問題描述
我有一個 pandas 數據框，其中包含根據兩列(A 和 B)的重復值:
I have a pandas dataframe which contains duplicates values according to two columns (A and B):
A B C
1 2 1
1 2 4
2 7 1
3 4 0
3 4 8
我想刪除在 C 列中保持最大值的行的重復項.這將導致:
I want to remove duplicates keeping the row with max value in column C. This would lead to:
A B C
1 2 4
2 7 1
3 4 8
我不知道該怎么做.我應該使用 drop_duplicates() 嗎?
I cannot figure out how to do that. Should I use drop_duplicates(), something else?
推薦答案
你可以使用 group by:
You can do it using group by:
c_maxes = df.groupby(['A', 'B']).C.transform(max)
df = df.loc[df.C == c_maxes]
c_maxes 是每個組中 C 最大值的Series，但長度和索引相同df.如果您還沒有使用過 .transform，那么打印 c_maxes 可能是一個好主意，看看它是如何工作的.
c_maxes is a Series of the maximum values of C in each group but which is of the same length and with the same index as df. If you haven't used .transform then printing c_maxes might be a good idea to see how it works.
使用 drop_duplicates 的另一種方法是
Another approach using drop_duplicates would be
df.sort('C').drop_duplicates(subset=['A', 'B'], take_last=True)
不確定哪個更有效，但我猜是第一種方法，因為它不涉及排序.
Not sure which is more efficient but I guess the first approach as it doesn't involve sorting.
從 pandas 0.18 開始，第二個解決方案是
From pandas 0.18 up the second solution would be
df.sort_values('C').drop_duplicates(subset=['A', 'B'], keep='last')
或者，或者，
df.sort_values('C', ascending=False).drop_duplicates(subset=['A', 'B'])
無論如何，groupby 解決方案的性能似乎要好得多:
In any case, the groupby solution seems to be significantly more performing:
%timeit -n 10 df.loc[df.groupby(['A', 'B']).C.max == df.C]
10 loops, best of 3: 25.7 ms per loop
%timeit -n 10 df.sort_values('C').drop_duplicates(subset=['A', 'B'], keep='last')
10 loops, best of 3: 101 ms per loop
這篇關于從數據框中刪除重復項，基于兩列 A，B，在另一列 C 中保持具有最大值的行的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！