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Python Decimal module not rounding as expected

I am trying to use the Decimal module to do some FX calculations (instead of using floats).
However when I do the following, I do not get the expected value output:
>>> from decimal import Decimal
>>> x = Decimal(1.3755)
>>> y = Decimal(1.2627)
>>> z = y/(1/x)
>>> print(z)
1.736843849999999839084452447
>>>
The output should be: 1.73684385
I thought using Decimals would correct this issue. How can I resolve this rounding issue?

from decimal import Decimal
x = Decimal('1.3755')
y = Decimal('1.2627')
print(y/(1/x))
# 1.736843850000000000000000000
Use str to construct Decimal instead of real number.

String decimal objects are required:
>>> from decimal import Decimal
>>> x = Decimal('1.3755')
>>> y = Decimal('1.2627')
>>> z = y/(1/x)
>>> print(z)
1.736843850000000000000000000
>>>
Otherwise they would still be treated as regular python floats.
As #schwobaseggl mentioned:
Note that the float is passed to the Decimal constructor, so the precision is lost there already. The Decimal constructor has no chance of knowing the desired precision. The result is still a decimal object, but with the precision of the passed float.

you can use 0:.nf to round the numbers like below
:
print(z)
# 1.736843849999999839084452447
print("{0:.8f}".format(z))
# 1.73684385

try this, this will helps you, set your precision level
>>> from decimal import *
>>> getcontext().prec = 6
>>> Decimal(1) / Decimal(7)
Decimal('0.142857')
>>> getcontext().prec = 28
>>> Decimal(1) / Decimal(7)
Decimal('0.1428571428571428571428571429')

Decimal fixed precision

I would like to use decimal in currency calculations, so I would like to work on exactly two numbers after comma. Initially I thought that prec of decimal's context refers to that property, but after a few experiments I feel a bit confused.
Experiment #1:
In [1]: import decimal
In [2]: decimal.getcontext().prec = 2
In [3]: a = decimal.Decimal('159.9')
In [4]: a
Out[4]: Decimal('159.9')
In [5]: b = decimal.Decimal('200')
In [6]: b
Out[6]: Decimal('200')
In [7]: b - a
Out[7]: Decimal('40')
Experiment #2:
In [8]: decimal.getcontext().prec = 4
In [9]: a = decimal.Decimal('159.9')
In [10]: a
Out[10]: Decimal('159.9')
In [11]: b = decimal.Decimal('200')
In [12]: b
Out[12]: Decimal('200')
In [13]: b - a
Out[13]: Decimal('40.1')
Experiment #3: (prec is still set to 4)
In [14]: a = decimal.Decimal('159999.9')
In [15]: a
Out[15]: Decimal('159999.9')
In [16]: b = decimal.Decimal('200000')
In [17]: b
Out[17]: Decimal('200000')
In [18]: b - a
Out[18]: Decimal('4.000E+4')
Why does it work like in my examples? How should I work with decimals in my (currency calculations) case?

The precision sets the number of significant digits, which is not equivalent to the number of digits after the decimal point.
So if you have a precision of 2, you'll have two significant digits, so a number with 3 significant digits like 40.1 will be reduced to the upper two significant digits giving 40.
There's no easy way to set the number of digits after the decimal point with Decimal. However you could use a high precision and always round your results to two decimals:
>>> from decimal import Decimal, getcontext
>>> getcontext().prec = 60 # use a higher/lower one if needed
>>> Decimal('200') - Decimal('159.9')
Decimal('40.1')
>>> r = Decimal('200') - Decimal('159.9')
>>> round(r, 2)
Decimal('40.10')
The decimal FAQ also includes a similar question and answer (using quantize):
Q. In a fixed-point application with two decimal places, some inputs have many places and need to be rounded. Others are not supposed to have excess digits and need to be validated. What methods should be used?
A. The quantize() method rounds to a fixed number of decimal places. If the Inexact trap is set, it is also useful for validation:
>>> TWOPLACES = Decimal(10) ** -2 # same as Decimal('0.01')
>>> # Round to two places
>>> Decimal('3.214').quantize(TWOPLACES)
Decimal('3.21')
>>> # Validate that a number does not exceed two places
>>> Decimal('3.21').quantize(TWOPLACES, context=Context(traps=[Inexact]))
Decimal('3.21')
>>> Decimal('3.214').quantize(TWOPLACES, context=Context(traps=[Inexact]))
Traceback (most recent call last):
...
Inexact: None
Q. Once I have valid two place inputs, how do I maintain that invariant throughout an application?
A. Some operations like addition, subtraction, and multiplication by an integer will automatically preserve fixed point. Others operations, like division and non-integer multiplication, will change the number of decimal places and need to be followed-up with a quantize() step:
>>> a = Decimal('102.72') # Initial fixed-point values
>>> b = Decimal('3.17')
>>> a + b # Addition preserves fixed-point
Decimal('105.89')
>>> a - b
Decimal('99.55')
>>> a * 42 # So does integer multiplication
Decimal('4314.24')
>>> (a * b).quantize(TWOPLACES) # Must quantize non-integer multiplication
Decimal('325.62')
>>> (b / a).quantize(TWOPLACES) # And quantize division
Decimal('0.03')
In developing fixed-point applications, it is convenient to define functions to handle the quantize() step:
>>> def mul(x, y, fp=TWOPLACES):
... return (x * y).quantize(fp)
>>> def div(x, y, fp=TWOPLACES):
... return (x / y).quantize(fp)
>>> mul(a, b) # Automatically preserve fixed-point
Decimal('325.62')
>>> div(b, a)
Decimal('0.03')

