Python 带Lambda的Pandas-Groupby及其算法
给定此数据帧:Python 带Lambda的Pandas-Groupby及其算法,python,pandas,lambda,group-by,Python,Pandas,Lambda,Group By,给定此数据帧: import pandas as pd import jenkspy f = pd.DataFrame({'BreakGroup':['A','A','A','A','A','A','B','B','B','B','B'], 'Final':[1,2,3,4,5,6,10,20,30,40,50]}) BreakGroup Final 0 A 1 1 A 2 2 A
import pandas as pd
import jenkspy
f = pd.DataFrame({'BreakGroup':['A','A','A','A','A','A','B','B','B','B','B'],
'Final':[1,2,3,4,5,6,10,20,30,40,50]})
BreakGroup Final
0 A 1
1 A 2
2 A 3
3 A 4
4 A 5
5 A 6
6 B 10
7 B 20
8 B 30
9 B 40
10 B 50
我想使用jenkspy来识别组,基于4个组(类)的自然中断,组“BreakGroup”中“Final”中的每个值都属于该组
我一开始是这样做的:
jenks=lambda x: jenkspy.jenks_breaks(f['Final'].tolist(),nb_class=4)
f['Group']=f.groupby(['BreakGroup'])['BreakGroup'].transform(jenks)
…这导致:
BreakGroup
A [1.0, 10.0, 20.0, 30.0, 50.0]
B [1.0, 10.0, 20.0, 30.0, 50.0]
Name: BreakGroup, dtype: object
这里的第一个问题,正如您所猜测的,是它将lambda函数应用于“最终”分数的整列,而不仅仅是属于Groupby中每个组的分数。第二个问题是,我需要一个列来指定正确的组(类)成员身份,可能是通过使用transform而不是apply
然后我试了一下:
jenks=lambda x: jenkspy.jenks_breaks(f['Final'].loc[f['BreakGroup']==x].tolist(),nb_class=4)
f['Group']=f.groupby(['BreakGroup'])['BreakGroup'].transform(jenks)
…但很快就被击退屈服:
ValueError: Can only compare identically-labeled Series objects
更新:
f.sort_values('BreakGroup',inplace=True)
f.reset_index(drop=True,inplace=True)
jenks = lambda x: jenkspy.jenks_breaks(x['Final'].tolist(),nb_class=4)
g = f.set_index('BreakGroup')
g['Groups'] = f.groupby(['BreakGroup']).apply(jenks)
g.reset_index(inplace=True)
groups= lambda x: [gp for gp in x['Groups']]
#'final' value should be > lower and <= upper
upper = lambda x: [gp for gp in x['Groups'] if gp >= x['Final']][0] # or gp == max(x['Groups'])
lower= lambda x: [gp for gp in x['Groups'] if gp < x['Final'] or gp == min(x['Groups'])][-1]
GroupIndex= lambda x: [x['Groups'].index(gp) for gp in x['Groups'] if gp < x['Final'] or gp == min(x['Groups'])][-1]
f['Groups']=g.apply(groups, axis=1)
f['Upper'] = g.apply(upper, axis=1)
f['Lower'] = g.apply(lower, axis=1)
f['Group'] = g.apply(GroupIndex, axis=1)
f['Group']=f['Group']+1
这是期望的结果。“结果”列包含每组“BreakGroup”的“Final”中相应值的组上限:
提前谢谢
我根据接受的解决方案略微修改了应用程序:
f.sort_values('BreakGroup',inplace=True)
f.reset_index(drop=True,inplace=True)
jenks = lambda x: jenkspy.jenks_breaks(x['Final'].tolist(),nb_class=4)
g = f.set_index('BreakGroup')
g['Groups'] = f.groupby(['BreakGroup']).apply(jenks)
g.reset_index(inplace=True)
groups= lambda x: [gp for gp in x['Groups']]
#'final' value should be > lower and <= upper
upper = lambda x: [gp for gp in x['Groups'] if gp >= x['Final']][0] # or gp == max(x['Groups'])
lower= lambda x: [gp for gp in x['Groups'] if gp < x['Final'] or gp == min(x['Groups'])][-1]
GroupIndex= lambda x: [x['Groups'].index(gp) for gp in x['Groups'] if gp < x['Final'] or gp == min(x['Groups'])][-1]
f['Groups']=g.apply(groups, axis=1)
f['Upper'] = g.apply(upper, axis=1)
f['Lower'] = g.apply(lower, axis=1)
f['Group'] = g.apply(GroupIndex, axis=1)
f['Group']=f['Group']+1
f.sort\u值('BreakGroup',inplace=True)
f、 重置索引(drop=True,inplace=True)
jenks=lambda x:jenkspy.jenks_breaks(x['Final'].tolist(),nb_class=4)
g=f.set_索引('BreakGroup')
g['Groups']=f.groupby(['BreakGroup'])。应用(jenks)
g、 重置索引(就地=真)
组=λx:[x['groups']中gp的gp]
#“final”值应大于等于x['final'][0]#或gp==max(x['Groups'])
下限=λx:[如果gp
这将返回:
当前,您正在将一个序列传递到
transform()
中,而不是像您希望的那样传递给过滤器条件的标量。考虑第一个值的索引,如<代码> x.Curry[0 ] < /代码>,因为所有代码在<代码> GROPPB/<代码>系列中都是相同的。您甚至可以运行min(x)
或max(x)
:
您将
jenks
定义为lambda变量x
中的常数,因此它不取决于使用apply
或transform
为其提供的内容。将jenks
的定义更改为
jenks = lambda x: jenkspy.jenks_breaks(x['Final'].tolist(),nb_class=4)
给予
从这一重新定义继续
g = f.set_index('BreakGroup')
g['Groups'] = f.groupby(['BreakGroup']).apply(jenks)
g.reset_index(inplace=True)
group = lambda x: [gp for gp in x['Groups'] if gp > x['Final'] or gp == max(x['Groups'])][0]
f['Result'] = g.apply(group, axis=1)
给予
你能发布目标输出吗?当然;看更新。太好了!顺便说一句,我用它来获得下限值:group2=lambda x:[gp for gp in x['Groups']如果gp@DanceParty2,因为您使用的是
如何查找值的索引而不是值本身(即,顶部行为0)?您的意思是,jenks产生的组列表中的哪个索引?是的,给定g行列表中元素的索引。
In [315]: f.groupby(['BreakGroup']).apply(jenks)
Out[315]:
BreakGroup
A [1.0, 2.0, 3.0, 4.0, 6.0]
B [10.0, 20.0, 30.0, 40.0, 50.0]
dtype: object
g = f.set_index('BreakGroup')
g['Groups'] = f.groupby(['BreakGroup']).apply(jenks)
g.reset_index(inplace=True)
group = lambda x: [gp for gp in x['Groups'] if gp > x['Final'] or gp == max(x['Groups'])][0]
f['Result'] = g.apply(group, axis=1)
In [323]: f
Out[323]:
BreakGroup Final Result
0 A 1 2.0
1 A 2 3.0
2 A 3 4.0
3 A 4 6.0
4 A 5 6.0
5 A 6 6.0
6 B 10 20.0
7 B 20 30.0
8 B 30 40.0
9 B 40 50.0
10 B 50 50.0