有没有一种方法可以对熊猫数据框中的某些项目进行排名并排除其他项目?

时间:2020-03-17 08:45:37

标签: python pandas filtering ranking

我有一个名为ranks的熊猫数据框,其中包含我的群集及其关键指标。我使用rank()对它们进行排名,但是我想将两个特定的集群与其他集群进行排名。

ranks = pd.DataFrame(data={'Cluster': ['0', '1', '2',
                                   '3', '4', '5','6', '7', '8', '9'],
                        'No. Customers': [145118, 
                                        2, 
                                        1236, 
                                        219847, 
                                        9837,
                                        64865,
                                        3855,
                                        219549,
                                        34171,
                                        3924120],  
                        'Ave. Recency': [39.0197, 
                                        47.0, 
                                        15.9716, 
                                        41.9736, 
                                        23.9330,
                                        24.8281,
                                        26.5647,
                                        17.7493,
                                        23.5205,
                                        24.7933],
                        'Ave. Frequency': [1.7264, 
                                        19.0, 
                                        24.9101, 
                                        3.0682, 
                                        3.2735,
                                        1.8599,
                                        3.9304,
                                        3.3356,
                                        9.1703,
                                        1.1684],
                        'Ave. Monetary': [14971.85, 
                                        237270.00, 
                                        126992.79, 
                                        17701.64, 
                                        172642.35,
                                        13159.21,
                                        54333.56,
                                        17570.67,
                                        42136.68,
                                        4754.76]})
ranks['Ave. Spend'] = ranks['Ave. Monetary']/ranks['Ave. Frequency']
   Cluster   No. Customers| Ave. Recency| Ave. Frequency| Ave. Monetary| Ave. Spend|
0    0           145118        39.0197       1.7264         14,971.85     8,672.07
1    1           2             47.0          19.0          237,270.00    12,487.89
2    2           1236          15.9716       24.9101       126,992.79     5,098.02
3    3           219847        41.9736       3.0682         17,701.64     5,769.23
4    4           9837          23.9330       3.2735        172,642.35    52,738.42
5    5           64865         24.8281       1.8599         13,159.21     7,075.19
6    6           3855          26.5647       3.9304         54,333.56    13,823.64
7    7           219549        17.7493       3.3356         17,570.67     5,267.52
8    8           34171         23.5205       9.1703         42,136.68     4,594.89
9    9           3924120       24.7933       1.1684          4,754.76     4,069.21 

然后我像这样应用rank()方法:

ranks['r_rank'] = ranks['Ave. Recency'].rank()
ranks['f_rank'] = ranks['Ave. Frequency'].rank(ascending=False)
ranks['m_rank'] = ranks['Ave. Monetary'].rank(ascending=False)
ranks['s_rank'] = ranks['Ave. Spend'].rank(ascending=False)
ranks['overall'] = ranks.apply(lambda row: row.r_rank + row.f_rank + row.m_rank + row.s_rank, axis=1)
ranks['overall_rank'] = ranks['overall'].rank(method='first')

哪个给我这个:

   Cluster  No. Customers|Ave. Recency|Ave. Frequency|Ave. Monetary|Ave. Spend|r_rank|f_rank|m_rank|s_rank|overall|overall_rank
0    0          145118       39.0197      1.7264        14,971.85    8,672.07     8     9       8      4      29        9     
1    1          2            47.0         19.0         237,270.00   12,487.89     10    2       1      3      16        3 
2    2          1236         15.9716      24.9101      126,992.79    5,098.02     1     1       3      8      13        1
3    3          219847       41.9736      3.0682        17,701.64    5,769.23     9     7       6      6      28        7
4    4          9837         23.9330      3.2735       172,642.35   52,738.42     4     6       2      1      13        2
5    5          64865        24.8281      1.8599        13,159.21    7,075.19     6     8       9      5      28        8
6    6          3855         26.5647      3.9304        54,333.56   13,823.64     7     4       4      2      17        4
7    7          219549       17.7493      3.3356        17,570.67    5,267.52     2     5       7      7      21        6
8    8          34171        23.5205      9.1703        42,136.68    4,594.89     3     3       5      9      20        5
9    9          3924120      24.7933      1.1684         4,754.76    4,069.21     5     10      10     10     35        10

这是应该做的事情,但是Ave. Spend最高的集群必须始终排在第1位,Ave. Recency最高的集群必须一直排在最后。

所以我修改了上面的代码,使其看起来像这样:

if(ranks['s_rank'].min() == 1):
    ranks['overall_rank_2'] = 1
elif(ranks['r_rank'].max() == len(ranks)):
    ranks['overall_rank_2'] = len(ranks)
else:
    ranks_2 = ranks.drop(ranks.index[[ranks[ranks['s_rank'] == ranks['s_rank'].min()].index[0],ranks[ranks['r_rank'] == ranks['r_rank'].max()].index[0]]])
    ranks_2['r_rank'] = ranks_2['Ave. Recency'].rank()
    ranks_2['f_rank'] = ranks_2['Ave. Frequency'].rank(ascending=False)
    ranks_2['m_rank'] = ranks_2['Ave. Monetary'].rank(ascending=False)
    ranks_2['s_rank'] = ranks_2['Ave. Spend'].rank(ascending=False)
    ranks_2['overall'] = ranks.apply(lambda row: row.r_rank + row.f_rank + row.m_rank + row.s_rank, axis=1)
    ranks['overall_rank_2'] = ranks_2['overall'].rank(method='first')

