获取数据帧每行的第n个列ID-Python/Pandas
我试图找到一种方法来查找排名第n的值并返回列名。例如,给定一个数据帧:获取数据帧每行的第n个列ID-Python/Pandas,python,python-2.7,pandas,ranking,Python,Python 2.7,Pandas,Ranking,我试图找到一种方法来查找排名第n的值并返回列名。例如,给定一个数据帧: df = pd.DataFrame(np.random.randn(5, 4), columns = list('ABCD')) # Return column name of "MAX" value, compared to other columns in any particular row. df['MAX1_NAMES'] = df.idxmax(axis=1) print df A
df = pd.DataFrame(np.random.randn(5, 4), columns = list('ABCD'))
# Return column name of "MAX" value, compared to other columns in any particular row.
df['MAX1_NAMES'] = df.idxmax(axis=1)
print df
A B C D MAX1_NAMES
0 -0.728424 -0.764682 -1.506795 0.722246 D
1 1.305500 -1.191558 0.068829 -1.244659 A
2 -0.175834 -0.140273 1.117114 0.817358 C
3 -0.255825 -1.534035 -0.591206 -0.352594 A
4 -2.408806 -1.925055 -1.797020 2.381936 D
这将在行中找到最高值,并返回出现该值的列名。但我需要这样的情况,我可以选择期望值的特定秩,并希望得到如下所示的数据帧:
A B C D MAX1_NAMES MAX2_NAMES
0 -0.728424 -0.764682 -1.506795 0.722246 D A
1 1.305500 -1.191558 0.068829 -1.244659 A C
2 -0.175834 -0.140273 1.117114 0.817358 C D
3 -0.255825 -1.534035 -0.591206 -0.352594 A D
4 -2.408806 -1.925055 -1.797020 2.381936 D C
其中,MAX2\u name
是行中的第二大值
谢谢。您可以对每行应用一个
argsort()
,反转索引并在第二个位置选取一个:
df['MAX2_NAMES'] = df.iloc[:,:4].apply(lambda r: r.index[r.argsort()[::-1][1]], axis = 1)
df
# A B C D MAX1_NAMES MAX2_NAMES
#0 -0.728424 -0.764682 -1.506795 0.722246 D A
#1 1.305500 -1.191558 0.068829 -1.244659 A C
#2 -0.175834 -0.140273 1.117114 0.817358 C D
#3 -0.255825 -1.534035 -0.591206 -0.352594 A D
#4 -2.408806 -1.925055 -1.797020 2.381936 D C
您希望只对特定的排名
n
执行排序,因此我建议您只对每行中排名最高的n个条目进行排序索引,而不是对所有元素进行排序。这是为了提高性能。在回答中详细讨论了性能优势,希望我们也能从中获益
因此,在函数格式中,我们将-
def rank_df(df,rank):
coln = 'MAX' + str(rank) + '_NAMES'
sortID = np.argpartition(-df[['A','B','C','D']].values,rank,axis=1)[:,rank-1]
df[coln] = df.columns[sortID]
样本运行-
In [84]: df
Out[84]:
A B C D
0 -0.124851 0.152432 1.436602 -0.391178
1 0.371932 1.732399 0.340876 -1.340609
2 -1.218608 0.444246 0.169968 -1.437259
3 -0.828132 0.821613 -0.556643 -0.407703
4 -0.390477 0.048824 -2.087323 1.597030
In [85]: rank_df(df,1)
In [86]: rank_df(df,2)
In [87]: df
Out[87]:
A B C D MAX1_NAMES MAX2_NAMES
0 -0.124851 0.152432 1.436602 -0.391178 C B
1 0.371932 1.732399 0.340876 -1.340609 B A
2 -1.218608 0.444246 0.169968 -1.437259 B C
3 -0.828132 0.821613 -0.556643 -0.407703 B D
4 -0.390477 0.048824 -2.087323 1.597030 D B
运行时测试
我正在对本文前面列出的基于np.argpartition
的方法进行计时,并对@Psidom在适当大小的数据帧上列出的基于np.argsort
的方法进行计时
In [92]: df = pd.DataFrame(np.random.randn(10000, 4), columns = list('ABCD'))
In [93]: %timeit rank_df(df,2)
100 loops, best of 3: 2.36 ms per loop
In [94]: df = pd.DataFrame(np.random.randn(10000, 4), columns = list('ABCD'))
In [95]: %timeit df['MAX2_NAMES'] = df.iloc[:,:4].apply(lambda r: r.index[r.argsort()[::-1][1]], axis = 1)
1 loops, best of 3: 3.32 s per loop
您可以通过组合rank、apply和idxmin来实现这一点 例如:
df = pd.util.testing.makeTimeDataFrame(5)
df
A B C D
2000-01-03 -1.814888 -0.709120 -0.134390 -0.906183
2000-01-04 0.459742 1.235481 0.109602 -0.226923
2000-01-05 -1.567867 0.562368 -1.185567 -2.176161
2000-01-06 0.747989 -0.160384 1.617100 0.242830
2000-01-07 -1.288061 -1.631342 -0.857830 -0.210695
df['rank_2_col'] = df.rank(1).apply(lambda r: r[r==2].idxmin(), axis=1)
df
A B C D rank_2_col
2000-01-03 -1.814888 -0.709120 -0.134390 -0.906183 D
2000-01-04 0.459742 1.235481 0.109602 -0.226923 C
2000-01-05 -1.567867 0.562368 -1.185567 -2.176161 A
2000-01-06 0.747989 -0.160384 1.617100 0.242830 D
2000-01-07 -1.288061 -1.631342 -0.857830 -0.210695 A
非常好,无论速度如何,我都更痴迷于获得任何解决方案,但是numpy排序性能提示将派上用场。