Python：如何对字典数据进行排序和组织

Question

我正在尝试按邮政编码对犯罪总数进行排序，并按犯罪类型对受害者计数进行排序。 我已经按报告号建立了词典。 这是我在打印字典时输出的一小部分数据示例：

{'100065070': ['64130', '18', 'VIC', 'VIC', 'VIC'], '20003319': ['64130', '13', 'VIC'], '60077156': ['64130', '18', 'VIC'], '100057708': ['99999', '17', 'VIC', 'VIC'], '40024161': ['64108', '17', 'VIC', 'VIC']}

该词典的结构如下：{Report_number：[邮政编码，犯罪类型，受害者人数]}

我是编码的新手，只是在学习字典。 我将如何对字典进行排序以将我的数据整理成这种格式？

 Zip Codes Crime totals 

====================

   64126 809
   64127 3983

   64128 1749
   64129 1037
   64130 4718
   64131 2080
   64132 2060
   64133 2005
   64134 2928

任何帮助将非常感激。 下面是到目前为止的代码。 我正在访问大约50,000行数据的两个文件，因此效率非常重要。

from collections import Counter

incidents_f =  open('incidents.csv', mode = "r")

crime_dict = dict()

for line in incidents_f:
    line_1st = line.strip().split(",")
    if line_1st[0].upper() != "REPORT_NO":
        report_no = line_1st[0]
        offense = line_1st[3]
        zip_code = line_1st[4]
        if len(zip_code) < 5:
            zip_code = "99999"

        if report_no in crime_dict:
            crime_dict[report_no].append(zip_code).append(offense)
        else:
            crime_dict[report_no] = [zip_code]+[offense]

#close File
incidents_f.close

details_f = open('details.csv',mode = 'r')
for line in details_f:
    line_1st = line.strip().split(",")
    if line_1st[0].upper() != "REPORT_NO":
        report_no = line_1st[0]
        involvement = line_1st[1]
        if involvement.upper() == 'VIC':
            victims = "VIC"

        if report_no in crime_dict:
            crime_dict[report_no].append(victims)
        else:
            continue


#close File
details_f.close



print(crime_dict)

Answer 1

这是一种比@Alexander解决方案更多代码的方法：

crime_dict ={
    '100065070': ['64130', '18', 'VIC', 'VIC', 'VIC'], 
    '20003319': ['64130', '13', 'VIC'], 
    '60077156': ['64130', '18', 'VIC'],
    '100057708': ['99999', '17', 'VIC', 'VIC'], 
    '40024161': ['64108', '17', 'VIC', 'VIC']
    }

crimes_by_zip = {}
for k, v in crime_dict.items():
    zip = v[0]
    if zip not in crimes_by_zip.keys():
        crimes_by_zip[zip] = 0
    crimes_by_zip[zip] += 1

for zip in sorted(crimes_by_zip.keys()):
    print(zip, crimes_by_zip[zip])

64108 1
64130 3
99999 1

Answer 2

D = {'100065070': ['64130', '18', 'VIC', 'VIC', 'VIC'], '20003319': ['64130', '13', 'VIC'], '60077156': ['64130', '18', 'VIC'], '100057708': ['99999', '17', 'VIC', 'VIC'], '40024161': ['64108', '17', 'VIC', 'VIC']}

data_with_zip_duplicate = [(D[key][0],key) for  key in sorted(D.keys(), key = lambda x:D[x][0] )]
print(*data_with_zip_duplicate, sep = "\n")