英文:
convert a column(string) in csv file to a tuple of ints
问题
要选择第三列，将字符串“2/23/2023”更改为元组(2, 23, 2023)，您可以在类型列表中将第三列的类型更改为一个函数，该函数将执行所需的转换。然后，您可以在创建字典的列表时应用此新的函数。以下是您的代码的修改部分：
import csvfile = open('csvfile.csv')lines = csv.reader(file)header = next(lines)# 修改第三列的类型为一个函数，该函数将执行转换types = [str,float,lambda date_str: tuple(map(int, date_str.split('/')))]# 创建包含字典的列表alist_of_dicts = [{name: func(val)for name, func, val in zip(header, types, line)}for line in lines]
这将使第三列的字符串“2/23/2023”转换为元组(2, 23, 2023)。不过请注意，要确保文件中的日期字符串的格式始终是“M/D/YYYY”，以便分割和转换工作正常。
英文:
Currently, I process a file using the csv module, which creates a list of dictionaries.
import csvfile = open('csvfile.csv')lines = csv.reader(file)header = next(lines) # ['name', 'price', 'date']# when I do the followingfor line in lines:print(line)# I get the following['xxxx', '5.00', '2/23/2023']# assigning types to the columns to do type conversion using a functiontypes = [str,float,str # this need to be a tuple# tried tuple(map(int, cannotchoosecolumn.split('/')))# did not work]# now to create a list of dictsalist_of_dicts = [{name: func(val)for name, func, val in zip(header, types, line)}for line in lines]
How would I select the third column str(2/23/2023) to change to a tuple(2, 21, 2007) using the format I am currently using?
答案1
得分: 2
你可以将一个函数传递给你的 types 列表：
import datetimedef read_date(s):d = datetime.datetime.strptime(s, "%m/%d/%Y")return (d.month, d.day, d.year)header = ["name", "price", "date"]lines = [["xxxx", "5.00", "2/23/2023"]]types = [str,float,read_date,]alist_of_dicts = [{name: func(val) for name, func, val in zip(header, types, line)} for line in lines]print(alist_of_dicts)# 输出: [{'name': 'xxxx', 'price': 5.0, 'date': (2, 23, 2023)}]
这段代码可能有点难以理解。相反，我建议你使用 csv.DictReader 来将 CSV 文件读取为字符串 -> 字符串的字典，然后进行列的转换。
英文:
You can pass a function to your types list:
import datetimedef read_date(s):d = datetime.datetime.strptime(s, "%m/%d/%Y")return (d.month, d.day, d.year)header = ["name", "price", "date"]lines = [["xxxx", "5.00", "2/23/2023"]]types = [str,float,read_date,]alist_of_dicts = [{name: func(val) for name, func, val in zip(header, types, line)} for line in lines]print(alist_of_dicts)# prints: [{'name': 'xxxx', 'price': 5.0, 'date': (2, 23, 2023)}]
This is hard to understand code though. Instead, I recommend you use csv.DictReader to read the csv as a dictionary of strings -> strings, then transforming the columns
答案2
得分: 2
使用csv.DictReader 并在读取时转换列：
import csvwith open('csvfile.csv', newline='') as file:a_list_of_dicts = []for line in csv.DictReader(file):line['price'] = float(line['price'])line['date'] = tuple(int(n) for n in line['date'].split('/'))a_list_of_dicts.append(line)print(a_list_of_dicts)
csvfile.csv
name,price,datexxxx,5.00,2/23/2023yyyy,6.75,2/24/2023
输出：
[{'name': 'xxxx', 'price': 5.0, 'date': (2, 23, 2023)}, {'name': 'yyyy', 'price': 6.75, 'date': (2, 24, 2023)}]
英文:
Use a csv.DictReader and convert the columns as you read them:
import csvwith open('csvfile.csv', newline='') as file:a_list_of_dicts = []for line in csv.DictReader(file):line['price'] = float(line['price'])line['date'] = tuple(int(n) for n in line['date'].split('/'))a_list_of_dicts.append(line)print(a_list_of_dicts)
csvfile.csv
name,price,datexxxx,5.00,2/23/2023yyyy,6.75,2/24/2023
Output:
[{'name': 'xxxx', 'price': 5.0, 'date': (2, 23, 2023)}, {'name': 'yyyy', 'price': 6.75, 'date': (2, 24, 2023)}]
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论