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In python, is there hidden rules to control how to display the precision of decimal number

For python, do read this link: https://docs.python.org/3/tutorial/floatingpoint.html, "Floating Point Arithmetic: Issues and Limitations"
I do understand that there is mismatch(tiny difference) between a binary-represented float & exact-decimal represented float, ex.
exact-decimal represented float:: 1.005
python binary-represented float:: 1.00499999999999989341858963598497211933135986328125
here is what I typed in python:
>>> 1.005
1.005
>>> from decimal import Decimal
>>> Decimal(1.005)
Decimal('1.00499999999999989341858963598497211933135986328125')
Here is my question:
why python showed 1.005 when I type in 1.005? why it is not 1.00499999999999989341858963598497211933135986328125?
if you tell me that python round result to some digits after decimal point, then what is rounding rule for my situation? it looks there is default rounding rule when start python, if this default rounding rule exists, how to change it?
Thanks

When asked to convert the float value 1.0049999999999999 to string, Python displays it with rounding:
>>> x = 1.0049999999999999; print(x)
1.005
According to the post that juanpa linked, Python uses the David Gay algorithm to decide how many digits to show when printing a float. Usually around 16 digits are shown, which makes sense, since 64-bit floats can represent 15 to 17 digits of significance.
If you want to print a float with some other number of digits shown, use an f-string or string interpolation with a precision specifier (see e.g. Input and Output - The Python Tutorial). For instance to print x with 20 digits:
>>> print(f'{x:.20}')
1.0049999999999998934
>>> print('%.20g' % x)
1.0049999999999998934

Python f-string surprising results on floats

I am trying to format float numbers in a fixed point notation: x.xxx, three digits following the decimal point regardless of the value of the number. I am getting surprising results. The first in particular would suggest that it is giving me three significant places rather than three digits after the decimal point. How do I tell it what I really want?
>>> print(f"{.0987:5.03}")
0.0987
*expected: 0.099*
>>> print(f"{0.0:05.03}")
000.0
*expected: 0.000*
>>> print(f"{0.0:5.3}")
0.0

# added "3f" to specify decimals places
print(f"{.0987:5.3f}")
#expected: 0.099*
print(f"{0.9687:05.3f}")
#expected: 0.000*
print(f"{0.0:5.3f}")

Floating point division vs integer division gives different answer

I had some issues with a piece of code and ended up doing the following command line snippet.This was just an experiment and I didn't store such large values in any variable in the real code(modulo 10**9 +7).
>>> a=1
>>> for i in range(1,101):
... a=a*i
...
>>> b=1
>>> for i in range(1,51):
... b=b*i
...
>>> c=pow(2,50)
>>> a//(b*c)
2725392139750729502980713245400918633290796330545803413734328823443106201171875
>>> a/(b*c)
2.7253921397507295e+78
>>> (a//(b*c))%(10**9 +7)
196932377
>>> (a/(b*c))%(10**9 +7)
45708938.0
>>>
I don't understand why integer divison gives the correct output while floating point divison fails.
Basically I calculated: ( (100!) / ((50!)*(2^50)) ) % (10**9 +7)

Because of precision.
Integers and floats are coded differently. In particular, in python 3, integers can be arbitrarily large - the one you gave, for example, is more than 250 bits large when you convert it to binary. They're stored in a way that can accommodate however large they are.
However, floating-point numbers are constrained to a certain size - usually 64 bits. These 64 bits are divided into a sign (1 bit), mantissa, and exponent - the number of bits in the mantissa limit how precise the number can be. Python's documentation contains a section on this limitation.
So, when you do
(a//(b*c))%(10**9 +7)
you're performing that calculation with integers, which, again, are arbitrarily large. However, when you do this:
(a/(b*c))%(10**9 +7)
you're performing that calculation with a number that only has 18 significant digits - it's already imprecise, and doing more calculations with it only further corrupts the answer.
What you can do to avoid this, if you need to use very large floating-point numbers, is use python's decimal module (which is part of the standard library), which will not have these problems.

The reason is that integers are precise, but floats are limited by the floating point precision: Python2.7 default float precision

sometimes missing a cent when translating euros to euro cents

I have to translate euro's (in a string) to euro cents (int):
Examples:
'12,1' => 1210
'14,51' => 1451
I use this python function:
int(round(float(amount.replace(',', '.')), 2) * 100)
But with this amount '1229,84' the result is : 122983
Update
I use the solution from Wim, bacause I use integers in both Python / Jinja and javascript for currency artitmetic. See also the answer from Chepner.
int(round(100 * float(amout.replace(',', '.')), 2))
My questions was anwered by Mr. Me, who explained the above result.

