The code gives an error because the value of "var" is very close to zero, less than 1e-80. I tried to fix this error using "Import decimal *", but it didn't really work. Is there a way to tell Python to round a number to zero when float number is very close to zero, i.e. < 1e-50? Or any other way to fix this issue?
Thank you
CODE:
import math
H=6.6260755e-27
K=1.3807e-16
C=2.9979E+10
T=100.0
x=3.07175e-05
cst=2.0*H*H*(C**3.0)/(K*T*T*(x**6.0))
a=H*C/(K*T*x)
var=cst*math.exp(a)/((math.exp(a)-1.0)**2.0)
print var
OUTPUT:
Traceback (most recent call last):
File "test.py", line 11, in <module>
var=cst*math.exp(a)/((math.exp(a)-1.0)**2.0)
OverflowError: (34, 'Numerical result out of range')
To Kevin:
The code was edited with following lines:
from decimal import *
getcontext().prec = 7
cst=Decimal(2.0*H*H*(C**3.0)/(K*T*T*(x**6.0)))
a=Decimal(H*C/(K*T*x))
The problem is that (math.exp(a)-1.0)**2.0 is too large to hold as an intermediate result.
>>> (math.exp(a) - 1.0)**2.0
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
OverflowError: (34, 'Result too large')
However, for the value of a you are using,
>>> math.exp(a)/(math.exp(a)-1.0) == 1.0
True
so you can essentially cancel that part of the fraction, leaving
var = cst/(math.exp(a)-1.0)
which evaluates nicely to
>>> cst/(math.exp(a)-1.0)
7.932672271698049e-186
If you aren't comfortable rewriting the formula to that extent, use the associativity of the operations to avoid the large intermediate value. The resulting product is the same.
>>> cst/(math.exp(a)-1.0)*math.exp(a)/(math.exp(a)-1.0)
7.932672271698049e-186
I solved this issue but that will work only for this particular problem, not in general. The main issue is the nature of this function:
math.exp(a)/(math.exp(a)-1.0)**2.0
which decays very rapidly.
Problem can be easily solved restricting the value of "a" (which won't make any significant change in calculation). i.e.
if a>200:
var=0.0
else:
var=cst*math.exp(a)/((math.exp(a)-1.0)**2.0)
Related
I have built a code for a codewars problem. I think it is correct but it shows me an error I don't understand.
Can you tell me what am I doing wrong?
import math
def waterbombs(fire, w):
s=""
countx=0
for i in fire:
if i=="x":
countx+=1
elif i=="Y":
countx=0
return sum(math.ceil(countx/w))
waterbombs("xxYxx", 3)
This is the error:
Traceback (most recent call last):
File "D:\Curso Python Pildorasinformaticas\Ejercicios Codewars\Aerial Firefighting.py", line 16, in <module>
waterbombs("xxYxx", 3)
File "D:\Curso Python Pildorasinformaticas\Ejercicios Codewars\Aerial Firefighting.py", line 13, in waterbombs
return sum(math.ceil(countx/w))
TypeError: 'int' object is not iterable
[Finished in 0.2s]
Why are you doing sum(math.ceil(countx/w)) ?
What is the objective of the sum method here, since there is only value returned by math.ceil ?
The sum would throw that error if you pass a single value to it. You're supposed to pass a list of values of the sum method.
For eg: sum(5) would give you the same error you see above, but sum([5]) would return you 5.
I have this for-loop:
for i in range(1000000000, 1000000030):
foo(i)
When I execute it this error is given:
Traceback (most recent call last):
File "/CENSORED/Activity.py", line 11, in <module>
for i in range(1000000000, 10000000030):
OverflowError: range() result has too many items.
As far as I know, this range-object should have exactly 30 elements...
Where is the problem?
Edit:
I have removed the extra zero, now I get this:
Traceback (most recent call last):
File "/CENSORED/Activity.py", line 12, in <module>
factorizeInefficient(i)
MemoryError
Edit 2:
def factorizeInefficient(n):
teiler = list()
for i in range(n):
if i != 0:
if (n%i)==0:
teiler.append(i)
print teiler
Just found the solution myself: There is a range(n) object in this as well and this causes the memory Error...
An extra question: How did you guys know this was python 2? (Btw you were right...)
Copy/pasting the range() part of your code:
>>> len(range(1000000000, 10000000030))
9000000030
So there are actually about 9 billion elements in the range. Your first argument is presumably missing a zero, or second argument has a zero too many ;-)
count your zeros once again ;) I'd say it's one too much.
