(Most other related questions in the web concern conversion between C's longdouble and python's. This question is different.)
I do not see why I cannot correctly get a longdouble in python like this:
In [72]: import numpy as np
In [73]: np.longdouble(1e3000)
Out[73]: inf
It seems that I need to let my python console know 1e3000 is a longdouble instead of double. How can I do that
The problem is that by using an expression like ...(1e3000), the Python parser has to calculate what is inside the parentheses first, and pass the result to the function call. Long double is not a native type, therefore, the value inside the parentheses is inf - which is passed to the longdouble constructor. The fact the string version fails could maybe be considered a bug in NumPy - it indicates the string is converted to a Python float (which is a "float64" or "double" in C) internally, possibly using the normal Python float constructor.
The workaround is to build the long double object first, with a value that is compatble with a Python float, and them multiply it to get to the desired value. If you need to do that with several values, use a NumPy array instead of a single value:
>>> x = np.longdouble(10)
>>> x
10.0
>>> x **= 3000
>>> x
9.9999999999999999999e+2999
Python doesn't have "long doubles". By using scientific notation, you are making a float literal. Those cannot represent 1e3000, so you get inf. If you use integers, you might be able to do what you need: 10**3000.
Related
i want adding and subtracting this type of data: $12,587.30.which returns answer in same format.how can do this ?
Here is my code example:
print(int(col_ammount2.lstrip('$'))-int(col_ammount.lstrip('$')))
I removed $ sign and convert it to int but it gives me base 10 error.
You mentioned you want to do arithmetic operations to the numbers (addition/subtraction) so you probably want them in float instead. The difference between an integer (int) and float is that integers do not carry decimal points.
Additionally, as #officialaimm mentioned you need to remove the commas too, for example
float('$3,333.33'.replace('$', '').replace(',', ''))
will give you
3333.33
So putting it into your code
print(float(col_ammount2.lstrip('$').replace(',', ''))
- float(col_ammount.lstrip('$').replace(',', '')))
An additional note for when you parse a floating point number (same applies to integers too), you may want to watch out for empty values, i.e.
float('')
is bad. One of the things u can do in case col_amount and col_amount2 may be empty at some point is default them to 0 if that happens
float(col_amount.lstrip(...).replace(...) or 0)
You also want to read this to know about workaround to problems you may face with floating point arithmetic https://docs.python.org/3/tutorial/floatingpoint.html
There are two things you are missing here. Firstly python int(...) cannot parse numbers with commas so you will need to remove commas as well by using .replace(',',''). Secondly int() cannot parse floating point values you will have to use float(...) first and after that maybe typecast it to int using int or math.ceil, math.floor appropriately as per your choice and needs.
Maybe something like this will solve your problem:
col_ammount2='$1,587.30'
col_ammount = '$2,567.67'
print(int(float(col_ammount2.lstrip('$').replace(',','')))-int(float(col_ammount.lstrip('$').replace(',',''))))
If you are doing these sorts of things quite often in your code, making a function as such might be handy:
integerify_currency = lambda x:int(float(x.lstrip('$').replace(',','')))
I tried to compute math.exp(9500) but encountered an OverflowError: math range error (it's roughly 6.3e4125). From this question it seems like it's due to a too large float, the accepted answer says "(...) is slightly outside of the range of a double, so it causes an overflow".
I know that Python can deal with arbitrarily large integers (long type), is there a way to deal with arbitrarily large floats in the same manner ?
Edit : my original question was about using integers for calculating exp(n) but as Eric Duminil said, the simplest way to do that would be 3**n which doesn't provide any useful result. I know realize this question might be similar to this one.
I don't think it's possible to approximate exp() with integers. If you use 3**n instead of 2.71828182845905**n, your calculations will be completely useless.
One possible solution would be to use Sympy. According to the documentation:
There is essentially no upper precision limit
>>> from sympy import *
>>> exp(9500)
exp(9500)
>>> exp(9500).evalf()
6.27448493490172e+4125
You can also specify the desired precision:
>>> exp(9500).evalf(1000)
6.274484934901720177929867046175406311474380389941415760684209191232450360090766458256588885184199320756050569665785657269735313171886975309933254563488343491718198237894473901620914303565550450204805537225888529509352754121292701357622411614860860409639719786022989336837263283678476008817556351031696366815467221836948040042378034720460820127399855873232167818091083005170669482845098735176209372328114732133251096196535355946589133977397512846130629857604295369747597459602137604440011394793443041829253598478244189078131130488653468669559814695095974271938947640276013215753183113041899037415404445478806695965167014404297848725756879184380559837391976534521522360723388582608454995349380217499779247330557664230806254642768796486899322646423713763772064068933790640394967085887914192401473425799354391464743910233873602389444180426155866237536459654917521713769608318128404177877383203786348495822099924812081683286880293701785567962687838594752986160305764297117036426951203418854463404773701882e+4125
With exp(9500).evalf(5000), you even get the integer closest to exp(9500).
Here's another way to calculate the result with Python:
exp(9500)
is too big.