What the Docs Say, and a simple explanation
I tried it out, and was surprised that this was happening. So I turned to the documentation, and there is a little note in there that says.
Note The behavior of round() for floats can be surprising: for
example, round(2.675, 2) gives 2.67 instead of the expected 2.68. This
is not a bug: it’s a result of the fact that most decimal fractions
can’t be represented exactly as a float.
Now what does that mean, most decimal fractions can't be represented as a float. Well the documentations follows up with a great link at explains this, but since you probably didn't come here to read a nerdy technical document, let me summarize what is going on.
Python uses the IEEE-754 floating point standard to represent floats. This standard compromises accuracy for speed. Some numbers cannot be accurately represented. For example .1 is actually represented as 0.1000000000000000055511151231257827021181583404541015625. Interestingly, .1 in binary is actually an infinitely repeating number, just like 1/3 is an infinitely repeating .333333.
An Under the Hood Case Study
Now on to your particular case. This was pretty fun to look into, and this is what I discovered.
first lets simplify what you where trying to do
>>> amount = '1229,84'
>>> int(round(float(amount.replace(',', '.')), 2) * 100)
>>> 122983
to
>>>int(1229.84 * 100)
>>> 122983
Sometimes Python1 is unable to 100% accurately display binary floating point numbers, for the same reason we are unable to display the fraction 1/3 as a decimal. When this happens Python hides any extra digits. .1 is actually stored as -0.100000000000000092, but Python will display it as .1 if you type it into the console. We can see those extra digits by doing int(1.1) - 1.13. we can apply this int(myNum) - myNum formula to most floating point numbers to see the extra hidden digits behind them.4. In your case we would do the following.
>>> int(1229.84) - 1229.84
-0.8399999999999181
1229.84 is actually 1229.8399999999999181. Continuing on.5
>>> 1229.84, 2) * 100
122983.99999999999 #there's all of our hidden digits showing up.
Now on to the last step. This is the part we are concerned about. Changing it back to an integer.
>>> int(122983.99999999999)
122983
It rounds downwards instead of upwards, however, if we never had multiplied it by 100, we would still have 2 more 9s at the end, and Python would round up.
>>> int(122983.9999999999999)
122984
??? Now what is going on. Why is Python rounding 122983.99999999999 down, but it rounds 122983.9999999999999 up? Well whenever Python turns a float into a integer it rounds down. However, you have to remember that to Python 122983.9999999999999 with the extra two 99s at the end is the same thing as 122984.0 For example.
>>> 122983.9999999999999
122984.0
>>> a = 122983.9999999999999
>>> int(a) - a
0.0
and without the two extra 99s on the end.
>>> 122983.99999999999
122983.99999999999
>>> a=122983.99999999999
>>> int(a) - a
-0.9999999999854481
Python is definitely treating 122983.9999999999999 as 122984.0 but not 122983.99999999999. Now back to casting 122983.99999999999 to an integer. Because we have created ourselves a decimal portion that is less than 122984 that Python sees as being a seperate number from 122984, and because casting to an integer always causes Python to round down, we get 122983 as a result.
Whew. That was a lot to go through, but I sure learned a lot writing this out, and I hope you did to. The solution to all of this is to use decimal numbers instead of floats which compromises speed for accuracy.
What about rounding? The original problem had some rounding in it as well -- it's useless. See appendix item 6.
The Solution
a) The easiest solution is to use the decimal module instead of floating point numbers. This is the preferred way of doing things in any finance or accounting program.
The documentation also mentioned the following solutions which I've summarized.
b) The exact value can be expressed and retrieved in a hexadecimal form via myFloat.hex() and float.fromhex(myHex)
c) The exact value can also be retrieved as a fraction through myFloat.as_integer_ratio()
d) The documentation briefly mentions using SciPy for floating point arithmitic, however this SO question mentions that SciPy's NumPy floats are nothing more than aliases to the built-in float type. The decimal module would be a better solution.
Appendix
1 - Even though I will often refer to Python's behavior, the things I talk about are part of the IEEE-754 floating point standard which is what the major programming languages use for their floating point numbers.
2 - int(1.1) - 1.1 gives me -0.10000000000000009, but according to the documentation .1 is really 0.1000000000000000055511151231257827021181583404541015625
3 - We used int(1.1) - 1.1 instead of int(.1) - .1 because int(.1) - .1 does not give us the hidden digits, but according to the documentation they should still be there for .1, hence I say int(someNum) -someNum works most of the time, but not all of the time.
4 - When we use the formula int(myNum) - myNum what is happening is that casting the number to an integer will round the number down so int(3.9) becomes 3, and when we minus 3 from 3.9 we are left with -.9. However, for some reason that I do not know, when we get rid of all the whole numbers, and we're just left with the decimal portion, Python decides to show us everything -- the whole mantissa.
5 - this does not really affect the outcome of our analysis, but when multiplying by 100, instead of the hidden digits being shifted over by 2 decimal places, they changed a little as well.
>>> a = 1229.84
>>> int(a) - a
-0.8399999999999181
>>> a = round(1229.84, 2) * 100
>>> int(a) - a
-0.9999999999854481 #I expected -0.9999999999918100?
6 - It may seem like we can get rid of all those extra digits by rounding to two decimal places.
>>> round(1229.84, 2) # which is really round(1229.8399999999999181, 2)
1229.84
But when we use our int(someNum) - someNum formula to see the hidden digits, they are still there.
>>> a = round(1229.84, 2)
>>> int(a) - a
-0.8399999999999181
This is because Python cannot store 1229.84 as a binary floating point number. It can't be done. So... rounding 1229.84 does absolutely nothing.