The best method I've found is to set a high prec and use Decimal.quantize to round your result:
decimal.getcontext().prec=100
a = Decimal('400000.123456789')
b = Decimal('200000.0')
a-b
>>> Decimal('2.0000E+5')
(a-b).quantize(Decimal('0.01'))
>>> Decimal('200000.12')

Floor of very large floats in python

As Python has no limit over integers but has some limit over its floats.
How do I go about calculating floor of very large floats?
I am trying to calculate floor(A*B), where A is a small irrational number, possibly sqrt(2), e, sqrt(5) etc and B is a very large number in range 10^1000.

You can use decimal module:
>>> from decimal import Decimal
>>> from math import floor, sqrt
>>>
>>> d1 = Decimal(sqrt(2))
>>> d2 = Decimal(10**1000)
>>>
>>> result = d1 * d2
>>> floor(result)
You can also set the precision for the decimal using getcontext().prec in order to get more precise result.
>>> from decimal import *
>>> getcontext().prec = 100
>>> d1 = Decimal(2).sqrt()
>>> d1
Decimal('1.414213562373095048801688724209698078569671875376948073176679737990732478462107038850387534327641573')

64 bit floats don't go above ~10^308. So your 10^1000 is definitely not going to fit, regardless of multiplication by any constant. (No 64-bit float can be small enough to make 10^1000 less than 10^308.) So your procedure isn't going to work for floats.
Consider using the decimal module. E.g.:
import decimal
import math
a = decimal.Decimal("10") ** 10000
b = decimal.Decimal("0.123")
math.floor(a*b)

Python decimal precision and rounding

I want to limit the precision of my float and use it as an exact value for next operations, but I'm doing something wrong with decimal module.
from decimal import *
factor = 350/float(255)
# Settings
getcontext().prec = 1
getcontext().rounding = ROUND_UP
print Decimal(factor)
Python output:
1.372549019607843145962533526471816003322601318359375
Expected and wanted result: 1.4

From Python Documentation: (https://docs.python.org/2/library/decimal.html)
The significance of a new Decimal is determined solely by the number of digits input. Context precision and rounding only come into play during arithmetic operations.
getcontext().prec = 6
Decimal('3.0')
Decimal('3.0')
Decimal('3.1415926535')
Decimal('3.1415926535')
Decimal('3.1415926535') + Decimal('2.7182818285')
Decimal('5.85987')
getcontext().rounding = ROUND_UP
Decimal('3.1415926535') + Decimal('2.7182818285')
Decimal('5.85988')
so you need to use decimal in operation, not when printing result.

Clarification on the Decimal type in Python

Everybody knows, or at least, every programmer should know, that using the float type could lead to precision errors. However, in some cases, an exact solution would be great and there are cases where comparing using an epsilon value is not enough. Anyway, that's not really the point.
I knew about the Decimal type in Python but never tried to use it. It states that "Decimal numbers can be represented exactly" and I thought that it meant a clever implementation that allows to represent any real number. My first try was:
>>> from decimal import Decimal
>>> d = Decimal(1) / Decimal(3)
>>> d3 = d * Decimal(3)
>>> d3 < Decimal(1)
True
Quite disappointed, I went back to the documentation and kept reading:
The context for arithmetic is an environment specifying precision [...]
OK, so there is actually a precision. And the classic issues can be reproduced:
>>> dd = d * 10**20
>>> dd
Decimal('33333333333333333333.33333333')
>>> for i in range(10000):
... dd += 1 / Decimal(10**10)
>>> dd
Decimal('33333333333333333333.33333333')
So, my question is: is there a way to have a Decimal type with an infinite precision? If not, what's the more elegant way of comparing 2 decimal numbers (e.g. d3 < 1 should return False if the delta is less than the precision).
Currently, when I only do divisions and multiplications, I use the Fraction type:
>>> from fractions import Fraction
>>> f = Fraction(1) / Fraction(3)
>>> f
Fraction(1, 3)
>>> f * 3 < 1
False
>>> f * 3 == 1
True
Is it the best approach? What could be the other options?

