python – 在特定列上的pandas上滚动平均值

stock  pop
Date
2016-01-04  325.316   82
2016-01-11  320.036   83
2016-01-18  299.169   79
2016-01-25  296.579   84
2016-02-01  295.334   82
2016-02-08  309.777   81
2016-02-15  317.397   75
2016-02-22  328.005   80
2016-02-29  315.504   81
2016-03-07  328.802   81
2016-03-14  339.559   86
2016-03-21  352.160   82
2016-03-28  348.773   84
2016-04-04  346.482   83
2016-04-11  346.980   80
2016-04-18  357.140   75
2016-04-25  357.439   77
2016-05-02  356.443   78
2016-05-09  365.158   78
2016-05-16  352.160   72
2016-05-23  344.540   74
2016-05-30  354.998   81
2016-06-06  347.428   77
2016-06-13  341.053   78
2016-06-20  363.515   80
2016-06-27  349.669   80
2016-07-04  371.583   82
2016-07-11  358.335   81
2016-07-18  362.021   79
2016-07-25  368.844   77
...             ...  ...

我想添加一个新的MA列,用于计算列pop的滚动平均值.我尝试了以下内容

df['MA']=data.rolling(5,on='pop').mean()

我收到一个错误

ValueError: Wrong number of items passed 2, placement implies 1

所以我想让我试试,如果它只是工作而不添加一列.我用了

data.rolling(5,on='pop').mean()

我得到了输出

stock  pop
Date
2016-01-04       NaN   82
2016-01-11       NaN   83
2016-01-18       NaN   79
2016-01-25       NaN   84
2016-02-01  307.2868   82
2016-02-08  304.1790   81
2016-02-15  303.6512   75
2016-02-22  309.4184   80
2016-02-29  313.2034   81
2016-03-07  319.8970   81
2016-03-14  325.8534   86
2016-03-21  332.8060   82
2016-03-28  336.9596   84
2016-04-04  343.1552   83
2016-04-11  346.7908   80
2016-04-18  350.3070   75
2016-04-25  351.3628   77
2016-05-02  352.8968   78
2016-05-09  356.6320   78
2016-05-16  357.6680   72
2016-05-23  355.1480   74
2016-05-30  354.6598   81
2016-06-06  352.8568   77
2016-06-13  348.0358   78
2016-06-20  350.3068   80
2016-06-27  351.3326   80
2016-07-04  354.6496   82
2016-07-11  356.8310   81
2016-07-18  361.0246   79
2016-07-25  362.0904   77
...              ...  ...

我似乎无法在列pop上应用Rolling mean.我究竟做错了什么？

df['new_col'] = data['column'].rolling(5).mean()

ac2001发布的答案并不是最有效的方法.他正在计算数据框中每一列的滚动平均值,然后他使用“pop”列分配“ma”列.以下第一种方法更有效：

%timeit df['ma'] = data['pop'].rolling(5).mean()
%timeit df['ma_2'] = data.rolling(5).mean()['pop']

1000 loops, best of 3: 497 µs per loop
100 loops, best of 3: 2.6 ms per loop

除非您需要在所有其他列上存储计算滚动方法,否则我不建议使用第二种方法.