Don't use floating-point arithmetic for currency; rounding error for values that cannot be represented exactly will cause the type of loss you are seeing. Instead, convert the string representation to an integer number of cents, which you can convert to euros-and-cents for display as needed.
euros, cents = '12,1'.split(',') # '12,1' -> ('12', '1')
cents = 100*int(euros) + int(cents * 10 if len(cents) == 1 else 1) # ('12', '1') -> 1210
(Notice you'll need a check to handle cents without a trailing 0.)
display_str = '%d,%d' % divMod(cents, 100) # 1210 -> (12, 10) -> '12.10'
You can also use the Decimal class from the decimal module, which essentially encapsulates all the logic for using integers to represent fractional values.

As #wim mentions in a comment, use the Decimal type from the stdlib decimal module instead of the built in float type. Decimal objects do not have the binary rounding behavior that floats have and also have a precision that can be user defined.
Decimal should be used anywhere you are doing financial calculations or anywhere you need floating point calculations that behave like the decimal math people learn in school (as opposed to the binary floating point behavior of the built in float type).

How to check if a float value is within a certain range and has a given number of decimal digits?

How to check if a float value is within a range (0.50,150.00) and has 2 decimal digits?
For example, 15.22366 should be false (too many decimal digits). But 15.22 should be true.
I tried something like:
data= input()
if data in range(0.50,150.00):
return True

Is that you are looking for?
def check(value):
if 0.50 <= value <= 150 and round(value,2)==value:
return True
return False
Given your comment:
i input 15.22366 it is going to return true; that is why i specified the range; it should accept 15.22
Simply said, floating point values are imprecise. Many values don't have a precise representation. Say for example 1.40. It might be displayed "as it":
>>> f = 1.40
>>> print f
1.4
But this is an illusion. Python has rounded that value in order to nicely display it. The real value as referenced by the variable f is quite different:
>>> from decimal import Decimal
>>> Decimal(f)
Decimal('1.399999999999999911182158029987476766109466552734375')
According to your rule of having only 2 decimals, should f reference a valid value or not?
The easiest way to fix that issue is probably to use round(...,2) as I suggested in the code above. But this in only an heuristic -- only able to reject "largely wrong" values. See my point here:
>>> for v in [ 1.40,
... 1.405,
... 1.399999999999999911182158029987476766109466552734375,
... 1.39999999999999991118,
... 1.3999999999999991118]:
... print check(v), v
...
True 1.4
False 1.405
True 1.4
True 1.4
False 1.4
Notice how the last few results might seems surprising at first. I hope my above explanations put some light on this.
As a final advice, for your needs as I guess them from your question, you should definitively consider using "decimal arithmetic". Python provides the decimal module for that purpose.

float is the wrong data type to use for your case, Use Decimal instead.
Check python docs for issues and limitations. To quote from there (I've generalised the text in Italics)
Floating-point numbers are represented in computer hardware as base 2 (binary) fractions.
no matter how many base 2 digits you’re willing to use, some decimal value (like 0.1) cannot be represented exactly as a base 2 fraction.
Stop at any finite number of bits, and you get an approximation
On a typical machine running Python, there are 53 bits of precision available for a Python float, so the value stored internally when you enter a decimal number is the binary fraction which is close to, but not exactly equal to it.
The documentation for the built-in round() function says that it rounds to the nearest value, rounding ties away from zero.
And finally, it recommends
If you’re in a situation where you care which way your decimal halfway-cases are rounded, you should consider using the decimal module.
And this will hold for your case as well, as you are looking for a precision of 2 digits after decimal points, which float just can't guarantee.
EDIT Note: The answer below corresponds to original question related to random float generation
Seeing that you need 2 digits of sure shot precision, I would suggest generating integer random numbers in range [50, 15000] and dividing them by 100 to convert them to float yourself.
import random
random.randint(50, 15000)/100.0

Why don't you just use round?
round(random.uniform(0.5, 150.0), 2)

Probably what you want to do is not to change the value itself. As said by Cyber in the comment, even if your round a floating point number, it will always store the same precision. If you need to change the way it is printed:
n = random.uniform(0.5, 150)
print '%.2f' % n # 58.03

The easiest way is to first convert the decimal to string and split with '.' and check if the length of the character. If it is >2 then pass on. i.e. Convert use input number to check if it is in a given range.
a=15.22366
if len(str(a).split('.')[1])>2:
if 0.50 <= value <= 150:
<do your stuff>>