计算 timedelta 列表的中位数(或均值)

calculating the median (or mean) of a timedelta list

我正在尝试查找从 PANDAS 数据帧生成的 timeDelta 对象列表的中位数。我试过这样使用统计库:

newList= list(DF.sort_values(['TimeDelta'])['TimeDelta'])
TDmedian = (st.median(newList))

st 是我导入的统计库。

但是我得到错误:

`TypeError: unsupported operand type(s) for /: 'str' and 'int'`

我试着做一个函数来计算它: `

def date_median(date_list):
    length = len(date_list)
    print(length)
//Checks if the length is odd cause median in odd numbered lists is the middle value
    if length % 2 != 0:
        return date_list[length//2]
    else:
//If it's even, it'll take the middle value and the one above it and generate the mean
        print((length//2), (length//2+1))
        lower = date_list[length//2]
        upper = date_list[(length//2) +1]
        return (lower + upper)/2`

我是这样使用的:

`TAmedian = date_median(newList)`

我得到这个错误:

`TypeError: unsupported operand type(s) for /: 'str' and 'int'`

有没有更简单的方法来做到这一点,如果没有,那我做错了什么?

示例数据:

DF['TimeDelta'] = [0 days 00:00:36.35700000,0 days 00:47:11.213000000]

好的。它应该 有效。著名的遗言吧?

我怀疑您的数据框的那一列中有一些元素不是数字。它应该像这样工作:

In [17]: import pandas as pd                                                                                    

In [18]: tds = [timedelta(t) for t in range(5)]                                                                 

In [19]: x = list(range(5))                                                                                     

In [20]: df = pd.DataFrame({'x': x, 'time delta': tds})                                                         

In [21]: df                                                                                                     
Out[21]: 
   x time delta
0  0     0 days
1  1     1 days
2  2     2 days
3  3     3 days
4  4     4 days

In [22]: import numpy as np                                                                                     

In [23]: np.median(df['time delta'])                                                                            
Out[23]: numpy.timedelta64(172800000000000,'ns')

那么,您是否测试过数据框以查看列中是否有一些非数值?最简单的就是使用 info() 命令。它看起来应该与此类似。如果显示"Object",你需要找出原因。

In [24]: df.info()                                                                                              
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 5 entries, 0 to 4
Data columns (total 2 columns):
 #   Column      Non-Null Count  Dtype          
---  ------      --------------  -----          
 0   x           5 non-null      int64          
 1   time delta  5 non-null      timedelta64[ns]
dtypes: int64(1), timedelta64[ns](1)
memory usage: 208.0 bytes

In [25]: df.describe()                                                                                          
Out[25]: 
              x              time delta
count  5.000000                       5
mean   2.000000         2 days 00:00:00
std    1.581139  1 days 13:56:50.394919
min    0.000000         0 days 00:00:00
25%    1.000000         1 days 00:00:00
50%    2.000000         2 days 00:00:00
75%    3.000000         3 days 00:00:00
max    4.000000         4 days 00:00:00

这是关于寻找非数字值的好post:

Finding non-numeric rows in dataframe in pandas?

为什么要转换为 listpandas.DataFrame 有你需要的一切:

import pandas as pd

DF = pd.DataFrame({'TimeDelta': pd.to_timedelta(['0 days 00:00:36.35700000', 
                                                 '0 days 00:47:11.213000000'])})

DF['TimeDelta'].mean()
# Timedelta('0 days 00:23:53.785000')
DF['TimeDelta'].median()
# Timedelta('0 days 00:23:53.785000')

当然,如果你一开始就没有df,你也可以不用,比如

pd.to_timedelta(['0 days 00:00:36.35700000', '0 days 00:47:11.213000000']).median()