Question

所以我正在进行一项任务，我正在试图找出一个步骤，我们将采用csv文件并将其转换为字典。我的一个限制是不导入csv。在这里搜索关于将CSV文件转换为字典的其他问题时，所有问题都使用import csv。例如，如果我的CSV文件包含以下内容（第一行是标题）：

"Headline","Newspaper", "Year Published", "Author"
"Crash", "New York Times", "2001", "Smith"
"Fire", "Washington Post", "2010", "Dudley"
"Addiction", "National Enquirer", "2008", "Kuhn"

如何在不导入csv的情况下将这些数据转换为Python中的字典？

我目前在做什么，我有一些伪代码，因为我知道我想要对这些部分执行什么，但我不是100％肯定如何这样做，因为这是我第一周使用文件和词典一起：

def read_file(filename):
    d={}
    with open(filename , 'r') as f:
        first_line = f.readline()
            for line in f:
                if line.strip().isdigit():
                # assign first string in line as key
                    #assign rest of strings in line as value to key in that line as a tuple

Answer 1

从标题中提取密钥后，您可以转置剩余的行：

with open("foo.csv") as f:
    # pull header line and split into keys
    keys = next(f).strip().split(",")
    # transpose remaining lines 
    values = zip(*(line.strip().split(",") for line in f))
    # zip keys and values and create a dict
    dct = dict(zip(keys, vals)

转置会逐列显示元素，因此它们将与每个标题键匹配。

Answer 2

您可以为每一行创建一个包含字典项的列表，这更像是JSON样式格式。

with open(filename) as f:
    keys = [val.strip().replace('"', '') for val in f.readline().split(',')]
    result = []
    for line in f:
        line = [val.strip().replace('"', '') for val in line.split(',')]
        result.append({key: val for key, val in zip(keys, line)})

>>> result
[{'Author': 'Smith',
  'Headline': 'Crash',
  'Newspaper': 'New York Times',
  'Year Published': '2001'},
 {'Author': 'Dudley',
  'Headline': 'Fire',
  'Newspaper': 'Washington Post',
  'Year Published': '2010'},
 {'Author': 'Kuhn',
  'Headline': 'Addiction',
  'Newspaper': 'National Enquirer',
  'Year Published': '2008'}]

CSV文件到dict

2 个答案: