Guillaume Thomas - 1 year ago 132

Python Question

Given a

`pandas.Series`

`In [14]: x`

Out[14]:

0 2015-11-03

1 2015-11-17

2 2015-12-08

3 2015-12-22

4 2016-01-05

dtype: datetime64[ns]

I want to

`In [14]: x`

Out[14]:

0 2015-11-02

1 2015-11-16

2 2015-12-07

3 2015-12-21

4 2016-01-04

dtype: datetime64[ns]

Answer Source

You can construct a `TimedeltaIndex`

using the `dt.dayofweek`

attribute as `Monday`

is day 0. Any day of week greater than this will result in timedeltas equivalent to the number of days to subtract:

```
In [49]:
s = s - pd.TimedeltaIndex(s.dt.dayofweek, unit='D')
s
Out[49]:
index
0 2015-11-02
1 2015-11-16
2 2015-12-07
3 2015-12-21
4 2016-01-04
Name: date, dtype: datetime64[ns]
```

E.g. for `2015-11-05`

, `2015-11-04`

, `2015-11-03`

, `2015-11-02`

the output of `pd.TimedeltaIndex(s.dt.dayofweek, unit='D')`

is `3 days`

, `2 days`

, `1 days`

, `0 days`

, respectively.