python之使用 Python 消除文件中的冗余

如何压缩，即从以下数据中消除冗余:

code: GB-ENG, jobs: 2673 
code: GB-ENG, jobs: 23 
code: GB-ENG, jobs: 459 
code: GB-ENG, jobs: 346 
code: RO-B, jobs: 9 
code: DE-NW, jobs: 4 
code: DE-BW, jobs: 3 
code: DE-BY, jobs: 9 
code: DE-HH, jobs: 34 
code: DE-BY, jobs: 11 
code: BE-BRU, jobs: 27 
code: GB-ENG, jobs: 20 

输出应该是这样的:

GB-ENG, 3521 
RO-B, 9 
DE-NW, 4 
DE-BW, 3 
DE-HH, 34 
DE-BY, 20 
BE-BRU, 27 

由每个代码的 1 个规范表示形式描述，即 DE-BY，表示与该代码的每个实例相关联的数字的总和，例如:

code: DE-BY, jobs: 11 
code: DE-BY, jobs: 9 

成为

DE-BY, 20 

目前我正在使用此 Python 脚本创建该输入:

import json 
import requests 
from collections import defaultdict 
from pprint import pprint 
 
def hasNumbers(inputString): 
    return any(char.isdigit() for char in inputString) 
 
# open up the output of 'data-processing.py' 
with open('job-numbers-by-location.txt') as data_file: 
 
    # print the output to a file 
    with open('phase_ii_output.txt', 'w') as output_file_: 
        for line in data_file: 
            identifier, name, coords, number_of_jobs = line.split("|") 
            coords = coords[1:-1] 
            lat, lng = coords.split(",") 
            # print("lat: " + lat, "lng: " + lng) 
            response = requests.get("http://api.geonames.org/countrySubdivisionJSON?lat="+lat+"&lng="+lng+"&username=s.matthew.english").json() 
 
 
            codes = response.get('codes', []) 
            for code in codes: 
                if code.get('type') == 'ISO3166-2': 
                    country_code = '{}-{}'.format(response.get('countryCode', 'UNKNOWN'), code.get('code', 'UNKNOWN')) 
                    if not hasNumbers( country_code ): 
                        # print("code: " + country_code + ", jobs: " + number_of_jobs) 
                        output_file_.write("code: " + country_code + ", jobs: " + number_of_jobs) 
    output_file_.close() 

将此功能作为该脚本的一部分包含在内可能是最有效的，但我还无法弄清楚如何做。

请您参考如下方法:

假设文本存储在文本文件中，这会起作用

infile = open('redundancy.txt','r') 
a= infile.readlines() 
print a 
d={} 
for item in a: 
    c=item.strip('\n')     
    b=c.split()     
    if b[1] in d : 
        d[b[1]]= int(d.get(b[1]))+eval((b[3])) 
    else: 
        d[b[1]]=b[3] 
print d 

它会给出一个结果:

{'DE-BY,': 20, 'DE-HH,': '34', 'DE-BW,': '3', 'DE-NW,': '4', 'RO-B,': '9', 'GB-ENG,': 3521, 'BE-BRU,': '27'}