7. 用链式方法重现idxmax

  1. #  和前面一样，只选出数值列
  2.  In[76]: college = pd.read_csv('data/college.csv', index_col='INSTNM')
  4.          cols = ['MD_EARN_WNE_P10', 'GRAD_DEBT_MDN_SUPP']
  5.          for col in cols:
  6.              college[col] = pd.to_numeric(college[col], errors='coerce')
  8.          college_n = college.select_dtypes(include=[np.number])
  9.          criteria = college_n.nunique() == 2
  10.          binary_cols = college_n.columns[criteria].tolist()
  11.          college_n = college_n.drop(labels=binary_cols, axis='columns')
  12.  In[77]: college_n.max().head()
  13. Out[77]: SATVRMID         765.0
  14.          SATMTMID         785.0
  15.          UGDS          151558.0
  16.          UGDS_WHITE         1.0
  17.          UGDS_BLACK         1.0
  18.          dtype: float64
  1. #  college_n.max()可以选出每列的最大值，用eq方法比较DataFrame的每个值和该列的最大值
  2.  In[78]: college_n.eq(college_n.max()).head()
  3. Out[78]:

7. 用链式方法重现idxmax - 图1

  1. #  用any方法，选出至少包含一个True值的行
  2.  In[79]: has_row_max = college_n.eq(college_n.max()).any(axis='columns')
  3.          has_row_max.head()
  4. Out[79]: INSTNM
  5.          Alabama A & M University               False
  6.          University of Alabama at Birmingham    False
  7.          Amridge University                     False
  8.          University of Alabama in Huntsville    False
  9.          Alabama State University               False
  10.          dtype: bool
  1. #  因为只有18列，has_row_max最多只能有18个True，来看下实际共有多少个
  2.  In[80]: college_n.shape
  3. Out[80]: (7535, 18)
  5.  In[81]: has_row_max.sum()
  6. Out[81]: 401
  1. #  结果很奇怪，这是因为许多百分比的列的最大值是1。转而使用cumsum()累积求和
  2.  In[82]: has_row_max.sum()
  3.  In[83]: college_n.eq(college_n.max()).cumsum()
  4. Out[83]:

7. 用链式方法重现idxmax - 图2

  1. #  一些列只有一个最大值，比如SATVRMID和SATMTMID，UGDS_WHITE列却有许多最大值。有109所学校的学生100%是白人。如果再使用一次cunsum，1在每列中就只出现一次，而且会是最大值首次出现的位置：
  2. >>> college_n.eq(college_n.max()).cumsum().cumsum()

7. 用链式方法重现idxmax - 图3

  1. #  现在就可以用eq方法去和1进行比较，然后用any方法，选出所有至少包含一个True值的行
  2.  In[84]: has_row_max2 = college_n.eq(college_n.max())\
  3.                                  .cumsum()\
  4.                                  .cumsum()\
  5.                                  .eq(1)\
  6.                                  .any(axis='columns')
  7.          has_row_max2.head()
  8. Out[84]: INSTNM
  9.          Alabama A & M University               False
  10.          University of Alabama at Birmingham    False
  11.          Amridge University                     False
  12.          University of Alabama in Huntsville    False
  13.          Alabama State University               False
  14.          dtype: bool
  1. #  查看有多少True值
  2.  In[85]: has_row_max2.sum()
  3. Out[85]: 16
  1. #  直接通过布尔索引选出这些学校
  2.  In[86]: idxmax_cols = has_row_max2[has_row_max2].index
  3.          idxmax_cols
  4. Out[86]: Index(['Thunderbird School of Global Management',
  5.                 'Southwest University of Visual Arts-Tucson', 'ABC Beauty College Inc',
  6.                 'Velvatex College of Beauty Culture',
  7.                 'California Institute of Technology',
  8.                 'Le Cordon Bleu College of Culinary Arts-San Francisco',
  9.                 'MTI Business College Inc', 'Dongguk University-Los Angeles',
  10.                 'Mr Leon's School of Hair Design-Moscow',
  11.                 'Haskell Indian Nations University', 'LIU Brentwood',
  12.                 'Medical College of Wisconsin', 'Palau Community College',
  13.                 'California University of Management and Sciences',
  14.                 'Cosmopolitan Beauty and Tech School', 'University of Phoenix-Arizona'],
  15.                 dtype='object', name='INSTNM')
  1. #  和idxmax方法的结果比较
  2.  In[87]: set(college_n.idxmax().unique()) == set(idxmax_cols)
  3. Out[87]: True

更多

  1. #  耗时比较
  2.  In[88]: %timeit college_n.idxmax().values
  3.          1.11 ms ± 50.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
  5. Out[89]: %timeit college_n.eq(college_n.max())\
  6.                                        .cumsum()\
  7.                                        .cumsum()\
  8.                                        .eq(1)\
  9.                                        .any(axis='columns')\
  10.                                        [lambda x: x].index
  11.          5.26 ms ± 35.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)