[英]Convert Python floating-point into C# decimal
我正在嘗試使用 pythonnet 將 Python 中的浮點數轉換為 C# 中的decimal
。
我嘗試了兩種方式:
from System import Decimal
from decimal import Decimal as PyDecimal
Decimal(PyDecimal("0.0234337688540165776476565071"))
# This lose every number under the floating-point
0
from System import Decimal from decimal import Decimal as PyDecimal Decimal(PyDecimal("0.0234337688540165776476565071")) # This lose every number under the floating-point 0
我應該怎么辦...?
從您問題中的示例來看,您似乎正在嘗試將字符串轉換為System.Decimal
。 為此, System.Decimal
有一個Parse
方法:
from System import Decimal
Decimal.Parse("0.0234337688540165776476565071")
注意:您可能還需要根據您的場景傳遞CultureInfo.InvariantCulture
。
如果您將字符串轉換為 python 浮點數,那么它將被截斷為 64 位,您將失去精度。 您將不得不為System.Decimal
使用不同的構造函數。 例如:
public Decimal (int lo, int mid, int hi, bool isNegative, byte scale);
跳過驗證,它可能看起來像
from System import Decimal
from System import Int32, Boolean, Byte
def str_to_csp_decimal(number: str) -> Decimal:
""" convert a string to a C# Decimal """
is_negative = number[0] == "-"
abs_value = number[1:] if is_negative else number
has_fraction = "." in abs_value
if has_fraction:
integer, fraction = abs_value.split(".")
scale = len(fraction) # denominator = 10 ** scale
numerator = int(integer + fraction) # no precision loss for integers
else:
scale = 0 # denominator == 1
numerator = int(abs_value)
assert numerator < (2 << 96), "Decimal only has 96 bits of precision"
# split the numerator into the lower, mid, and high 32 bits
mask = 0xFFFFFFFF
low = Int32(numerator & mask)
mid = Int32((numerator >> 32) & mask)
high = Int32((numerator >> 64) & mask)
return Decimal(low, mid, high, Boolean(is_negative), Byte(scale))
str_to_csp_decimal("0.0234337688540165776476565071").ToString()
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