那我明白了

   Cluster  No. Customers|Ave. Recency|Ave. Frequency|Ave. Monetary|Ave. Spend|r_rank|f_rank|m_rank|s_rank|overall|overall_rank|overall_rank_2
0    0          145118       39.0197      1.7264        14,971.85    8,672.07     8     9       8      4      29        9             1     
1    1          2            47.0         19.0         237,270.00   12,487.89     10    2       1      3      16        3             1 
2    2          1236         15.9716      24.9101      126,992.79    5,098.02     1     1       3      8      13        1             1
3    3          219847       41.9736      3.0682        17,701.64    5,769.23     9     7       6      6      28        7             1
4    4          9837         23.9330      3.2735       172,642.35   52,738.42     4     6       2      1      13        2             1
5    5          64865        24.8281      1.8599        13,159.21    7,075.19     6     8       9      5      28        8             1
6    6          3855         26.5647      3.9304        54,333.56   13,823.64     7     4       4      2      17        4             1
7    7          219549       17.7493      3.3356        17,570.67    5,267.52     2     5       7      7      21        6             1
8    8          34171        23.5205      9.1703        42,136.68    4,594.89     3     3       5      9      20        5             1
9    9          3924120      24.7933      1.1684         4,754.76    4,069.21     5     10      10     10     35        10            1

请帮助我修改以上if语句,或者完全建议使用其他方法。这当然需要尽可能地动态。

1 个答案:

答案 0 :(得分:2)

因此,您需要在数据帧上进行自定义排名,其中Ave. Spend最高的集群(/行)始终排名1,Ave. Recency最高的集群(/行)始终排名最后。

解决方案是五行。注意:

  • 您对DataFrame.drop()的想法是正确的,只需使用idxmax()即可获取需要特殊处理的两行的索引并将其存储,因此您无需费劲drop中的逻辑过滤器表达式。
  • 无需创建太多临时列或临时副本ranks_2 = ranks.drop(...);只需将drop()的结果传递到rank() ...
  • ...通过所需列上的.sum(axis=1),无需定义Lambda或将其输出保存在临时列'overall'中。
  • ...然后,我们将这些排名总和输入rank(),这将为我们提供1..8的值,因此我们加1以抵消rank()的结果为2..9。 (您可以对此进行概括)。
  • 然后我们为Ave. SpendAve. Recency行手动设置'overall_rank'。
  • (是的,您还可以将所有这些实现为自定义函数,其输入为四个Ave.列,否则为四个*_rank列。)

代码:(请参见底部的样板,以在您的数据框中读取,下次请制作示例MCVE,以帮助我们为您提供帮助

# Compute raw ranks like you do
ranks['r_rank'] = ranks['Ave. Recency'].rank()
ranks['f_rank'] = ranks['Ave. Frequency'].rank(ascending=False)
ranks['m_rank'] = ranks['Ave. Monetary'].rank(ascending=False)
ranks['s_rank'] = ranks['Ave. Spend'].rank(ascending=False)

# Find the indices of both the highest AveSpend and AveRecency    
ismax = ranks['Ave. Spend'].idxmax()
irmax = ranks['Ave. Recency'].idxmax()

# Get the overall ranking for every row other than these... add 1 to offset for excluding the max-AveSpend row:
ranks['overall_rank'] = 1 + ranks.drop(index = [ismax,irmax]) [['r_rank','f_rank','m_rank','s_rank']].sum(axis=1).rank(method='first')

# (Note: in .loc[], can't mix indices (ismax) with column-names)
ranks.loc[ ranks['Ave. Spend'].idxmax(), 'overall_rank' ] = 1 
ranks.loc[ ranks['Ave. Recency'].idxmax(), 'overall_rank' ] = len(ranks)

这是提取数据的样板:

import pandas as pd

from io import StringIO

# """Cluster   No. Customers| Ave. Recency| Ave. Frequency| Ave. Monetary| Ave. Spend|
dat = """
0           145118        39.0197       1.7264         14,971.85     8,672.07
1           2             47.0          19.0          237,270.00    12,487.89
2           1236          15.9716       24.9101       126,992.79     5,098.02
3           219847        41.9736       3.0682         17,701.64     5,769.23
4           9837          23.9330       3.2735        172,642.35    52,738.42
5           64865         24.8281       1.8599         13,159.21     7,075.19
6           3855          26.5647       3.9304         54,333.56    13,823.64
7           219549        17.7493       3.3356         17,570.67     5,267.52
8           34171         23.5205       9.1703         42,136.68     4,594.89
9           3924120       24.7933       1.1684          4,754.76     4,069.21 """

# Remove the comma thousands-separator, to prevent your floats being read in as string
dat = dat.replace(',', '')

ranks = pd.read_csv(StringIO(dat), sep='\s+', names=
    "Cluster|No. Customers|Ave. Recency|Ave. Frequency|Ave. Monetary|Ave. Spend".split('|'))