以pythonic方式从制表符分隔文件构建字典

Question

我编写了以下函数,该函数采用制表符分隔文件(作为字符串)并将其转换为字典,其中包含整数作为键和两个浮点数的列表以及值:

def parseResults(self, results):
    """
    Build a dictionary of the SKU (as key), current UK price and current Euro price
    """
    lines = results.split('\n')
    individual_results = []
    for i in range(1,len(lines)-1):
        individual_results.append(lines[i].split('\t'))
    results_dictionary = {}
    for i in range(len(individual_results)):
        results_dictionary[int(individual_results[i][0])] = [float(individual_results[i][1]), float(individual_results[i][2])]
    return results_dictionary

我一直在阅读有关使用列表理解和字典理解的内容,但我真的不知道构建它的最佳方法是什么.

我想我可以使用以下方法简化第一个列表构建:

individual_results = [results.split('\t') for results in lines[1:]]

但我不知道创建字典的最佳方法.我甚至没有创建中间列表,我觉得这可能是一种整洁的方式.

谢谢,

马特

Answer 1

像这样:

import csv
import StringIO
results = "sku\tdelivered-price-gbp\tdelivered-price-euro\tid\n32850238\t15.53\t35.38\t258505\n"

data = list(csv.DictReader(StringIO.StringIO(results), delimiter='\t'))
print(data)

输出:

[{'sku': '32850238', 'delivered-price-euro': '35.38', 'delivered-price-gbp': '15.53', 'id': '258505'}]

当然,如果您可以从实际文件中读取,则可以跳过stringIO部分.

要构建所需的字典类型,您可以这样做:

data = {}
for entry in csv.DictReader(StringIO.StringIO(results), delimiter='\t'):
    data[entry['sku']] = [entry['delivered-price-gbp'], entry['delivered-price-euro']]

或者甚至作为字典理解:

import csv
import StringIO
results = "sku\tdelivered-price-gbp\tdelivered-price-euro\tid\n32850238\t15.53\t35.38\t258505\n10395850\t35.21\t46.32\t3240582\n"

data = {entry['sku']: [entry['delivered-price-gbp'], entry['delivered-price-euro']] 
        for entry in csv.DictReader(StringIO.StringIO(results), delimiter='\t')}
print(data)

但这现在变得非常难以阅读.

在最后两种情况下的输出是:

{'32850238': ['15.53', '35.38'], '10395850': ['35.21', '46.32']}