Question

我有一个像这样的数据集：

year          artist           track  time date.entered  wk1  wk2
2000           Pac           Baby   4:22   2000-02-26   87   82
2000         Geher           The    3:15   2000-09-02   91   87
2000    three_DoorsDown      Kryptonite  3:53   2000-04-08   81   70
2000         ATeens          Dancing_Queen  3:44   2000-07-08   97   97
2000         Aaliyah         I_Dont_Wanna  4:15   2000-01-29   84   62
2000         Aaliyah         Try_Again  4:03   2000-03-18   59   53
2000         Yolanda         Open_My_Heart  5:30   2000-08-26   76   76

我想要的输出是这样的：

   year   artist          track  time       date  week  rank
0  2000      Pac           Baby  4:22 2000-02-26     1    87
1  2000      Pac           Baby  4:22 2000-03-04     2    82
6  2000   ATeens  Dancing_Queen  3:44 2000-07-08     1    97
7  2000   ATeens  Dancing_Queen  3:44 2000-07-15     2    97
8  2000  Aaliyah   I_Dont_Wanna  4:15 2000-01-29     1    84

基本上，我正在整理给定的广告牌数据。没有熊猫链，我可以很容易地做到这一点：

df = pd.read_clipboard()
df1 = (pd.wide_to_long(df, 'wk', i=df.columns.values[:5], j='week')
 .reset_index()
 .rename(columns={'date.entered': 'date', 'wk': 'rank'}))

df1['date'] = pd.to_datetime(df1['date']) + pd.to_timedelta((df1['week'] - 1) * 7, 'd')
df1 = df1.sort_values(by=['track', 'date'])

print(df1.head())

问题
有没有办法链接df1['date'] = pd.to_datetime(...)部分？这样整个操作才能融入单链？

Answer 1

使用assign：

df1 = (pd.wide_to_long(df, 'wk', i=df.columns.values[:5], j='week')
         .reset_index()
         .rename(columns={'date.entered': 'date', 'wk': 'rank'})
         .assign(date = lambda x: pd.to_datetime(x['date']) + 
                                  pd.to_timedelta((x['week'] - 1) * 7, 'd'))
         .sort_values(by=['track', 'date'])
         )

更改熊猫链中列的值

1 个答案: