Question

我对python pandas和numpy有一些奇怪的问题。

>>> np.float64(1) * np.float64(85000) * np.float64(7.543709)
641215.26500000001

>>> round( np.float64(1) * np.float64(85000) * np.float64(7.543709), 2 )
641215.26000000001

>>> np.round( np.float64(1) * np.float64(85000) * np.float64(7.543709), 2 )
641215.26000000001

如何舍入以获得正确的结果641215.27？

Answer 1

Numpy的圆形方法支持偶数，看看精简的numpy源代码：

def round_(a, decimals=0, out=None):
    return around(a, decimals=decimals, out=out)

def around(a, decimals=0, out=None):
    """
    Evenly round to the given number of decimals.

    Notes
    -----
    For values exactly halfway between rounded decimal values, NumPy
    rounds to the nearest even value. Thus 1.5 and 2.5 round to 2.0,
    -0.5 and 0.5 round to 0.0, etc. Results may also be surprising due
    to the inexact representation of decimal fractions in the IEEE
    floating point standard [1]_ and errors introduced when scaling
    by powers of ten.

    Examples
    --------
    >>> np.around([0.37, 1.64])
    array([ 0.,  2.])
    >>> np.around([0.37, 1.64], decimals=1)
    array([ 0.4,  1.6])
    >>> np.around([.5, 1.5, 2.5, 3.5, 4.5]) # rounds to nearest even value
    array([ 0.,  2.,  2.,  4.,  4.])
    >>> np.around([1,2,3,11], decimals=1) # ndarray of ints is returned
    array([ 1,  2,  3, 11])
    >>> np.around([1,2,3,11], decimals=-1)
    array([ 0,  0,  0, 10])

    """

实施例

如果您需要打印字符串，可以对其进行格式化以给出正确的答案：

import numpy as np

num = np.float64(1) * np.float64(85000) * np.float64(7.543709)
print(num)
print(float("{0:.2f}".format(num)))
print(np.round(num, 2))
print()

num += 0.02
print(num)
print(float("{0:.2f}".format(num)))
print(np.round(num, 2))

给你

Answer 2

是的，但在使用数据框时无法使用round( float(num), 2 )：

为考试：df.first * df.second * df.third 那个案子怎么回事？你无法进行float(dt.first)？

这是一个解决方案：df.first.apply(lambda x: round(float(x), 2)) 但我觉得不快......

在numpy四舍五入？

2 个答案:

实施例