如何解决错误-名称未在python中的函数中定义

Question

假设我有一个包含列、电影 ID、标题、年份和评级的数据集。 以下是我的数据的示例子集：

| movieId |   title  | rating | year |
|:-------:|:--------:|:------:|:----:|
| 1       | abc      | 3.5    | 1995 |
| 1       | abc      | 3      | 1995 |
| 1       | abc      | 4      | 1995 |
| 1       | abc      | 3      | 1995 |
| 1       | abc      | 5      | 1995 |
| 1       | abc      | 3.5    | 1995 |
| 1       | abc      | 4.5    | 1995 |
| 1       | abc      | 0.5    | 1995 |
| 1       | abc      | 3.5    | 1995 |
| 1       | abc      | 4.5    | 1995 |
| 1       | abc      | 4      | 1995 |
| 1       | abc      | 5      | 1995 |
| 1       | abc      | 4.5    | 1995 |
| 1       | abc      | 4      | 1995 |
| 1       | abc      | 4      | 1995 |
| 1       | abc      | 4      | 1995 |
| 1       | abc      | 4      | 1995 |
| 1       | abc      | 3      | 1995 |
| 1       | abc      | 4      | 1995 |
| 1       | abc      | 3.5    | 1995 |
| 1       | abc      | 3      | 1995 |
| 1       | abc      | 4      | 1995 |
| 1       | abc      | 5      | 1995 |
| 1       | abc      | 4.5    | 1995 |
| 1       | abc      | 5      | 1995 |
| 2       | xyz      | 3      | 2000 |
| 2       | xyz      | 2      | 2000 |
| 2       | xyz      | 3.5    | 2000 |
| 2       | xyz      | 4      | 2000 |
| 2       | xyz      | 3.5    | 2000 |
| 2       | xyz      | 5      | 2000 |
| 2       | xyz      | 3.5    | 2000 |
| 2       | xyz      | 3      | 2000 |
| 2       | xyz      | 3      | 2000 |
| 2       | xyz      | 2      | 2000 |
| 2       | xyz      | 3.5    | 2000 |
| 2       | xyz      | 3      | 2000 |
| 2       | xyz      | 3      | 2000 |
| 2       | xyz      | 4      | 2000 |
| 2       | xyz      | 2      | 2000 |
| 2       | xyz      | 3.5    | 2000 |
| 2       | xyz      | 1      | 2000 |
| 3       | pqr      | 3      | 1997 |
| 3       | pqr      | 2      | 1997 |
| 3       | pqr      | 3.5    | 1997 |
| 3       | pqr      | 3.5    | 1997 |
| 3       | pqr      | 3      | 1997 |
| 3       | pqr      | 3      | 1997 |
| 3       | pqr      | 3      | 1997 |
| 3       | pqr      | 3      | 1997 |
| 3       | pqr      | 4.5    | 1997 |
| 3       | pqr      | 3.5    | 1997 |
| 3       | pqr      | 4      | 1997 |
| 3       | pqr      | 1.5    | 1997 |
| 3       | pqr      | 2      | 1997 |
| 3       | pqr      | 2      | 1997 |
| 3       | pqr      | 2.5    | 1997 |
| 4       | def      | 3      | 1999 |
| 4       | def      | 2.5    | 1999 |
| 4       | def      | 2.5    | 1999 |
| 4       | def      | 0.5    | 1999 |
| 4       | def      | 2      | 1999 |
| 4       | def      | 3      | 1999 |
| 5       | movie123 | 4      | 2006 |
| 5       | movie123 | 4      | 2006 |
| 5       | movie123 | 3      | 2006 |
| 5       | movie123 | 1.5    | 2006 |
| 5       | movie123 | 3      | 2006 |
| 5       | movie123 | 2      | 2006 |
| 5       | movie123 | 2.5    | 2006 |
| 5       | movie123 | 3      | 2006 |
| 5       | movie123 | 4      | 2006 |
| 5       | movie123 | 0.5    | 2006 |
| 5       | movie123 | 1      | 2006 |
| 5       | movie123 | 3.5    | 2006 |
| 5       | movie123 | 2      | 2006 |
| 5       | movie123 | 3      | 2006 |
| 5       | movie123 | 1.5    | 2006 |
| 5       | movie123 | 2.5    | 2006 |
| 5       | movie123 | 4      | 2006 |
| 5       | movie123 | 4      | 2006 |
| 5       | movie123 | 3.5    | 2006 |
| 5       | movie123 | 3      | 2006 |
| 6       | movie456 | 4      | 2012 |
| 6       | movie456 | 3.5    | 2012 |
| 6       | movie456 | 3.5    | 2012 |
| 6       | movie456 | 4      | 2012 |
| 6       | movie456 | 5      | 2012 |
| 6       | movie456 | 2.5    | 2012 |
| 6       | movie456 | 4      | 2012 |
| 6       | movie456 | 4      | 2012 |
| 6       | movie456 | 3.5    | 2012 |
| 6       | movie456 | 5      | 2012 |
| 6       | movie456 | 2      | 2012 |
| 6       | movie456 | 4      | 2012 |

