从带有关键年份和值列表的 txt 文件创建字典

Question

I have the txt file below

2014,STAR,105,234
2014,COMET,2877,1426
2014,ASTEROID,73,68
2014,PLANET,134,143
2014,"ACTIVE STARS,DEADSTARS",166,125
2015,STAR,69,24
2015,ASTEROID,59,32
2015,PLANET,42,13
2015,STAR,79,33
2015,BLACK HOLES,8,3
2015,"ACTIVE STARS,DEADSTARS",19,16
2015,ASTEROID,12,0
2016,STAR,120,47
2016,"ACTIVE STARS,DEADSTARS",4,1
2016,PLANET,14,12
2016,ASTEROID,21,1

and i need to write a function that reads the file and creates a dictionary with key the year, and values, a list with values the sum of 3 column and the sum of 4 column of txt file for the year.

def file_read(fname):
  with open(fname,'r') as document:
    
    zip_lists=[]
    dictionary1 ={}
    tempYear=[]
    tempSum3=[]
    tempSum4=[]
    for line in document:
        #print(line)
        line=line.strip()
        alist=line.split(',')
        tempYear.append(alist[0])
        #print(alist[2])
        if len(alist)==5:
          tempSum3.append(int(alist[3]))
          tempSum4.append(int(alist[4]))
          
        else:
          tempSum3.append(int(alist[2]))
          tempSum4.append(int(alist[3]))

    
    zip_lists=zip(tempSum3,tempSum4)
    

    dictionary1=dict(zip(tempYear,zip_lists))
   
    return dictionary1

Answer 1

I don't know exactly what you want. Maybe this work for you:

import pandas as pd

def file_read(fname):
    df = pd.read_csv(fname, sep=',')
    df.columns = ['Year', 'col2', 'col3', 'col4']

    sum_col3 = df.groupby('Year')['col3'].sum().to_list()
    sum_col4 = df.groupby('Year')['col4'].sum().to_list()

    n1 = zip(sum_col3, sum_col4)
    n2 = zip(df['Year'].unique(), n1)
    dictionary = {x[0]: list(x[1]) for x in n2}

    return dictionary

The output:

{2014: [3250, 1762], 2015: [288, 121], 2016: [159, 61]}

Answer 2

You can try something like this:

import csv 
from itertools import groupby

def conv(s):
    try:
        return int(s)
    except ValueError:
        return s
    
result={}

with open(ur_file) as f_in:
    r=csv.reader(f_in)
    for k,g in groupby(r, key=lambda x:x[0]):
        sl=list(g)
        result[k]=(sum(conv(e[2]) for e in sl), sum(conv(e[3]) for e in sl))

>>> result
{2014: (3355, 1996), 2015: (288, 121), 2016: (159, 61)}