C++与 Python 精度对比
尝试查找 num^num 的前 k 位数字的问题我用 C++ 和 Python
C++
long double intpart,num,f_digit,k;
cin>>num>>k;
f_digit= pow(10.0,modf(num*log10(num),&intpart)+k-1);
cout<<f_digit;
Python
(a,b) = modf(num*log10(num))
f_digits = pow(10,b+k-1)
print f_digits
输入
19423474 9
输出
C++ > 163074912
Python > 163074908
编写了相同的程序，我检查了结果，C++ 解决方案是准确的。 在http://www.wolframalpha.com/input/?i=19423474^19423474检查它
知道如何在Python中获得相同的精度吗???
编辑：我知道外部库包可以获得这种精度，但有任何原生解决方案吗？？？
Trying out a problem of finding the first k digits of a num^num I wrote the same program in C++ and Python
C++
long double intpart,num,f_digit,k;
cin>>num>>k;
f_digit= pow(10.0,modf(num*log10(num),&intpart)+k-1);
cout<<f_digit;
Python
(a,b) = modf(num*log10(num))
f_digits = pow(10,b+k-1)
print f_digits
Input
19423474 9
Output
C++ > 163074912
Python > 163074908
I checked the results the C++ solution is the accurate one.
Checked it at http://www.wolframalpha.com/input/?i=19423474^19423474
Any idea how can I get the same precision in Python ???
EDIT : I know about the external library packages to obtain this precision but any NATIVE solution ???
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Decimal 是一个内置的 Python 类，可以正确处理浮点（以 10 为基数，而不是 IEEE 7somethingsomething 标准）。我不知道它是否支持对数等等。
编辑：它确实支持对数“以及所有这些”。
你可以也设置它的精度。默认为 28 个位置，但可以根据需要设置任意大小。将其视为小数的 BigInt。
Decimalis a built in python class that handles floating points correctly (as base 10, not as IEEE 7somethingsomething standard). I don't know if it supports logarithms and all that though.Edit: It does indeed support logarithms "and all that".
You can set the precision of it as well. Default is 28 places, but it can be as large as you want. Think of it as a BigInt for decimals.
正如您所发现的，Python 浮点数在底层是双精度的。您将不得不求助于 C 代码或外部库，以获得更好的浮点精度。
GMP 库是一个很好的库，它有一个名为“GMPY”的 Python 包装器，可以在 PyPI 上找到
Python floats are doubles under the hood, as you discovered. You will have to resort to C code, or an external library, to get better floating-point precision.
The GMP library is a good one, and it has a python wrapper called 'GMPY', available on PyPI
一般来说，我会这样做。然而，对于您的示例数字来说，它的执行速度似乎不够快。
In general, I would do it this way. However, it doesn't seem to perform anywhere near fast enough for your example numbers.