I am currently creating a compiler and would also like to implement the division of zero. I noticed the decimal module in python and thought it could be useful. The code below shows what I am trying to get at. Is there anyway to split up the expression and check for the division of 0 for both negative and positive numbers? thanks in advance.
if input negative int/0 = -infin
if pos int/0 = infin
if 0/0 = null
ect.
The python documentation says the default behaviour is to raise an exception.
>>> import decimal
>>> D = decimal.Decimal
>>> a = D("12")
>>> b = D("0")
>>> a/b
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python3.4/decimal.py", line 1350, in __truediv__
return context._raise_error(DivisionByZero, 'x / 0', sign)
File "/usr/lib/python3.4/decimal.py", line 4050, in _raise_error
raise error(explanation)
decimal.DivisionByZero: x / 0
>>>
But also, instead of raising an exception:
If this signal is not trapped, returns Infinity or -Infinity with the
sign determined by the inputs to the calculation.
As of the zero by zero division, it is not mathematically defined, so I don't know what you may want to do with it. Python would return another exception:
>>> b/b
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python3.4/decimal.py", line 1349, in __truediv__
return context._raise_error(DivisionUndefined, '0 / 0')
File "/usr/lib/python3.4/decimal.py", line 4050, in _raise_error
raise error(explanation)
decimal.InvalidOperation: 0 / 0
>>>
and if the signal is not trapped, it returns NaN. (Might not be useful to know whether the operation was a zero divided by zero or not).
I am trying to read financial data and store it. The place I get the financial data from stores the data with incredible precision, however I am only interested in 5 figures after the decimal point. Therefore, I have decided to use t = .quantize(cdecimal.Decimal('.00001'), rounding=cdecimal.ROUND_UP) on the Decimal I create, but I keep getting an InvalidOperation exception. Why is this?
>>> import cdecimal
>>> c = cdecimal.getcontext()
>>> c.prec = 5
>>> s = '45.2091000080109'
>>> # s = '0.257585003972054' works!
>>> t = cdecimal.Decimal(s).quantize(cdecimal.Decimal('.00001'), rounding=cdecimal.ROUND_UP)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
cdecimal.InvalidOperation: [<class 'cdecimal.InvalidOperation'>]
Why is there an invalid operation here? If I change the precision to 7 (or greater), it works. If I set s to be '0.257585003972054' instead of the original value, that also works! What is going on?
Thanks!
decimal version gives a better description of the error:
Python 2.7.2+ (default, Feb 16 2012, 18:47:58)
>>> import decimal
>>> s = '45.2091000080109'
>>> decimal.getcontext().prec = 5
>>> decimal.Decimal(s).quantize(decimal.Decimal('.00001'), rounding=decimal.ROUND_UP)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python2.7/decimal.py", line 2464, in quantize
'quantize result has too many digits for current context')
File "/usr/lib/python2.7/decimal.py", line 3866, in _raise_error
raise error(explanation)
decimal.InvalidOperation: quantize result has too many digits for current context
>>>
Docs:
Unlike other operations, if the length of the coefficient after the
quantize operation would be greater than precision, then an
InvalidOperation is signaled. This guarantees that, unless there is an
error condition, the quantized exponent is always equal to that of the
right-hand operand.
But i must confess i don't know what this means.
What is the easiest way to determine the maximum match length of a regular expression?
Specifically, I am using Python's re module.
E.g. for foo((bar){2,3}|potato) it would be 12.
Obviously, regexes using operators like * and + have theoretically unbounded match lengths; in those cases returning an error or something is fine. Giving an error for regexes using the (?...) extensions is also fine.
I would also be ok with getting an approximate upper bound, as long as it is always greater than the actual maximum length, but not too much greater.
Using pyparsing's invRegex module:
import invRegex
data='foo(bar{2,3}|potato)'
print(list(invRegex.invert(data)))
# ['foobarr', 'foobarrr', 'foopotato']
print(max(map(len,invRegex.invert(data))))
# 9
Another alternative is to use ipermute from this module.
import inverse_regex
data='foo(bar{2,3}|potato)'
print(list(inverse_regex.ipermute(data)))
# ['foobarr', 'foobarrr', 'foopotato']
print(max(map(len,inverse_regex.ipermute(data))))
# 9
Solved, I think. Thanks to unutbu for pointing me to sre_parse!
import sre_parse
def get_regex_max_match_len(regex):
minlen, maxlen = sre_parse.parse(regex).getwidth()
if maxlen >= sre_parse.MAXREPEAT: raise ValueError('unbounded regex')
return maxlen
Results in:
>>> get_regex_max_match_len('foo((bar){2,3}|potato)')
12
>>> get_regex_max_match_len('.*')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 3, in get_regex_max_match_len
ValueError: unbounded regex