熊猫：计算行和所有其他行之间的差异

Question

I am trying to create new columns corresponding to each date in my date column. Each column must contain the difference between the date and all other dates.

Input:

date
11-Sep-18
8-Jun-18
12-Sep-17
12-Jun-17

Output must be:

date        Diff_date1  Diff_date2  Diff_date3  Diff_date4
11-Sep-18   0             -95           -364       -456
8-Jun-18    95             0            -269       -361
12-Sep-17   364           269             0         -92
12-Jun-17   456           361             92         0

EDIT : Sub-question How can I do this with groupby? Each group has different number of rows in it.

ID  date        Diff_date1  Diff_date2  Diff_date3  Diff_date4
A  11-Sep-18    0             -95           -364       -456
A   8-Jun-18    95             0            -269       -361
A  12-Sep-17    364           269             0         -92
A  12-Jun-17    456           361             92         0


ID  date        Diff_date1  Diff_date2  Diff_date3  Diff_date4  Diff_date5
B  11-Jun-18    0             -123          -395       -456        -730
B   8-Feb-18    123            0            -272       -333        -607
B  12-May-17    395           272             0         -61        -335
B  12-Mar-17    456           333            274         0         -274
B  11-Jun-16    730           607            398        274         0

Answer 1

Setup

df['date'] = pd.to_datetime(df['date'])

---

You can make use of numpy here with subtract.outer

d = np.subtract.outer(df.date, df.date)

u = pd.DataFrame(d, index=df.date).rename(columns=lambda x: f'Diff_date{x+1}')

           Diff_date1 Diff_date2 Diff_date3 Diff_date4
date
2018-09-11     0 days    95 days   364 days   456 days
2018-06-08   -95 days     0 days   269 days   361 days
2017-09-12  -364 days  -269 days     0 days    92 days
2017-06-12  -456 days  -361 days   -92 days     0 days

If you want the opposite relationship (which you might based on your desired output), just multiply by -1

Answer 2

Using numpy broadcast

s=pd.to_datetime(df.date)

df=pd.concat([df,pd.DataFrame((s.values[:,None]-s.values)/np.timedelta64(1, 'D'),index=df.index)],axis=1)
df
Out[193]: 
        date      0      1      2      3
0  11-Sep-18    0.0   95.0  364.0  456.0
1   8-Jun-18  -95.0    0.0  269.0  361.0
2  12-Sep-17 -364.0 -269.0    0.0   92.0
3  12-Jun-17 -456.0 -361.0  -92.0    0.0

Answer 3

Another way using numpy,

df['date'] = pd.to_datetime(df['date'])
columns = ['Diff_date_' + str(i) for i in np.arange(len(df['date']))]
pd.DataFrame(df['date'].values - df['date'].values[:,np.newaxis], index = df['date'], columns = columns)


            Diff_date_0 Diff_date_1 Diff_date_2 Diff_date_3
date                
2018-09-11  0 days      -95 days    -364 days   -456 days
2018-06-08  95 days     0 days      -269 days   -361 days
2017-09-12  364 days    269 days    0 days      -92 days
2017-06-12  456 days    361 days    92 days     0 days