创建一个字典，提取平均数字

Question

我正在尝试做回年，每年的平均分数。 我想要做的是创建一个符合年份的字典：年级，然后得到另一个年份字典：sum_of_grade然后等等。

它从csv文件，它有两个头是：一年级

Year Grade 
2001 100 
2002 99 
2001 88 
2003 11 
2005 55 

还有很多，但我不认为这是需要有完整的数据。

def construct_values(file): 
    """ 
    Construct the values needed to graph the average grade of the class over time 

    Parameters 
    ---------- 
    file_path: A string. Absolute path to file. 

    Returns 
    ------- 
    years: array of integers 
    average_grades: array of floats 
    """ 
    years, average_grades = [], [] 
    grades = [] 
    d = {} 
    with open(file,'r') as f: 
     next(f) 
     for line in f: 
      year, grade = (s.strip() for s in line.split(',')) 
      years.append(year) # array year 
      grades.append(grade) # array grade 
      d = dict(zip(years,grades)) # dict year:grade 

     for i,j in d: 
      # i for count frequencies of years 
      # j for summation of grades 
      # j/i for average grade and extract as array 


     return years, average_grades 

我试图说清楚，但如果仍不清楚，请告诉我。

Answer 1

当你使用这个有个问题：

d = dict(zip(years,grades)) # dict year:grade

把你的输入数据作为例子，它会生成一个字典，如：

{2001: 88, 2002: 99, 2003:11, 2005: 55} 

因为在构造字典中存在重复键时，该值将被覆盖。

那么，要实现这一点，我建议用另一种字典生成方法，做这样的事情：

def construct_values(file): 
    """ 
    Construct the values needed to graph the average grade of the class over time 

    Parameters 
    ---------- 
    file_path: A string. Absolute path to file. 

    Returns 
    ------- 
    years: array of integers 
    average_grades: array of floats 
    """ 
    years, average_grades = [], [] 
    # grades = []  This variable don't need anymore 
    d = {} 
    with open(file,'r') as f: 
     next(f) 
     for line in f: 
      year, grade = (s.strip() for s in line.split(',')) 

      # here is the begin line difference from your code 
      if year not in d: 
       d[year] = [int(grade), 1] 
      else: 
       d[year][0] += int(grade) 
       d[year][1] += 1 

     for year, grade_info in d.items(): 
      years.append(year) 
      average_grades.append(grade_info[0]/grade_info[1]) 
      # end difference from your code 

     return years, average_grades 

在中间字典d，价值节省约[sum_of_grade，times_appeared_in_the_year信息]，因此当迭代字典时，可以使用sum_of_grade/times_appeared_in_the_year轻松地使用来计算平均值。

所以，当你看到一个表（CSV文件是一个），你应该觉得熊猫（我认为）你并不需要尽快使用额外的变量等级

Answer 2

虽然创建为dict(zip(years,grades))重复密钥将不允许在字典中。所以最好使用字典以外的替代方法。

有些事情是这样的。

from itertools import groupby 
combined = zip(year,grade)  
for n,g in groupby(sorted(combined, key = lambda x:x[0]),key=lambda x:x[0]): 
    grades = [int(i[1])for i in g] 
    print 'year : %s average : %s' %(n,sum(grades)/len(grades)) 

结果：

year : 2001 average : 94 
year : 2002 average : 99 
year : 2003 average : 11 
year : 2005 average : 55 

Answer 3

。

这里是一个大熊猫溶液：

import pandas as pd 
import io 

csv = """Year,Grade 
2001,100 
2002,99 
2001,88 
2003,11 
2005,55""" 

df = pd.read_csv(io.StringIO(csv)) 

year_grade = {k: list(v) for k,v in df.groupby("Year")["Grade"]} 
year_avg_grade = df.groupby("Year")["Grade"].mean().to_dict() 

year_grade：

{2001: [100, 88], 2002: [99], 2003: [11], 2005: [55]} 

year_avg_grade：

{2001: 94, 2002: 99, 2003: 11, 2005: 55}