查找pandas df中timedelta对象的均值和标准差

Question

我想从下面显示的两列中计算一个mean和standard deviation一个timedelta银行dataframe.当我运行代码(也显示如下)时,我得到以下错误:

pandas.core.base.DataError: No numeric types to aggregate

我的数据帧:

   bank                          diff
   Bank of Japan                 0 days 00:00:57.416000
   Reserve Bank of Australia     0 days 00:00:21.452000
   Reserve Bank of New Zealand  55 days 12:39:32.269000
   U.S. Federal Reserve          8 days 13:27:11.387000

我的代码:

means = dropped.groupby('bank').mean()
std = dropped.groupby('bank').std()

Answer 1

Pandasmean()和其他聚合方法支持numeric_only=False参数。

dropped.groupby('bank').mean(numeric_only=False)

在这里找到：Python DataFrame 中 Timedelta 值的聚合

Answer 2

您需要转换timedelta为某个数值,例如int64通过values最准确的值,因为转换ns为以下数字表示的数字timedelta:

dropped['new'] = dropped['diff'].values.astype(np.int64)

means = dropped.groupby('bank').mean()
means['new'] = pd.to_timedelta(means['new'])

std = dropped.groupby('bank').std()
std['new'] = pd.to_timedelta(std['new'])

另一种解决方案是值转换seconds的total_seconds,但是这是不准确的:

dropped['new'] = dropped['diff'].dt.total_seconds()

means = dropped.groupby('bank').mean()

Answer 3

无需timedelta来回转换。Numpy和Pandas可以更快地为您无缝完成此操作。使用您的dropped DataFrame：

import numpy as np

grouped = dropped.groupby('bank')['diff']

mean = grouped.apply(lambda x: np.mean(x))
std = grouped.apply(lambda x: np.std(x))