ID values
111 reason1
111 reason2
111 reason3
222 reason2
222 reason4
222 reason5
df.drop_duplicates(["ID"], keep='???', inplace=True)
我知道的方法是使用drop_duplicates,但它只给我选项first
,last
。我想检查是否有reason2,然后保留记录与reason2,否则检查reason3等。基本上,有特定的顺序,如reason2,reason3,reason4等。
答案 0 :(得分:4)
根据评论,这可以是其中一个实现:(实现@brittenb的想法。)
priority_dict = {
'reason1':1,
'reason2':2,
'reason3':3,
'reason4':4,
'reason5':5
}
df['priority'] = df['values'].map(priority_dict)
df = df.sort_values(by=['ID', 'priority'])
df.drop_duplicates(['ID'], keep='first')
输出:
ID values priority
0 111 reason1 1
3 222 reason2 2
答案 1 :(得分:0)
使用定义顺序和排序的'category'dtype:
df['values'] = df['values'].astype('category', ordered=True)\
.cat.reorder_categories(['reason2',
'reason3',
'reason1',
'reason4',
'reason5'])
df.sort_values('values').drop_duplicates('ID', keep='first')
输出:
ID values
1 111 reason2
3 222 reason2