下面的答案是一个古老的答案。它现在还能用。无论如何，已经发布了另一个非常优雅的解决方案(如下所示)，使用key参数。

我刚刚发现Pandas15.1可以用范畴序列(https://pandas.pydata.org/docs/user_guide/categorical.html)

对于您的示例，让我们定义相同的 Dataframe 和排序器：

import pandas as pd

data = {
    'id': [2967, 5335, 13950, 6141, 6169],
    'Player': ['Cedric Hunter', 'Maurice Baker', 
               'Ratko Varda' ,'Ryan Bowen' ,'Adrian Caldwell'],
    'Year': [1991, 2004, 2001, 2009, 1997],
    'Age': [27, 25, 22, 34, 31],
    'Tm': ['CHH', 'VAN', 'TOT', 'OKC', 'DAL'],
    'G': [6, 7, 60, 52, 81]
}

# Create DataFrame

df = pd.DataFrame(data)

# Define the sorter

sorter = ['TOT', 'ATL', 'BOS', 'BRK', 'CHA', 'CHH', 'CHI', 'CLE', 'DAL', 'DEN',
          'DET', 'GSW', 'HOU', 'IND', 'LAC', 'LAL', 'MEM', 'MIA', 'MIL',
          'MIN', 'NJN', 'NOH', 'NOK', 'NOP', 'NYK', 'OKC', 'ORL', 'PHI',
          'PHO', 'POR', 'SAC', 'SAS', 'SEA', 'TOR', 'UTA', 'VAN', 'WAS', 'WSB']

有了 Dataframe 和分类器，这是一个分类顺序，我们可以在Pandas15.1中做以下事情：

# Convert Tm-column to category and in set the sorter as categories hierarchy

# Youc could also do both lines in one just appending the cat.set_categories()

df.Tm = df.Tm.astype("category")
df.Tm = f.Tm.cat.set_categories(sorter)

print(df.Tm)
Out[48]: 
0    CHH
1    VAN
2    TOT
3    OKC
4    DAL
Name: Tm, dtype: category
Categories (38, object): [TOT < ATL < BOS < BRK ... UTA < VAN < WAS < WSB]

df.sort_values(["Tm"])  ## 'sort' changed to 'sort_values'
Out[49]: 
   Age   G           Player   Tm  Year     id
2   22  60      Ratko Varda  TOT  2001  13950
0   27   6    Cedric Hunter  CHH  1991   2967
4   31  81  Adrian Caldwell  DAL  1997   6169
3   34  52       Ryan Bowen  OKC  2009   6141
1   25   7    Maurice Baker  VAN  2004   5335

下面是一个对 Dataframe 执行词典排序的示例。其想法是基于特定的排序创建一个数字索引。然后根据索引执行数值排序。为此，在 Dataframe 中添加一列，然后将其删除。

import pandas as pd

# Create DataFrame

df = pd.DataFrame(
{'id':[2967, 5335, 13950, 6141, 6169],
    'Player': ['Cedric Hunter', 'Maurice Baker',
               'Ratko Varda' ,'Ryan Bowen' ,'Adrian Caldwell'],
    'Year': [1991, 2004, 2001, 2009, 1997],
    'Age': [27, 25, 22, 34, 31],
    'Tm': ['CHH' ,'VAN' ,'TOT' ,'OKC', 'DAL'],
    'G': [6, 7, 60, 52, 81]})

# Define the sorter

sorter = ['TOT', 'ATL', 'BOS', 'BRK', 'CHA', 'CHH', 'CHI', 'CLE', 'DAL','DEN',
          'DET', 'GSW', 'HOU', 'IND', 'LAC', 'LAL', 'MEM', 'MIA', 'MIL',
          'MIN', 'NJN', 'NOH', 'NOK', 'NOP', 'NYK', 'OKC', 'ORL', 'PHI',
          'PHO', 'POR', 'SAC', 'SAS', 'SEA', 'TOR', 'UTA', 'VAN',
          'WAS', 'WSB']

# Create the dictionary that defines the order for sorting

sorterIndex = dict(zip(sorter, range(len(sorter))))

# Generate a rank column that will be used to sort

# the dataframe numerically

df['Tm_Rank'] = df['Tm'].map(sorterIndex)

# Here is the result asked with the lexicographic sort

# Result may be hard to analyze, so a second sorting is

# proposed next

## NOTE:

## Newer versions of pandas use 'sort_values' instead of 'sort'

df.sort_values(['Player', 'Year', 'Tm_Rank'],
        ascending = [True, True, True], inplace = True)
df.drop('Tm_Rank', 1, inplace = True)
print(df)

