pandas 中所有 NaN 的总和返回零? [英] Sum across all NaNs in pandas returns zero?
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)
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
如果你说 min_count=1 那么总和的结果将是 nan
If you say min_count=1 then the result of the sum will be a nan
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