获取数据帧每行的第 n 个排名列 ID - Python/Pandas

Question

我正在尝试寻找一种方法来查找排名第 n 的值并 returning 列名。因此，例如，给定一个数据框：

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

这会找到行中的最大值，return 出现它的列名。但我需要这样一种情况，我可以选择所需值的特定等级，并希望得到如下数据框：

          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_NAMES 是该行中的第二大值。

谢谢。

Answer 1

您可以每行应用一个 argsort()，反转索引并在第二个位置选择一个：

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

Answer 2

您只想对特定排名 n 进行排名，所以我想建议 np.argpartition that would get sorted indices just for the highest n-ranked entries at each row rather than sorting all elements. This is aimed at improved performance. The performance benefits are discussed in length in answers to A fast way to find the largest N elements in an numpy array 希望我们也能从中受益。

因此，在函数格式中，我们将有 -

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

我正在计时基于 np.argpartition 的方法，正如前面列出的 post 和基于 np.argsort 的方法，如@Psidom 在大小合适的数据帧上的另一个解决方案中所列。

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

Answer 3

您可以通过组合 rank、apply 和 idxmin 来做到这一点。

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

获取数据帧每行的第 n 个排名列 ID - Python/Pandas

Getting n-th ranked column IDs per row of a dataframe - Python/Pandas

python

ranking

python-2.7

pandas