Truncating a float value to a specific number of decimal places, without rounding, can be achieved in Python using the 'truncate()' function. This function takes two arguments: the float value to be truncated and the number of decimal places to retain. Here's an implementation of the function that uses intelligent rounding:
def truncate(f, n): '''Truncates/pads a float f to n decimal places without rounding''' s = '{}'.format(f) if 'e' in s or 'E' in s: return '{0:.{1}f}'.format(f, n) i, p, d = s.partition('.') return '.'.join([i, (d+'0'*n)[:n]])
The function converts the float to a string, ensuring that the full precision is preserved. It then divides the string into three parts: the integer part, the decimal point, and the decimal digits. It retains the decimal digits, padding with zeros if necessary, and rejoins the parts to form the truncated string.
In certain cases, it's not possible to exactly represent a float in decimal notation. For example, 0.3 exactly, can be represented accurately as an IEEE float but not as a decimal. In such cases, the function uses the default string formatting operation provided by Python 2.7 and higher. This operation intelligently selects the closest decimal representation of the float, without rounding.
If you're using Python versions earlier than 2.7 or 3.0, you can truncate the float and round it to a large number of decimal places (e.g., 12) before formatting the result as a string. This approach is less precise but works well in most cases:
def truncate(f, n): '''Truncates/pads a float f to n decimal places without rounding''' s = '%.12f' % f i, p, d = s.partition('.') return '.'.join([i, (d+'0'*n)[:n]])
Note that the intelligent rounding may produce false positives, where numbers are rounded that were not intended to be. To avoid this, a fixed precision can be used. Additionally, earlier Python versions may produce inaccurate decimal representations. In such cases, using the 12-digit workaround is recommended.
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