But log10(exp(9500)) isn't. You cannot calculate it this way in Python, but log10(exp(9500)) is log(exp(9500))/ln(10), which is 9500/ln(10):
>>> from math import log
>>> 9500/log(10)
4125.797578080892
>>> int(9500/log(10))
4125
>>> 10**(9500/log(10) % 1)
6.274484934896202
This way, you can calculate that exp(9500) is 6.27448493 * 10**4125 in plain Python, without any library!
try long type.
int type has been remove from python since 3.0 version.
so far this is what i found:
from random import randint
randint(100000000000000000000, 999999999999999999999)
the output is:
922106555361958347898L
but i do not want that L there..
i can only use this as an int if there is no "L" there at the end of it.
UPDATE
would it be a better idea to generate two small numbers and then combine them if the goal is to simply have a random number that is 30 digits long ?
The reason there's an L there is because this is too large to fit into an int,* so it's a long. See Numeric Types — int, long, float, complex in the docs for more details.
So, why do you get that L, and how do you get rid of it?
Python has two different ways to turn a value into a string representation:
repr either returns the canonical source-code representation, or something like <__main__.Eggs at 0x10fbd8908>).
str returns a human-friendly representation.
For example, for strings, str('abc') is abc, while repr('abc') is 'abc'.
And for longs, str(1L) is 1, while repr(1L) is 1L.
When you just type an expression at the interactive prompt, it uses repr. But when you use the print command, it uses str. So, if you want to see the value without the L, just print it:
print randint(100000000000000000000, 999999999999999999999)
If you want to, e.g., save the string in a variable or write it to a file, you have to call str explicitly.
But if you just want to use it as a number, you don't have to worry about this at all; it's a number, and int and long values can be intermixed freely (as of Python 2.3 or so).
And if you're trying to store it in a MySQL database, whichever MySQL interface you use won't care whether you're giving it int values or long, as long as they fit into the column type.**
Or you could upgrade to Python 3.x, where there is no separate long type anymore (all integers are int, no matter how big) and no L suffix.
* The exact cutoff isn't documented anywhere, but at least for CPython, it's whatever fits into a C long on your platform. So, on most 64-bit platforms, the max value is (1<<63)-1; on the other 64-bit platforms, and all 32-bit platforms, it's (1<<31)-1. You can see for yourself on your platform by printing sys.maxint. At any rate, your number takes 70 bits, so unless someone ports Python 2.x to a platform with 128-bit C longs, it won't fit.
** Note that your values are too big to fit into even a MySQL BIGINT, so you're going to be using either DECIMAL or NUMERIC. Depending on which interface you're using, and how you've set things up, you may have to convert to and from strings manually. But you can do that with the str and int functions, without worrying about which values fit into the int type and which don't.)
If you're on the interactive prompt, explicitly print the value. The repr of the value has an L, but the str of the value doesn't.
>>> 922106555361958347898
922106555361958347898L
>>> print 922106555361958347898
922106555361958347898
The output in the REPL has an L suffixed; if you print the value, it is not displayed.
>>> from random import randint
>>> print randint(100000000000000000000, 999999999999999999999)
106315199286113607384
>>>
What is the proper/accepted way to print and convert a numpy.float64 to a string? I've noticed just using print or str() will lose some precision. However, repr maintains the full precision. For example:
>>> import numpy
>>> print numpy.float64('6374.345407799015')
6374.3454078
>>> print repr(numpy.float64('6374.345407799015'))
6374.3454077990154
I assume that just calling print turns into calling str() on the float64 object. So is __str__() for numpy.float64 implemented with something like '%s' % (float(self)) or somehow casts the float64 with Python's built-in float()? I tried to quickly look around the numpy source for this but wasn't immediately obvious what was happening.
I've always thought repr() should return valid Python code that could be used by eval() to re-create the object. Is this an accepted convention? Luckily in this case numpy does not follow this convention because repr() returns just the raw number as a string instead of something like "numpy.float64('6374.345407799015')".
So, all of this confuses me. What is the correct way to convert a numpy.float64 to a string and/or print it while guaranteeing you always have the same, full precision?
The astype method works well:
>>> numpy.float64('6374.345407799015').astype(str)
'6374.345407799015'
Look into numpy.set_printoptions. Specifically,
numpy.set_printoptions(precision=15)
I want to put numerics and strings into the same numpy array. However, I very rarely (difficult to replicate, but sometimes) run into an error where the numeric to string conversion results in a value that cannot back-translate into a decimal (ie, I get "9.8267567e", as opposed to "9.8267567e-5" in the array). This is causing problems after writing files. Here is an example of what I am doing (though on a much smaller scale):
import numpy as np
x = np.array(.94749128494582)
y = np.array(x, dtype='|S100')
My understanding is that this should allow 100 string characters, but sometimes I am seeing a cut-off after ~10. Is there another type that I should be assigning, or a way to limit the number of characters in my array (x)?
First of all, x = np.array(.94749128494582) may not be doing what you think because the argument passed into np.array should be some kind of sequence or something with the array interface. Perhaps you meant x = np.array([.94749128494582])?
Now, as for preserving the strings properly, you could solve this by using
y = np.array(x, dtype=object)
However, as Joe has mentioned in his comment, it's not very numpythonic and you may as well be using plain old python lists.
I would recommend to examine carefully why you seem to have this requirement to hold strings and numbers in the same array, it smells to me like you might have inappropriate data structures set up and could benefit from redesigning/refactoring. numpy arrays are for fast numerical operations, they are not really suited to be used for string manipulations or as some kind of storage/database.