I have a dataframe like this:
Date PlumeO Distance
2014-08-13 13:48:00 754.447905 5.844577
2014-08-13 13:48:00 754.447905 6.888653
2014-08-13 13:48:00 754.447905 6.938860
2014-08-13 13:48:00 754.447905 6.977284
2014-08-13 13:48:00 754.447905 6.946430
2014-08-13 13:48:00 754.447905 6.345506
2014-08-13 13:48:00 754.447905 6.133567
2014-08-13 13:48:00 754.447905 5.846046
2014-08-13 16:59:00 754.447905 6.345506
2014-08-13 16:59:00 754.447905 6.694847
2014-08-13 16:59:00 754.447905 5.846046
2014-08-13 16:59:00 754.447905 6.977284
2014-08-13 16:59:00 754.447905 6.938860
2014-08-13 16:59:00 754.447905 5.844577
2014-08-13 16:59:00 754.447905 6.888653
2014-08-13 16:59:00 754.447905 6.133567
2014-08-13 16:59:00 754.447905 6.946430
I'm trying to keep the date with the smallest distance, so drop the duplicates dates and keep the with the smallest distance.
Is there a way to achieve this in pandas' df.drop_duplicates
or am I stuck using if statements to find the smallest distance?
Sort by distances and drop by dates:
df.sort_values('Distance').drop_duplicates(subset='Date', keep='first')
Out:
Date PlumeO Distance
0 2014-08-13 13:48:00 754.447905 5.844577
13 2014-08-13 16:59:00 754.447905 5.844577
The advantage of these approaches is that it does not require a sort.
Option 1
You can identify the index values for the minimum values with idxmin
and you can use it within a groupby
. Use these results to slice your dataframe.
df.loc[df.groupby('Date').Distance.idxmin()]
Date PlumeO Distance
0 2014-08-13 13:48:00 754.447905 5.844577
13 2014-08-13 16:59:00 754.447905 5.844577
Option 2
You can use pd.DataFrame.nsmallest
to return the rows associated with the smallest distance.
df.groupby('Date', group_keys=False).apply(
pd.DataFrame.nsmallest, n=1, columns='Distance'
)
Date PlumeO Distance
0 2014-08-13 13:48:00 754.447905 5.844577
13 2014-08-13 16:59:00 754.447905 5.844577
我会说先对数据排序,然后删除重复的日期:
stripped_data = df.sort_values('distance').drop_duplicates('date', keep='first')
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