熊猫Groupby返回平均值但！ 排除NaN
[英]Pandas Groupby Return Average BUT! exclude NaN
问题描述
So Im trying to make sense of the pandas groupby function and to reduce a large data frame I have. 
因此，我试图利用pandas groupby函数并减少我拥有的大数据框。 Here is an example: 这是一个例子：
A B
2016-09-23 19:36:08+00:00 NaN 34.0
2016-09-23 19:36:11+00:00 NaN 33.0
2016-09-23 19:36:12+00:00 24.1 NaN
2016-09-23 19:36:14+00:00 NaN 34.0
2016-09-23 19:36:17+00:00 NaN 34.0
2016-09-23 19:36:20+00:00 NaN 34.0
2016-09-23 19:36:22+00:00 24.2 NaN
2016-09-23 19:36:23+00:00 NaN 34.0
2016-09-23 19:36:26+00:00 NaN 34.0
2016-09-23 19:36:29+00:00 NaN 34.0
2016-09-23 19:36:32+00:00 24.1 NaN
2016-09-23 19:36:33+00:00 NaN 34.0
2016-09-23 19:37:00+00:00 NaN 34.0
2016-09-23 19:37:02+00:00 24.1 NaN
So I have 2 data series "A" and "B" that were sampled at different rates with their sampling time as the index of the original data frame. 因此，我有两个数据系列“ A”和“ B”，它们以不同的采样率作为原始数据帧的索引以不同的速率采样。
I would like to now group the rows of the data frame by date/hour/minute and return the average of the data per minute. 我现在想按日期/小时/分钟对数据框的行进行分组，并返回每分钟数据的平均值。 Here the average should ignore the missing values in the data frame. 在此，平均值应忽略数据帧中的缺失值。
So for example, I would return something like this: 因此，例如，我将返回以下内容：
A B
2016-09-23 19:36:00+00:00 24 34.0
2016-09-23 19:37:00+00:00 24.1 33.0
Is it possible to do this with a built in pandas function? 是否可以使用内置的熊猫函数来做到这一点？
1 个解决方案
解决方案1
2 已采纳 2016-10-03 18:49:17
I think you need resample with Resampler.mean , which compute mean of groups, excluding missing values: 我认为您需要使用Resampler.mean resample Resampler.mean ，该Resampler.mean将计算组的平均值，不包括缺失值：
print (df.resample('1Min').mean())
A B
2016-09-23 19:36:00 24.133333 33.888889
2016-09-23 19:37:00 24.100000 34.000000
Another solution with groupby : groupby另一种解决方案：
print (df.groupby([pd.TimeGrouper('1Min')]).mean())
A B
2016-09-23 19:36:00 24.133333 33.888889
2016-09-23 19:37:00 24.100000 34.000000
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