pandas 中所有 NaN 的总和返回零?

编程入门行业动态更新时间:2024-10-13 22:19:34

本文介绍了 pandas 中所有 NaN 的总和返回零?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！问题描述

我正在尝试对 Pandas 数据框的列求和，当我在每一列中都有 NaN 时，总和 = 零；根据文档，我希望 sum = NaN .这是我所拥有的:

I'm trying to sum across columns of a Pandas dataframe, and when I have NaNs in every column I'm getting sum = zero; I'd expected sum = NaN based on the docs. Here's what I've got:

In [136]: df = pd.DataFrame() In [137]: df['a'] = [1,2,np.nan,3] In [138]: df['b'] = [4,5,np.nan,6] In [139]: df Out[139]: a b 0 1 4 1 2 5 2 NaN NaN 3 3 6 In [140]: df['total'] = df.sum(axis=1) In [141]: df Out[141]: a b total 0 1 4 5 1 2 5 7 2 NaN NaN 0 3 3 6 9

pandas.DataFrame.sum 文档说如果整个行/列都是 NA，结果将是 NA"，所以我不明白为什么索引 2 的total"= 0 而不是 NaN.我是什么不见了?

The pandas.DataFrame.sum docs say "If an entire row/column is NA, the result will be NA", so I don't understand why "total" = 0 and not NaN for index 2. What am I missing?

推荐答案

pandas 文档 » API 参考 » DataFrame » pandas.DataFrame »

DataFrame.sum(self, axis=None, skipna=None, level=None, numeric_only=None, min_count=0, **kwargs)

min_count:整数，默认为 0

所需的有效值数量执行操作.如果少于 min_count 的非 NA 值是呈现结果将是 NA.

The required number of valid values to perform the operation. If fewer than min_count non-NA values are present the result will be NA.

0.22.0 新版:新增，默认为 0.这意味着全 NA 或空系列的总和为 0，并且全 NA 或空系列为 1.

New in version 0.22.0: Added with the default being 0. This means the sum of an all-NA or empty Series is 0, and the product of an all-NA or empty Series is 1.

引用 pandas 的最新文档，它说 min_count 对于所有 NA 系列都是 0

Quoting from pandas latest docs it says the min_count will be 0 for all-NA series