关于Python的CSV到dictionary

CSV to dictionary

我需要一些协助:

编写一个函数,打开文件Exports2012.csv并返回爱沙尼亚十大出口产品的地图。地图应该将产品名称与其对应的美元价值关联起来。

为了方便起见,您应该将字符串'$2,268,911,208.49'转换为浮动值。

CSV实例:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
#,HS,Name,Value (USD),Percent
1,8517,Telephones,"$2,823,450,843.60",15.38%
2,2710,Refined Petroleum,"$2,124,413,818.52",11.57%
3,8703,Cars,"$371,092,090.84",2.02%
4,7204,Scrap Iron,"$331,463,406.48",1.81%
5,8544,Insulated Wire,"$319,352,873.32",1.74%
6,4011,Rubber Tires,"$242,977,533.70",1.32%
7,8708,Vehicle Parts,"$241,059,109.78",1.31%
8,8429,Large Construction Vehicles,"$239,589,588.65",1.31%
9,4407,Sawn Wood,"$238,358,904.17",1.30%
10,4418,Wood Carpentry,"$237,521,163.59",1.29%
11,7210,Coated Flat-Rolled Iron,"$213,137,606.81",1.16%
12,9404,Mattresses,"$208,042,615.08",1.13%
13,4403,Rough Wood,"$206,112,209.11",1.12%
14,9403,Other Furniture,"$202,900,185.49",1.11%
15,8504,Electrical Transformers,"$202,856,149.28",1.10%

我知道如何提取2。3。纵队,但我被困在这一点上。

1
2
3
4
5
6
import csv
f= open('EstonianExports2011.csv', 'rb')
archive = csv.reader(f, delimiter=',')
arch_dict = {}
arch_dict = {row[2]: row[3]for row in archive}
print arch_dict

我很感谢你的帮助。


您的文件已经从高到低排序,所以您只需要在头后取前十行,还需要去掉$号并替换,号:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
import csv
with open('EstonianExports2011.csv', 'rb')as f:
    archive = list(csv.reader(f, delimiter=','))[1:11] # get lines 1 to 10
    arch_dict = {row[2]: float(row[3].strip("$").replace(",","")) for row in archive}

arch_dict
{'Rubber Tires': 242977533.7, 'Cars': 371092090.84, 'Vehicle Parts': 241059109.78, 'Insulated Wire': 319352873.32, 'Scrap Iron': 331463406.48, 'Telephones': 2823450843.6, 'Sawn Wood': 238358904.17, 'Large Construction Vehicles': 239589588.65, 'Wood Carpentry': 237521163.59, 'Refined Petroleum': 2124413818.52}

In [2]: s ="$213,137,606.81"

In [3]: s.strip("$") # strips from the ends of a string
Out[3]: '213,137,606.81'

In [5]: s.strip("$").replace(",","") # combine with replace to remove the commas
Out[5]: '213137606.81'

使其成为一个函数应该是非常直接的。


由于这是一个分配,我不会(至少在最初)提供显式代码,而是提供一个算法建议:

  • product = row[2]valuestring = row[3]放在你的档案里;
  • 只取第一个字符和最后一个字符之间的valuestring部分;
  • 从修剪过的valuestring中删除逗号;
  • 转换为浮动;
  • 将其保存到成对列表(产品和值),可能使用zip函数;
  • 在字典理解中使用sorted(zipped_list, key=lambda l:l[1]函数编写字典,就像当前的一样。