【发布时间】:2020-06-21 19:59:02
【问题描述】:
我有一本像下面这样的字典:
{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}
想要得到这样的输出数据帧
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
我的失败尝试: 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 以及它们的值都被当作 dict.values()。如何将这些字母部分分隔为 Dataframe 列和数值部分是它们自己的列值?
【问题讨论】:
标签: python pandas dictionary