我的数据框大致如下:
>>> data
price currency
id
2 1050 EU
5 1400 EU
4 1750 EU
8 4000 EU
7 630 GBP
1 1000 GBP
9 1400 GBP
3 2000 USD
6 7000 USD
我需要为每种货币提供包含n
最高价产品的新数据框,其中n
取决于货币,并在另一个数据框中给出:
>>> select_number
number_to_select
currency
GBP 2
EU 2
USD 1
如果我必须选择相同数量的最高价格元素,我可以使用pandas.groupby
按货币对数据进行分组,然后使用分组对象的head
方法。
但是,head
只接受一个数字,而不是数组或某个表达式。
当然,我可以编写一个for
循环,但这样做我们会非常尴尬和低效。
如何以良好的方式做到这一点?
答案 0 :(得分:9)
您可以使用:
data = pd.DataFrame({'id': {0: 2, 1: 5, 2: 4, 3: 8, 4: 7, 5: 1, 6: 9, 7: 3, 8: 6}, 'price': {0: 1050, 1: 1400, 2: 1750, 3: 4000, 4: 630, 5: 1000, 6: 1400, 7: 2000, 8: 7000}, 'currency': {0: 'EU', 1: 'EU', 2: 'EU', 3: 'EU', 4: 'GBP', 5: 'GBP', 6: 'GBP', 7: 'USD', 8: 'USD'}})
select_number = pd.DataFrame({'number_to_select': {'USD': 1, 'GBP': 2, 'EU': 2}})
print (data)
currency id price
0 EU 2 1050
1 EU 5 1400
2 EU 4 1750
3 EU 8 4000
4 GBP 7 630
5 GBP 1 1000
6 GBP 9 1400
7 USD 3 2000
8 USD 6 7000
print (select_number)
number_to_select
EU 2
GBP 2
USD 1
按dict
映射的解决方案:
d = select_number.to_dict()
d1 = d['number_to_select']
print (d1)
{'USD': 1, 'EU': 2, 'GBP': 2}
print (data.groupby('currency').apply(lambda dfg: dfg.nlargest(d1[dfg.name],'price'))
.reset_index(drop=True))
currency id price
0 EU 8 4000
1 EU 4 1750
2 GBP 9 1400
3 GBP 1 1000
4 USD 6 7000
溶液2:
print (data.groupby('currency')
.apply(lambda dfg: (dfg.nlargest(select_number
.loc[dfg.name, 'number_to_select'], 'price')))
.reset_index(drop=True))
id price currency
0 8 4000 EU
1 4 1750 EU
2 9 1400 GBP
3 1 1000 GBP
4 6 7000 USD
说明:
我认为调试是f
的最佳使用函数print
:
def f(dfg):
#dfg is DataFrame
print (dfg)
#name of group
print (dfg.name)
#select value from select_number
print (select_number.loc[dfg.name, 'number_to_select'])
#return top rows per groups
print (dfg.nlargest(select_number.loc[dfg.name, 'number_to_select'], 'price'))
return (dfg.nlargest(select_number.loc[dfg.name, 'number_to_select'], 'price'))
print (data.groupby('currency').apply(f))
currency id price
0 EU 2 1050
1 EU 5 1400
2 EU 4 1750
3 EU 8 4000
currency id price
0 EU 2 1050
1 EU 5 1400
2 EU 4 1750
3 EU 8 4000
EU
2
currency id price
3 EU 8 4000
2 EU 4 1750
currency id price
4 GBP 7 630
5 GBP 1 1000
6 GBP 9 1400
GBP
2
currency id price
6 GBP 9 1400
5 GBP 1 1000
currency id price
7 USD 3 2000
8 USD 6 7000
USD
1
currency id price
8 USD 6 7000
currency id price
currency
EU 3 EU 8 4000
2 EU 4 1750
GBP 6 GBP 9 1400
5 GBP 1 1000
USD 8 USD 6 7000
答案 1 :(得分:3)
这是一个解决方案:
select_number = select_number['number_to_select'] # easier to select from series
df.groupby('currency').apply(
lambda dfg: dfg.nlargest(select_number[dfg.name], columns='price')
)
修改 - 我从jezrael's answer得到了答案:我将dfg.currency.iloc[0]
替换为dfg.name
。
第二次修改 - 正如评论中所指出的,select_number
是一个数据框,因此我首先将其转换为系列。
MaxU和jezrael,感谢您的评论!
答案 2 :(得分:1)
你可以这样做:
df['rn'] = (df.sort_values(['price'], ascending=False)
.groupby('currency').cumcount() + 1
)
qry = (select_number
.reset_index()
.astype(str)
.apply(lambda x: '((currency=="{0[0]}") & (rn<={0[1]}))'.format(x), axis=1)
.str.cat(sep=' | ')
)
print(df.query(qry))
输出
In [147]: df.query(qry)
Out[147]:
price currency rn
id
4 1750 EU 2
8 4000 EU 1
1 1000 GBP 2
9 1400 GBP 1
6 7000 USD 1
说明:
rn
- 是一个帮助列 - 每个分区/组的row_number,按price
降序排列(在该组内)
qry
- 是动态生成的查询
In [149]: qry
Out[149]: '((currency=="EU") & (rn<=2)) | ((currency=="GBP") & (rn<=2)) | ((currency=="USD") & (rn<=1))'