我有一本字典，如下所示：
{1: ds yhat yhat_lower yhat_upper
30 2015-08-09 49.908927 31.632462 66.742083
31 2015-08-16 49.750056 34.065527 67.069122
32 2015-08-23 49.591185 32.620258 67.403908
33 2015-08-30 49.432314 32.257891 67.541757
34 2015-09-06 72.395618 55.612973 89.711030
35 2015-09-13 49.114572 32.199945 66.255518
36 2015-09-20 48.955701 30.759960 66.118051,
2: ds yhat yhat_lower yhat_upper
30 2015-08-09 38.001931 23.583157 51.291784
31 2015-08-16 37.922999 25.370967 50.504328
32 2015-08-23 37.844068 23.743860 51.143868
33 2015-08-30 37.765136 24.903955 50.309284
34 2015-09-06 39.227773 25.089493 52.719935
35 2015-09-13 37.607273 24.370609 51.313454
36 2015-09-20 37.528341 23.395560 50.499454}
想要获取类似此输出的数据帧
ProductCode ds yhat yhat_lower yhat_upper
1 2015-08-09 49.908927 31.632462 66.742083
1 2015-08-16 49.750056 34.065527 67.069122
1 2015-08-23 49.591185 32.620258 67.403908
1 2015-08-30 49.432314 32.257891 67.541757
1 2015-09-06 72.395618 55.612973 89.711030
1 2015-09-13 49.114572 32.199945 66.255518
1 2015-09-20 48.955701 30.759960 66.118051,
2 2015-08-09 38.001931 23.583157 51.291784
2 2015-08-16 37.922999 25.370967 50.504328
2 2015-08-23 37.844068 23.743860 51.143868
2 2015-08-30 37.765136 24.903955 50.309284
2 2015-09-06 39.227773 25.089493 52.719935
2 2015-09-13 37.607273 24.370609 51.313454
2 2015-09-20 37.528341 23.395560 50.499454
我失败的尝试： new_df = pd。DataFrame(df.items((， columns=['ProductCode'， 'yhat']( 打印(new_df(
ProdutCode yhat
0 1 ds yhat yhat_lower yhat_upp...
1 2 ds yhat yhat_lower yhat_upp...
在字典中，所有的 ds yhat yhat_lower 和 yhat_upper head 及其值都被视为 dict.values((。如何将这些字母部分分隔为数据帧列，而数值部分是它们自己的列值？
让我们试试pd.concat
yourdf=pd.concat(d).reset_index(level=0)