The Decimal class is best for financial type addition, subtraction multiplication, division type problems:
>>> (1.1+2.2-3.3)*10000000000000000000
4440.892098500626 # relevant for government invoices...
>>> import decimal
>>> D=decimal.Decimal
>>> (D('1.1')+D('2.2')-D('3.3'))*10000000000000000000
Decimal('0.0')
The Fraction module works well with the rational number problem domain you describe:
>>> from fractions import Fraction
>>> f = Fraction(1) / Fraction(3)
>>> f
Fraction(1, 3)
>>> f * 3 < 1
False
>>> f * 3 == 1
True
For pure multi precision floating point for scientific work, consider mpmath.
If your problem can be held to the symbolic realm, consider sympy. Here is how you would handle the 1/3 issue:
>>> sympy.sympify('1/3')*3
1
>>> (sympy.sympify('1/3')*3) == 1
True
Sympy uses mpmath for arbitrary precision floating point, includes the ability to handle rational numbers and irrational numbers symbolically.
Consider the pure floating point representation of the irrational value of √2:
>>> math.sqrt(2)
1.4142135623730951
>>> math.sqrt(2)*math.sqrt(2)
2.0000000000000004
>>> math.sqrt(2)*math.sqrt(2)==2
False
Compare to sympy:
>>> sympy.sqrt(2)
sqrt(2) # treated symbolically
>>> sympy.sqrt(2)*sympy.sqrt(2)==2
True
You can also reduce values:
>>> import sympy
>>> sympy.sqrt(8)
2*sqrt(2) # √8 == √(4 x 2) == 2*√2...
However, you can see issues with Sympy similar to straight floating point if not careful:
>>> 1.1+2.2-3.3
4.440892098500626e-16
>>> sympy.sympify('1.1+2.2-3.3')
4.44089209850063e-16 # :-(
This is better done with Decimal:
>>> D('1.1')+D('2.2')-D('3.3')
Decimal('0.0')
Or using Fractions or Sympy and keeping values such as 1.1 as ratios:
>>> sympy.sympify('11/10+22/10-33/10')==0
True
>>> Fraction('1.1')+Fraction('2.2')-Fraction('3.3')==0
True
Or use Rational in sympy:
>>> frac=sympy.Rational
>>> frac('1.1')+frac('2.2')-frac('3.3')==0
True
>>> frac('1/3')*3
1
You can play with sympy live.

So, my question is: is there a way to have a Decimal type with an infinite precision?
No, since storing an irrational number would require infinite memory.
Where Decimal is useful is representing things like monetary amounts, where the values need to be exact and the precision is known a priori.
From the question, it is not entirely clear that Decimal is more appropriate for your use case than float.

is there a way to have a Decimal type with an infinite precision?
No; for any non-empty interval on the real line, you cannot represent all the numbers in the set with infinite precision using a finite number of bits. This is why Fraction is useful, as it stores the numerator and denominator as integers, which can be represented precisely:
>>> Fraction("1.25")
Fraction(5, 4)

If you are new to Decimal, this post is relevant: Python floating point arbitrary precision available?
The essential idea from the answers and comments is that for computationally tough problems where precision is needed, you should use the mpmath module https://code.google.com/p/mpmath/. An important observation is that,
The problem with using Decimal numbers is that you can't do much in the way of math functions on Decimal objects

Just to point out something that might not be immediately obvious to everyone:
The documentation for the decimal module says
... The exactness carries over into arithmetic. In decimal floating point, 0.1 + 0.1 + 0.1 - 0.3 is exactly equal to zero.
(Also see the classic: Is floating point math broken?)
However, if we use decimal.Decimal naively, we get the same "unexpected" result
>>> Decimal(0.1) + Decimal(0.1) + Decimal(0.1) == Decimal(0.3)
False
The problem in the naive example above is the use of float arguments, which are "losslessly converted to [their] exact decimal equivalent," as explained in the docs.
The trick (implicit in the accepted answer) is to construct the Decimal instances using e.g. strings, instead of floats
>>> Decimal('0.1') + Decimal('0.1') + Decimal('0.1') == Decimal('0.3')
True
or, perhaps more convenient in some cases, using tuples (<sign>, <digits>, <exponent>)
>>> Decimal((0, (1,), -1)) + Decimal((0, (1,), -1)) + Decimal((0, (1,), -1)) == Decimal((0, (3,), -1))
True
Note: this does not answer the original question, but it is closely related, and may be of help to people who end up here based on the question title.