函数使用规则返回最高计数值

时间:2018-02-08 08:14:29

标签: python pandas

我有两个列,如下所示,并尝试返回第二列的最高计数,但它只是给我最高的评分数而不考虑性别

数据:

print(df)

   AGE GENDER rating
0   10      M     PG
1   10      M      R
2   10      M      R
3    4      F   PG13
4    4      F   PG13

代码:

 s = (df.groupby(['AGE', 'GENDER'])['rating']
       .apply(lambda x: x.value_counts().head(2))
       .rename_axis(('a','b', 'c'))
       .reset_index(level=2)['c'])

输出:

print (s[F])
('PG')

print(s[M]

('PG', 'R')

2 个答案:

答案 0 :(得分:2)

以下是此文件的标准库解决方案:

%%file "test.txt"
gender  rating
M   PG
M   R
F   NR
M   R
F   PG13
F   PG13

<强>鉴于

import collections as ct


def read_file(fname):
    with open(fname, "r") as f:
        header = next(f)
        for line in f:
            gender, rating = line.strip().split()
            yield gender, rating

<强>代码

filename = "test.txt"

dd = ct.defaultdict(ct.Counter)
for k, v in sorted(read_file(filename), key=lambda x: x[0]):
    dd[k][v] += 1 

{k: v.most_common(1) for k, v in dd.items()}
# {'F': [('PG13', 2)], 'M': [('R', 2)]}

<强>详情

解析文件的每一行并将其添加到defaultdict。键是性别,但每个性别的每个评级的值为Counter个对象。调用Counter.most_common()来检索最常出现的事件。

由于数据按性别分组,因此您可以浏览更多信息。例如,每个性别的唯一评分:

{k: set(v.elements()) for k, v in dd.items()}
# {'F': {'NR', 'PG13'}, 'M': {'PG', 'R'}}

答案 1 :(得分:1)

我认为您需要使用groupby + value_counts + head来计算类别和评分:

df1 = (df.groupby('gender')['rating']
         .apply(lambda x: x.value_counts().head(1))
         .rename_axis(('gender','rating'))
         .reset_index(name='val'))
print (df1)
  gender rating  val
0      F   PG13    2
1      M      R    2

如果只想获得最高评级,请选择每组索引的第一个值:

s = df.groupby('gender')['rating'].apply(lambda x: x.value_counts().index[0])
print (s)
gender
F    PG13
M       R
Name: rating, dtype: object

print (s['M'])
R
print (s['F'])
PG13

或者只有最高计数选择每组Series的第一个值:

s = df.groupby('gender')['rating'].apply(lambda x: x.value_counts().iat[0])
print (s)
gender
F    2
M    2
Name: rating, dtype: int64

print (s['M'])
2
print (s['F'])
2

编辑:

s = df.groupby('gender')['rating'].apply(lambda x: x.value_counts().index[0])

def gen_mpaa(gender):
    return s[gender]

print (gen_mpaa('M'))

print (gen_mpaa('F'))

编辑:

解决方案,如果genre id值是字符串:

print (type(df.loc[0, 'genre id']))
<class 'str'>

df = df.set_index('gender')['genre id'].str.split(',', expand=True).stack()
print (df)
gender   
M       0    11
        1    22
        2    33
        0    22
        1    44
        2    55
        0    33
        1    44
        2    55
F       0    11
        1    22
        0    22
        1    55
        0    55
        1    44
dtype: object

d = df.groupby(level=0).apply(lambda x: x.value_counts().index[0]).to_dict()
print (d)
{'M': '55', 'F': '55'}

EDIT1:

print (df)
   AGE GENDER rating
0   10      M     PG
1   10      M      R
2   10      M      R
3    4      F   PG13
4    4      F   PG13

s = (df.groupby(['AGE', 'GENDER'])['rating']
       .apply(lambda x: x.value_counts().head(2))
       .rename_axis(('a','b', 'c'))
       .reset_index(level=2)['c'])
print (s)

a   b
4   F    PG13
10  M       R
    M      PG
Name: c, dtype: object