Iloc: Difference between revisions
Jump to navigation
Jump to search
Created page with "Find the latest value per ID for a dataframe like this: {| class="wikitable" style="width:60%" |- !Index !ID !date !value |- |0 |1 |2019-06-14 23:00:03 |7550.00 |- |0 |2 |201..." |
mNo edit summary |
||
| (2 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
Find the latest value per ID for a dataframe like this | Function for a [[Pandas#DataFrame]] locating an index(row) | ||
;Find the latest value per ID for a dataframe like this | |||
{| class="wikitable" style="width:60%" | {| class="wikitable" style="width:60%" | ||
| Line 13: | Line 15: | ||
|7550.00 | |7550.00 | ||
|- | |- | ||
| | |1 | ||
| | |1 | ||
|2019-06-15 00:00:03 | |2019-06-15 00:00:03 | ||
|7690.00 | |7690.00 | ||
| Line 30: | Line 32: | ||
|} | |} | ||
<code> | <code>df.groupby('ID').apply(lambda d: d.iloc[-1])</code> | ||
Latest revision as of 14:19, 30 December 2019
Function for a Pandas#DataFrame locating an index(row)
- Find the latest value per ID for a dataframe like this
| Index | ID | date | value |
|---|---|---|---|
| 0 | 1 | 2019-06-14 23:00:03 | 7550.00 |
| 1 | 1 | 2019-06-15 00:00:03 | 7690.00 |
| 23560 | 17 | 2019-12-15 18:10:43 | 25.50 |
| 23561 | 17 | 2019-12-16 18:10:45 | 25.52 |
df.groupby('ID').apply(lambda d: d.iloc[-1])