:f}'.format(y*10**93))

Result:

2172533123329948244974667712184302503862484572004323002958790801469409104613465879037786193920.000000
2172533123329948250187989505303380345960860577775669342927459052809311382170807040229342819817270425458795653742290299639999303458637786346385752271584468621244918497826695717627392688128.000000

To remove zeros after dot - '{:.0f}'

print('{:.0f}'.format(y))
print('{:.0f}'.format(y*10**93))

Result:

2172533123329948244974667712184302503862484572004323002958790801469409104613465879037786193920
2172533123329948250187989505303380345960860577775669342927459052809311382170807040229342819817270425458795653742290299639999303458637786346385752271584468621244918497826695717627392688128

To display with e - '{:e}'

print('{:e}'.format(x))

Result:

9.333697e+01

More about string formating on pyformat.info

Code:

import math

a = 2172533123329917620172861889109867406049975387320169322608751844460406222231626416365984808960
x = math.log10(a)
y = 10**x 

print('{:e}'.format(x))
print('{:.0f}'.format(y))
print('{:.0f}'.format(y*10**93))

prefect sir, thank you for explain everything, but the value is not same as i compare with the variable a , you can take look into the results . thank you again — hani, Sep 30 '19 at 07:22
because `float` is not ideal - it can't keep all real values and there is always problem like this `0.1 + 0.2 == 0.3` which gives `False` but we expect `True`. To get better result you should use method mentioned in other answers. — furas, Sep 30 '19 at 07:28

score 0 · Answer 3 · answered Sep 30 '19 at 07:22

0

As furas says, if you want to avoid the "engineering notation" when printing the result specify a precision in the format string.

If, however, you want a higher precision for the calculation you can use the decimal library:

from decimal import Decimal, getcontext

a = Decimal(2172533123329917620172861889109867406049975387320169322608751844460406222231626416365984808960)

# Increase the precision
getcontext().prec = 100

x = a.log10()
y = 10 ** x
print(y)

answered Sep 30 '19 at 07:22

Joe Halliwell

1,155
6
21

thank you sir, your code is help me and work great after use print('{:.0f}'.format(y)) give me prefect answer ,from @ furas – hani Sep 30 '19 at 07:37
Thanks! @L3viathan's style of answer may also be useful, if you need `y` to be exactly equal to `a` for some reason. – Joe Halliwell Sep 30 '19 at 07:40

score 0 · Answer 4 · answered Sep 30 '19 at 07:38

0

thank you everybody for perfect answers. the final code as shown below

from decimal import Decimal, getcontext

a = Decimal(2172533123329917620172861889109867406049975387320169322608751844460406222231626416365984808960)

# Increase the precision
getcontext().prec = 100

x = a.log10()
y = 10 ** x
print(y)
print('{:.0f}'.format(y))

answered Sep 30 '19 at 07:38

hani

19
6

stackoverflow is not normal forum and we rather don't use place for answers/solutions to show final code but we append it in question - sometimes with text "EDIT:" – furas Sep 30 '19 at 07:48

How i can display the value of suffix e+number in python?

4 Answers4