如何替换组中的空值?
How could I replace null value In a group?
我创建了这个数据框,我计算了我正在寻找的差距,但问题是有些公寓价格相同,我得到的价格差为 0。我如何用差值替换值 0同组最低价
例如:
neighboorhood:a, bed:1, bath:1, price:5
neighboorhood:a, bed:1, bath:1, price:5
neighboorhood:a, bed:1, bath:1, price:3
neighboorhood:a, bed:1, bath:1, price:2
我得到 0,2,1,nan 的差价,我正在寻找 2,2,1,nan(简而言之,我不想比较价格相同的 2 个单位)
提前致谢,祝您愉快。
data=[
[1,'a',1,1,5],[2,'a',1,1,5],[3,'a',1,1,4],[4,'a',1,1,2],[5,'b',1,2,6],[6,'b',1,2,6],[7,'b',1,2,3]
]
df = pd.DataFrame(data, columns = ['id','neighborhoodname', 'beds', 'baths', 'price'])
df['difference_price'] = ( df.dropna()
.sort_values('price',ascending=False)
.groupby(['city','beds','baths'])['price'].diff(-1) )
我认为您可以先删除用于 groupby
和 diff
的所有列的重复项,在过滤后的数据中创建新列,最后使用左连接合并到原始数据:
df1 = (df.dropna()
.sort_values('price',ascending=False)
.drop_duplicates(['neighborhoodname','beds','baths', 'price']))
df1['difference_price'] = df1.groupby(['neighborhoodname','beds','baths'])['price'].diff(-1)
df = df.merge(df1[['neighborhoodname','beds','baths','price', 'difference_price']], how='left')
print (df)
id neighborhoodname beds baths price difference_price
0 1 a 1 1 5 1.0
1 2 a 1 1 5 1.0
2 3 a 1 1 4 2.0
3 4 a 1 1 2 NaN
4 5 b 1 2 6 3.0
5 6 b 1 2 6 3.0
6 7 b 1 2 3 NaN
或者您可以使用 lambda 函数回填每组的 0
值以避免错误输出,如果一个行组(数据从另一组移动):
df['difference_price'] = (df.sort_values('price',ascending=False)
.groupby(['neighborhoodname','beds','baths'])['price']
.apply(lambda x: x.diff(-1).replace(0, np.nan).bfill()))
print (df)
id neighborhoodname beds baths price difference_price
0 1 a 1 1 5 1.0
1 2 a 1 1 5 1.0
2 3 a 1 1 4 2.0
3 4 a 1 1 2 NaN
4 5 b 1 2 6 3.0
5 6 b 1 2 6 3.0
6 7 b 1 2 3 NaN
我创建了这个数据框,我计算了我正在寻找的差距,但问题是有些公寓价格相同,我得到的价格差为 0。我如何用差值替换值 0同组最低价
例如:
neighboorhood:a, bed:1, bath:1, price:5
neighboorhood:a, bed:1, bath:1, price:5
neighboorhood:a, bed:1, bath:1, price:3
neighboorhood:a, bed:1, bath:1, price:2
我得到 0,2,1,nan 的差价,我正在寻找 2,2,1,nan(简而言之,我不想比较价格相同的 2 个单位)
提前致谢,祝您愉快。
data=[
[1,'a',1,1,5],[2,'a',1,1,5],[3,'a',1,1,4],[4,'a',1,1,2],[5,'b',1,2,6],[6,'b',1,2,6],[7,'b',1,2,3]
]
df = pd.DataFrame(data, columns = ['id','neighborhoodname', 'beds', 'baths', 'price'])
df['difference_price'] = ( df.dropna()
.sort_values('price',ascending=False)
.groupby(['city','beds','baths'])['price'].diff(-1) )
我认为您可以先删除用于 groupby
和 diff
的所有列的重复项,在过滤后的数据中创建新列,最后使用左连接合并到原始数据:
df1 = (df.dropna()
.sort_values('price',ascending=False)
.drop_duplicates(['neighborhoodname','beds','baths', 'price']))
df1['difference_price'] = df1.groupby(['neighborhoodname','beds','baths'])['price'].diff(-1)
df = df.merge(df1[['neighborhoodname','beds','baths','price', 'difference_price']], how='left')
print (df)
id neighborhoodname beds baths price difference_price
0 1 a 1 1 5 1.0
1 2 a 1 1 5 1.0
2 3 a 1 1 4 2.0
3 4 a 1 1 2 NaN
4 5 b 1 2 6 3.0
5 6 b 1 2 6 3.0
6 7 b 1 2 3 NaN
或者您可以使用 lambda 函数回填每组的 0
值以避免错误输出,如果一个行组(数据从另一组移动):
df['difference_price'] = (df.sort_values('price',ascending=False)
.groupby(['neighborhoodname','beds','baths'])['price']
.apply(lambda x: x.diff(-1).replace(0, np.nan).bfill()))
print (df)
id neighborhoodname beds baths price difference_price
0 1 a 1 1 5 1.0
1 2 a 1 1 5 1.0
2 3 a 1 1 4 2.0
3 4 a 1 1 2 NaN
4 5 b 1 2 6 3.0
5 6 b 1 2 6 3.0
6 7 b 1 2 3 NaN