Question

请指导我。我有以下数据框：

df
    dob         date                                         
1   8/11/1966   3/1/1990, 5/1/2000, 8/1/2010
2   6/13/1970   4/1/2014, 3/1/2016, 4/1/2017
3   10/10/2010  4/13/2017

我的目标是生成一个列，显示＆＃39; dob＆＃39;之间的年份差异。和＆＃39; date＆＃39;像这样的列：

df
    dob         date                            difference                  
1   8/11/1966   3/1/1990, 5/1/2000, 8/1/2010    23.570, 23.740, 23.992
2   6/13/1970   4/1/2014, 3/1/2016, 4/1/2017    43.833, 45.751, 46.836
3   10/10/2010  4/13/2017                       6.512

使用以下代码，

diff = (df['date'].sub(df['dob']))/365
diff = (diff / np.timedelta64(1, 'D')).astype(float)
df['difference'] = diff.round(3)

我能够计算单个日期存在时的差异，但是当一个单元格中有多个值用逗号分隔时，它不起作用。我怎样才能实现目标？先感谢您。

Answer 1

考虑数据框df

df = pd.DataFrame(dict(
        dob=['8/11/1996', '6/13/1970'],
        date=[['3/1/1990', '5/1/2000', '8/1/2010'],
              ['4/1/2014', '3/1/2016', '4/1/2017']]
    )).reindex_axis(['dob', 'date'], 1)

l = df.date.str.len()
ilvl0 = df.index.repeat(l)
ilvl1 = np.concatenate(l.apply(np.arange))
date = pd.Series(
    pd.to_datetime(np.concatenate(df.date.values)),
    [ilvl0, ilvl1]
)

difference = date.sub(
    dob, level=0).dt.days.div(365.25).groupby(level=0).apply(list)
df.assign(difference=difference)

         dob                            date            difference
0  8/11/1996  [3/1/1990, 5/1/2000, 8/1/2010]  [-6.45, 3.72, 13.97]
1  6/13/1970  [4/1/2014, 3/1/2016, 4/1/2017]   [43.8, 45.72, 46.8]

OLD ANSWER

date_df = pd.to_datetime(
    pd.DataFrame(df.date.values.tolist(), df.index).stack()
 ).unstack()

有点神奇......

df.assign(
    difference=date_df.sub(
        pd.to_datetime(df.dob), 0
    ).stack().dt.days.groupby(level=0).apply(list)
)

         dob                            date             difference
0 1996-08-11  [3/1/1990, 5/1/2000, 8/1/2010]    [-2355, 1359, 5103]
1 1970-06-13  [4/1/2014, 3/1/2016, 4/1/2017]  [15998, 16698, 17094]

如果你想要它多年而不是几天

df.assign(
    difference=date_df.sub(
        pd.to_datetime(df.dob), 0
    ).stack().apply(lambda x: x.days / 365.25).round(2).groupby(level=0).apply(list)
)

         dob                            date            difference
0  8/11/1996  [3/1/1990, 5/1/2000, 8/1/2010]  [-6.45, 3.72, 13.97]
1  6/13/1970  [4/1/2014, 3/1/2016, 4/1/2017]   [43.8, 45.72, 46.8]

当一个单元格中存在多个值时，计算列之间的年份差异

1 个答案: