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What is Decimal(0.70) in python?

+3 votes
What is Decimal(0.70) in python?
posted Feb 8, 2015 by Gnanendra Reddy

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can not understand the question looks like you are looking for the decimal module in python.

The decimal module provides support for decimal floating point arithmetic. Python has a complete documentation on it please look at

If you are looking something different please comment :)

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+5 votes

Using 1/3 as an example,

 >>> 1./3
 >>> print "%.50f" % (1./3)
 >>> print "%.50f" % (10./3)
 >>> print "%.50f" % (100./3)

which seems to mean real (at least default) decimal precision is limited to "double", 16 digit precision (with rounding error). Is there a way to increase the real precision, preferably as the default?
For instance, UBasic uses a "Words for fractionals", f, "Point(f)" system, where Point(f) sets the decimal display precision, .1^int(ln(65536^73)/ln(10)), with the last few digits usually garbage.
Using "90*(pi/180)*180/pi" as an example to highlight the rounding error (4 = UBasic's f default value):

 Point(2)=.1^09: 89.999999306
 Point(3)=.1^14: 89.9999999999944
 Point(4)=.1^19: 89.9999999999999998772
 Point(5)=.1^24: 89.999999999999999999999217
 Point(7)=.1^33: 89.999999999999999999999999999999823
 Point(10)=.1^48: 89.999999999999999999999999999999999999999999997686
 Point(11)=.1^52: 89.9999999999999999999999999999999999999999999999999632

If not in the core program, is there a higher decimal precision module that can be added?

+1 vote

I need to get 32 bit binary equivalent of a decimal and need to change the 0's to 1's and 1's to 0's

For Example
if the input is 2 
Output should be:
the 32bit equivalent of 2 :0000 0000 0000 0000 0000 0000 0000 0010
and the 1's compliment is:1111 1111 1111 1111 1111 1111 1111 1101

is there any pre-defined function to get the above results in python??

0 votes

I want to create a random float array of size 100, with the values in the array ranging from 0 to 5. I have tried random.sample(range(5),100) but that does not work. How can i get what i want to achieve?