# Here is an example where 'Tm' is sorted first, that will

# give the first row of the DataFrame df to contain TOT as 'Tm'

df['Tm_Rank'] = df['Tm'].map(sorterIndex)

## NOTE:

## Newer versions of pandas use 'sort_values' instead of 'sort'

df.sort_values(['Tm_Rank', 'Player', 'Year'],
        ascending = [True , True, True], inplace = True)
df.drop('Tm_Rank', 1, inplace = True)
print(df)

df1 = df.set_index('Tm')
df1.loc[sorter]

根据Pandas 1.1.0文档，可以使用key参数进行排序，就像在sorted函数中一样(终于！)下面是我们如何按Tm进行排序

import pandas as pd

data = {
    'id': [2967, 5335, 13950, 6141, 6169],
    'Player': ['Cedric Hunter', 'Maurice Baker', 
               'Ratko Varda' ,'Ryan Bowen' ,'Adrian Caldwell'],
    'Year': [1991, 2004, 2001, 2009, 1997],
    'Age': [27, 25, 22, 34, 31],
    'Tm': ['CHH', 'VAN', 'TOT', 'OKC', 'DAL'],
    'G': [6, 7, 60, 52, 81]
}

# Create DataFrame

df = pd.DataFrame(data)

def tm_sorter(column):
    """Sort function"""
    teams = ['TOT', 'ATL', 'BOS', 'BRK', 'CHA', 'CHH', 'CHI', 'CLE', 'DAL', 'DEN',
       'DET', 'GSW', 'HOU', 'IND', 'LAC', 'LAL', 'MEM', 'MIA', 'MIL',
       'MIN', 'NJN', 'NOH', 'NOK', 'NOP', 'NYK', 'OKC', 'ORL', 'PHI',
       'PHO', 'POR', 'SAC', 'SAS', 'SEA', 'TOR', 'UTA', 'VAN',
       'WAS', 'WSB']
    correspondence = {team: order for order, team in enumerate(teams)}
    return column.map(correspondence)

df.sort_values(by='Tm', key=tm_sorter)

遗憾的是，我们似乎只能在按1列排序时使用此功能(不接受包含keys的列表)。它可以通过groupby来规避

df.sort_values(['Player', 'Year']) 
  .groupby(['Player', 'Year']) 
  .apply(lambda x: x.sort_values(by='Tm', key=tm_sorter)) 
  .reset_index(drop=True)

如果你知道如何使用sort_values中的key，请告诉我

这几行代码就完成了这项工作

# Create a dummy df with the required list and the col name to sort on

dummy = pd.Series(sort_list, name = col_name).to_frame()

# Use left merge on the dummy to return a sorted df

sorted_df = pd.merge(dummy, df, on = col_name, how = 'left')

当您需要按单个定制列表排序时，将索引设置为DataFrame.loc非常有用。因为loc将为sorter中不在DataFrame中的值创建NaN行，所以我们将首先找到交叉点。这可以防止任何不必要的向上投射。任何值不在列表中的行都将被删除。

true_sort = [s for s in sorter if s in df.Tm.unique()]
df = df.set_index('Tm').loc[true_sort].reset_index()

    Tm     id           Player  Year  Age   G
0  TOT  13950      Ratko Varda  2001   22  60
1  CHH   2967    Cedric Hunter  1991   27   6
2  DAL   6169  Adrian Caldwell  1997   31  81
3  OKC   6141       Ryan Bowen  2009   34  52
4  VAN   5335    Maurice Baker  2004   25   7

起始数据：

print(df)
      id           Player  Year  Age   Tm   G
0   2967    Cedric Hunter  1991   27  CHH   6
1   5335    Maurice Baker  2004   25  VAN   7
2  13950      Ratko Varda  2001   22  TOT  60
3   6141       Ryan Bowen  2009   34  OKC  52
4   6169  Adrian Caldwell  1997   31  DAL  81

sorter = ['TOT', 'ATL', 'BOS', 'BRK', 'CHA', 'CHH', 'CHI', 'CLE', 'DAL', 'DEN',
          'DET', 'GSW', 'HOU', 'IND', 'LAC', 'LAL', 'MEM', 'MIA', 'MIL',
          'MIN', 'NJN', 'NOH', 'NOK', 'NOP', 'NYK', 'OKC', 'ORL', 'PHI',
          'PHO', 'POR', 'SAC', 'SAS', 'SEA', 'TOR', 'UTA', 'VAN', 'WAS', 'WSB']

