前往小程序,Get更优阅读体验!
立即前往
首页
学习
活动
专区
圈层
工具
发布
首页
学习
活动
专区
圈层
工具
MCP广场
社区首页 >专栏 >numpy.allclose

numpy.allclose

作者头像
狼啸风云
修改2022-09-03 20:00:20
修改2022-09-03 20:00:20
86500
代码可运行
举报
运行总次数:0
代码可运行

numpy.allclose(a, b, rtol=1e-05, atol=1e-08, equal_nan=False)[source]

Returns True if two arrays are element-wise equal within a tolerance.

The tolerance values are positive, typically very small numbers. The relative difference (rtol * abs(b)) and the absolute difference atol are added together to compare against the absolute difference between a and b.

If either array contains one or more NaNs, False is returned. Infs are treated as equal if they are in the same place and of the same sign in both arrays.

Parameters:

a, b : array_like Input arrays to compare. rtol : float The relative tolerance parameter (see Notes). atol : float The absolute tolerance parameter (see Notes). equal_nan : bool Whether to compare NaN’s as equal. If True, NaN’s in a will be considered equal to NaN’s in b in the output array. New in version 1.10.0.

Returns:

allclose : bool Returns True if the two arrays are equal within the given tolerance; False otherwise.

See also

isclose, all, any, equal

Notes

If the following equation is element-wise True, then allclose returns True.

absolute(a - b) <= (atol + rtol * absolute(b))

The above equation is not symmetric in a and b, so that allclose(a, b) might be different from allclose(b, a) in some rare cases.

The comparison of a and b uses standard broadcasting, which means that a and b need not have the same shape in order for allclose(a, b) to evaluate to True. The same is true for equal but not array_equal.

Examples:

代码语言:javascript
代码运行次数:0
运行
复制
>>> np.allclose([1e10,1e-7], [1.00001e10,1e-8])
False
>>> np.allclose([1e10,1e-8], [1.00001e10,1e-9])
True
>>> np.allclose([1e10,1e-8], [1.0001e10,1e-9])
False
>>> np.allclose([1.0, np.nan], [1.0, np.nan])
False
>>> np.allclose([1.0, np.nan], [1.0, np.nan], equal_nan=True)
True
本文参与 腾讯云自媒体同步曝光计划,分享自作者个人站点/博客。
原始发表:2020/01/12 ,如有侵权请联系 cloudcommunity@tencent.com 删除

本文分享自 作者个人站点/博客 前往查看

如有侵权,请联系 cloudcommunity@tencent.com 删除。

本文参与 腾讯云自媒体同步曝光计划  ,欢迎热爱写作的你一起参与!

评论
登录后参与评论
0 条评论
热度
最新
推荐阅读
领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档