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 csv
file = 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 csv
file = open('csvfile.csv')
lines = csv.reader(file)
header = next(lines) # ['name', 'price', 'date']
# when I do the following
for 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 function
types = [
str,
float,
str # this need to be a tuple
# tried tuple(map(int, cannotchoosecolumn.split('/')))
# did not work
]
# now to create a list of dicts
alist_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 datetime
def 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 datetime
def 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 csv
with 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,date
xxxx,5.00,2/23/2023
yyyy,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 csv
with 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,date
xxxx,5.00,2/23/2023
yyyy,6.75,2/24/2023
Output:
[{'name': 'xxxx', 'price': 5.0, 'date': (2, 23, 2023)}, {'name': 'yyyy', 'price': 6.75, 'date': (2, 24, 2023)}]
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