从1.1.0版开始，您可以使用key属性对值进行排序：

df.sort_values(by="Tm", key=lambda column: column.map(lambda e: sorter.index(e)), inplace=True)

仅对按分类列排序感兴趣的用户的部分解决方案：

您可以使用从定制列表创建排序顺序Map器的帮助器函数来实现这一点。

此示例仅包括一列中的值，但是可以通过创建包含所有列中出现的值的自定义顺序列表来扩展它以包括其他列。自然，由于您必须在排序字段中使用所有可能的值构建自定义列表，所以这主要适用于分类排序，而不适合连续变量(除非事先知道可能的值)和基数非常高的列。

import pandas as pd

# set up a dummy dataframe

df = pd.DataFrame({'a':list('abcde'), 'b':range(5)})

# helper function

def make_sorter(l):
    """
    Create a dict from the list to map to 0..len(l)
    Returns a mapper to map a series to this custom sort order
    """
    sort_order = {k:v for k,v in zip(l, range(len(l)))}
    return lambda s: s.map(lambda x: sort_order[x])

# define a custom sort order

my_order = list('bdeca')

df.sort_values('a', key=make_sorter(my_order))

   a b
1  b 1
3  d 3
4  e 4
2  c 2
0  a 0

使用OP的数据：

df = pd.DataFrame({
    'id':[2967, 5335, 13950, 6141, 6169],
    'Player': ['Cedric Hunter', 'Maurice Baker',
               'Ratko Varda' ,'Ryan Bowen' ,'Adrian Caldwell'],
    'Year': [1991, 2004, 2001, 2009, 1997],
    'Age': [27, 25, 22, 34, 31],
    'Tm': ['CHH' ,'VAN' ,'TOT' ,'OKC', 'DAL'],
    'G': [6, 7, 60, 52, 81]
})

# Define the sorter

sorter = [
    'TOT', 'ATL', 'BOS', 'BRK', 'CHA', 'CHH', 'CHI', 'CLE', 'DAL',
    'DEN', 'DET', 'GSW', 'HOU', 'IND', 'LAC', 'LAL', 'MEM', 'MIA',
    'MIL', 'MIN', 'NJN', 'NOH', 'NOK', 'NOP', 'NYK', 'OKC', 'ORL',
    'PHI', 'PHO', 'POR', 'SAC', 'SAS', 'SEA', 'TOR', 'UTA', 'VAN',
    'WAS', 'WSB'
]

df.sort_values('Tm', key=make_sorter(sorter))

      id           Player  Year  Age   Tm   G
2  13950      Ratko Varda  2001   22  TOT  60
0   2967    Cedric Hunter  1991   27  CHH   6
4   6169  Adrian Caldwell  1997   31  DAL  81
3   6141       Ryan Bowen  2009   34  OKC  52
1   5335    Maurice Baker  2004   25  VAN   7

我的想法是根据索引生成排序号，然后将排序号合并到原始 Dataframe 中

import pandas as pd

df = pd.DataFrame(
{'id':[2967, 5335, 13950, 6141, 6169],
 'Player': ['Cedric Hunter', 'Maurice Baker' ,
            'Ratko Varda' ,'Ryan Bowen' ,'Adrian Caldwell'],
 'Year': [1991 ,2004 ,2001 ,2009 ,1997],
 'Age': [27 ,25 ,22 ,34 ,31],
 'Tm':['CHH' ,'VAN' ,'TOT' ,'OKC' ,'DAL'],
 'G':[6 ,7 ,60 ,52 ,81]})

sorter = ['TOT', 'ATL', 'BOS', 'BRK', 'CHA', 'CHH', 'CHI', 'CLE', 'DAL', 'DEN',
   'DET', 'GSW', 'HOU', 'IND', 'LAC', 'LAL', 'MEM', 'MIA', 'MIL',
   'MIN', 'NJN', 'NOH', 'NOK', 'NOP', 'NYK', 'OKC', 'ORL', 'PHI',
   'PHO', 'POR', 'SAC', 'SAS', 'SEA', 'TOR', 'UTA', 'VAN',
   'WAS', 'WSB']

x = pd.DataFrame({'Tm': sorter})
x.index = x.index.set_names('number')
x = x.reset_index()

df = pd.merge(df, x, how='left', on='Tm')

df.sort_values(['Player', 'Year', 'number'], 
        ascending = [True, True, True], inplace = True)
df.drop('number', 1, inplace = True)