Python：我无法从函数中的stdin解析csv

Question

无法从函数中的stdin解析csv

[你好，我在Mac上尝试使用Python的csv模块来解析通过stdin传递的以空格分隔的文件：

printf "2020-01-01 Ben 2\n2020-02-01 Jenny 4\n" | ./tmp.py

当我在结尾处调用working函数运行代码时，得到了预期的结果：

$ printf "2020-01-01 Ben 2\n2020-02-01 Jenny 4\n" | ./tmp.py
function: working
['2020-01-01', 'Ben', '2']
['2020-02-01', 'Jenny', '4']

当我使用最后调用的not_working函数运行它时，出现错误：

$ printf "2020-01-01 Ben 2\n2020-02-01 Jenny 4\n" | ./tmp.py
function: not_working
Traceback (most recent call last):
  File "./tmp.py", line 36, in <module>
    not_working() # if working() was here, it would work
  File "./tmp.py", line 27, in not_working
    print_csv(args.infile, delimiter=' ')
  File "./tmp.py", line 20, in print_csv
    reader = csv.reader(infile, delimiter)
_csv.Error: unknown dialect

这是在最后调用not_working的最小示例

#!/usr/bin/env python3
# -*- coding: utf-8 -*-

import argparse
import csv
import sys

parser = argparse.ArgumentParser()
parser.add_argument(
    "infile",
    nargs="?",
    type=argparse.FileType("r"),
    default=sys.stdin,
)
args = parser.parse_args()

def print_csv(infile, delimiter):
    reader = csv.reader(infile, delimiter)
    for row in reader:
        print(row)

def not_working():
    print("function: not_working")
    with args.infile:
        print_csv(args.infile, delimiter=' ')

def working():
    print("function: working")
    with args.infile:
        reader = csv.reader(args.infile, delimiter=" ")
        for row in reader:
            print(row)

not_working() # if working() was here, it would work
# working()

为什么会这样？为什么我可以在函数外部而不是函数内部解析CSV？

Answer 1

您忘记了在delimiter中将print_csv作为关键字参数传递；这表示您实际上是在呼叫csv.reader(infile, dialect=delimiter)，而''是未知方言（通常仅限于'excel'，'excel-tab'或'unix'。

有关更多详细信息，请参阅csv.reader方法签名的文档。