ex_array = [-8.23294593e-02, -4.07239507e-02, 6.08131029e-02, 2.72433402e-02,
-4.73587631e-02, 5.15452252e-02, 1.32902476e-01, 1.22322232e-01,
2.71845990e-02, -1.16927038e-01, -2.62239877e-01, -1.46526396e-01,
-1.82859136e-01, -1.02089602e-01, -1.91863501e-04, -5.42572200e-02,
-1.41798506e-01, 2.32538185e-02, 1.44525705e-01, 1.33945461e-01,
5.01618120e-02, -1.32664337e-01, -2.97395262e-01, -1.02531532e-01,
-7.80204566e-02, -5.46991495e-02, 1.05868862e-01, 7.25526818e-03,
5.04192997e-02, 7.41281286e-02, 1.75069159e-01, 1.64488914e-01,
7.55396024e-02, -6.23800645e-02, -1.76950023e-01, -5.91491004e-02,
-4.00535768e-02, 6.59473071e-04, 5.98125666e-02, -1.49608356e-02,
-1.45519585e-02, 1.49876707e-01, 1.92880709e-01, 2.33158881e-01,
7.59751625e-02, -2.46659059e-02, -1.40025102e-01, -3.02416639e-02]
I need to compute the median for every 12 values.每个值表示一个月(从1月到12月),因此我想获得一年中每个月的中值.像这样:
方法:
-
我可以将数组转换为数据帧,并添加一个表示每个月的新列.然后,按月份分组并计算中位数.但我觉得这必须是一个更简单的解决方案.
-
我想的另一个解决方案是转换为数据帧,每12个值切片一次,每次都从不同的值开始.它可以工作,但我在获取可行的数组时遇到了问题.为前三个月添加示例:
'''
sol_array = []
sol_array.append(pd.DataFrame(ex_array).iloc[0::12].median().to_string())
sol_array.append(pd.DataFrame(ex_array).iloc[1::12].median().to_string())
sol_array.append(pd.DataFrame(ex_array).iloc[2::12].median().to_string())
但这就是结果.0和撇号不应该在那里.
['0 -0.075844',
'0 -0.089111',
'0 0.042705',
'0 0.002147',
'0 -0.010528',
'0 0.109443',
'0 0.198334',
'0 0.20983',
'0 0.075139',
'0 -0.062405']
所以,做you know another way to obtain the same outcome.我只有120个值,所以手动排列组(只有10组)仍然可行,但我觉得这不是一个理想的解决方案.
Or, do you know how to correct the above method I and obtain a workable array?