我创建了这个dataframe,并需要将我的数据按价格分组,将我的数据按相同数量的床位、城市、浴室和排序(降序)分类。第二,我需要找出每一种价格之间的差异,把排在后面的价格划入同一组。例如,结果应该是:
1张床,1间浴室,马德里,10张
1张床,1个浴缸,马德里,8
1张床,1个浴缸,马德里,5
1张床,1间浴室,马德里,1张
我应该得到2,3,4.
我试过一些代码,似乎比我预期的要远.
data=[[1,'paris',1,2,'10'],[2,'madrid',2,2,8],[3,'madrid',2,2,11],[4,'paris',1,2,6],[5,'paris',1,2,5],[6,'madrid',2,1,7],[7,'paris',2,1,7],[8,'paris',2,1,7],[9,'madrid',1,4],[10,'paris',2,1,3],[11,'madrid',2,2,7],[12,'paris',2,3,12],[13,'madrid',2,3,7],[14,'madrid',1,1,3],[15,'paris',1,1,3],[16,'madrid',1,1,4],[17,'paris',1,1,5]]
df=pd.DataFrame(data, columns=['id','city','beds','baths','price'])
df
df['gap'] = df.sort_values('price',ascending=False).groupby(['city','beds','baths'])['price'].diff()
print (df)在此之前,非常感谢您。
发布于 2020-01-15 13:30:20
我将使用pd.to_numeric和errors = 'coerce'消除price列中的字符串,然后计算差额,而不考虑价格未知的房间(使用DataFrame.dropna)。然后,我将在DataFrame中显示结果排序,而不需要排序:
df['price']=pd.to_numeric(df['price'],errors = 'coerce')df['difference_price'] = ( df.dropna()
.sort_values('price',ascending=False)
.groupby(['city','beds','baths'])['price'].diff(-1) )或者使用GroupBy.shift
df['difference_price'] = df['price'].sub( df.dropna()
.sort_values('price',ascending=False)
.groupby(['city','beds','baths'])
.price
.shift(-1) )显示结果
print(df,'\n'*3,'Sorted DatFrame: ')
print(df.sort_values(['city','beds','baths','price'],ascending = [True,True,True,False]))输出
id city beds baths price difference_price
0 1 paris 1 2 10.0 4.0
1 2 madrid 2 2 8.0 1.0
2 3 madrid 2 2 11.0 3.0
3 4 paris 1 2 6.0 1.0
4 5 paris 1 2 5.0 NaN
5 6 madrid 2 1 7.0 NaN
6 7 paris 2 1 7.0 0.0
7 8 paris 2 1 7.0 4.0
8 9 madrid 1 4 NaN NaN
9 10 paris 2 1 3.0 NaN
10 11 madrid 2 2 7.0 NaN
11 12 paris 2 3 12.0 NaN
12 13 madrid 2 3 7.0 NaN
13 14 madrid 1 1 3.0 NaN
14 15 paris 1 1 3.0 NaN
15 16 madrid 1 1 4.0 1.0
16 17 paris 1 1 5.0 2.0
Sorted DatFrame:
id city beds baths price difference_price
15 16 madrid 1 1 4.0 1.0
13 14 madrid 1 1 3.0 NaN
8 9 madrid 1 4 NaN NaN
5 6 madrid 2 1 7.0 NaN
2 3 madrid 2 2 11.0 3.0
1 2 madrid 2 2 8.0 1.0
10 11 madrid 2 2 7.0 NaN
12 13 madrid 2 3 7.0 NaN
16 17 paris 1 1 5.0 2.0
14 15 paris 1 1 3.0 NaN
0 1 paris 1 2 10.0 4.0
3 4 paris 1 2 6.0 1.0
4 5 paris 1 2 5.0 NaN
6 7 paris 2 1 7.0 0.0
7 8 paris 2 1 7.0 4.0
9 10 paris 2 1 3.0 NaN
11 12 paris 2 3 12.0 NaN发布于 2020-01-15 13:47:53
如果我理解正确的话:
将我的数据按相同数量的床位、城市、浴室和排序分类(降序)。
所有不满足值的数据都应该删除吗?(在床和浴室不同的地方)。这是我的代码,给出你的问题的答案:
import numpy as np
import pandas as pd
data=[[1,'paris',1,2,'10'],[2,'madrid',2,2,8],[3,'madrid',2,2,11],[4,'paris',1,2,6],[5,'paris',1,2,5],[6,'madrid',2,1,7],[7,'paris',2,1,7],[8,'paris',2,1,7],[9,'madrid',1,4],[10,'paris',2,1,3],[11,'madrid',2,2,7],[12,'paris',2,3,12],[13,'madrid',2,3,7],[14,'madrid',1,1,3],[15,'paris',1,1,3],[16,'madrid',1,1,4],[17,'paris',1,1,5]]
df=pd.DataFrame(data, columns=['id','city','beds','baths','price'])
df_new = df[df['beds'] == df['baths']]
df_new = df_new.sort_values(['city','price'],ascending=[False,False]).reset_index(drop=True)
df_new['diff_price'] = df_new.groupby(['city','beds','baths'])['price'].diff(-1)
print(df_new)输出:
id city beds baths price diff_price
0 17 paris 1 1 5 NaN
1 15 paris 1 1 3 -2
2 3 madrid 2 2 11 NaN
3 2 madrid 2 2 8 -3
4 11 madrid 2 2 7 -1
5 16 madrid 1 1 4 NaN
6 14 madrid 1 1 3 -1https://stackoverflow.com/questions/59751923
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