我想通过定义一个函数来计算整个数据集的平均值、计数、最小评分和平均评分。 所以，我首先计算每部电影的评分数量和平均值。

avg_rating = df.groupby(['movieId','year'])['ratings'].agg([('Count','size'), ('Mean','mean')]).sort_values(by='Mean',ascending=False)

由于某些电影的评论数量可能较少但评级较高，而其他电影的评论数量较多且评级较高，因此分析可能会出现偏差。 所以，我想计算加权平均值，为此我定义了一个函数。

# R = average for the movie (mean) = (Rating)
# v = number of ratings/reviews for the movie = (votes)
# m = minimum reviews required to be listed in the Top 250 movie list
# C = the mean rating across the whole report

def weighted_rating(R, v, m, C): 
  return (v/(v+m))*R + (m/(v+m))*C

avg_rating= avg_rating.assign(wr = weighted_rating(mean, count, 500, mean(mean)))

当我运行上面的最后一行代码时，我收到一个错误：name 'mean' is not defined

列计数仍然存在相同的错误：未定义名称“计数”

如何解决此错误，以便我的最终输出包含列 movieId、year、count、mean 和 wr？

Answer 1

您需要使用df的字段：

avg_rating = avg_rating.assign(
        wr=weighted_rating(avg_rating['Mean'], avg_rating['Count'], 500, mean(avg_rating['Mean'])))

所以最终的代码可能是这样的：

#!/usr/bin/env python
# -*- coding: utf-8 -*-

from __future__ import (division, absolute_import, print_function,
                        unicode_literals)

import pandas as pd
from statistics import mean


def weighted_rating(R, v, m, C):
    return (v / (v + m)) * R + (m / (v + m)) * C


def main():

    df = pd.read_csv('movies.csv')

    avg_rating = df.groupby(['movieId', 'year'])['rating'].agg(
        [('Count', 'size'), ('Mean', 'mean')]).sort_values(by='Mean', ascending=False)

    avg_rating = avg_rating.assign(
        wr=weighted_rating(avg_rating['Mean'], avg_rating['Count'], 500, mean(avg_rating['Mean'])))

    print(avg_rating)


if __name__ == '__main__':
    main()

使用 CSV：

movieId,title,rating,year
1,abc,3.5,1995
1,abc,3,1995
1,abc,4,1995
1,abc,3,1995
1,abc,5,1995
1,abc,3.5,1995
1,abc,4.5,1995
1,abc,0.5,1995
1,abc,3.5,1995
1,abc,4.5,1995
1,abc,4,1995
1,abc,5,1995
1,abc,4.5,1995
1,abc,4,1995
1,abc,4,1995
1,abc,4,1995
1,abc,4,1995
1,abc,3,1995
1,abc,4,1995
1,abc,3.5,1995
1,abc,3,1995
1,abc,4,1995
1,abc,5,1995
1,abc,4.5,1995
1,abc,5,1995
2,xyz,3,2000
2,xyz,2,2000
2,xyz,3.5,2000
2,xyz,4,2000
2,xyz,3.5,2000
2,xyz,5,2000
2,xyz,3.5,2000
2,xyz,3,2000
2,xyz,3,2000
2,xyz,2,2000
2,xyz,3.5,2000
2,xyz,3,2000
2,xyz,3,2000
2,xyz,4,2000
2,xyz,2,2000
2,xyz,3.5,2000
2,xyz,1,2000
3,pqr,3,1997
3,pqr,2,1997
3,pqr,3.5,1997
3,pqr,3.5,1997
3,pqr,3,1997
3,pqr,3,1997
3,pqr,3,1997
3,pqr,3,1997
3,pqr,4.5,1997
3,pqr,3.5,1997
3,pqr,4,1997
3,pqr,1.5,1997
3,pqr,2,1997
3,pqr,2,1997
3,pqr,2.5,1997
4,def,3,1999
4,def,2.5,1999
4,def,2.5,1999
4,def,0.5,1999
4,def,2,1999
4,def,3,1999
5,movie12,4,2006
5,movie12,4,2006
5,movie12,3,2006
5,movie12,1.5,2006
5,movie12,3,2006
5,movie12,2,2006
5,movie12,2.5,2006
5,movie12,3,2006
5,movie12,4,2006
5,movie12,0.5,2006
5,movie12,1,2006
5,movie12,3.5,2006
5,movie12,2,2006
5,movie12,3,2006
5,movie12,1.5,2006
5,movie12,2.5,2006
5,movie12,4,2006
5,movie12,4,2006
5,movie12,3.5,2006
5,movie12,3,2006
6,movie45,4,2012
6,movie45,3.5,2012
6,movie45,3.5,2012
6,movie45,4,2012
6,movie45,5,2012
6,movie45,2.5,2012
6,movie45,4,2012
6,movie45,4,2012
6,movie45,3.5,2012
6,movie45,5,2012
6,movie45,2,2012
6,movie45,4,2012