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不带编码的字符串参数

data-analysis numpy python-3.x python-2.7 python

johschnei · 技术社区 · 7 年前

有人能告诉我为什么在python 3上收到以下错误吗。以下是回溯:

TypeError
Traceback (most recent call last) <ipython-input-24-a81d4875414b> in <module>()
      7     filename = [("id"), ("name"), ("email"), ("amount"),("sent")]
      8     writer= csv.DictWriter(temp_file, fieldnames = fieldnames)
----> 9     writer.writeheader()
     10 
     11     for row in reader:
C:\Users\johsc_001\AppData\Local\conda\conda\envs\ipykernel_py3\lib\csv.py in writeheader(self)
    142     def writeheader(self):
    143         header = dict(zip(self.fieldnames, self.fieldnames))
--> 144         self.writerow(header)
    145 
    146     def _dict_to_list(self, rowdict):
C:\Users\johsc_001\AppData\Local\conda\conda\envs\ipykernel_py3\lib\csv.py in writerow(self, rowdict)
    153 
    154     def writerow(self, rowdict):
--> 155         return self.writer.writerow(self._dict_to_list(rowdict))
    156 
    157     def writerows(self, rowdicts):
C:\Users\johsc_001\AppData\Local\conda\conda\envs\ipykernel_py3\lib\tempfile.py in func_wrapper(*args, **kwargs)
    481             @_functools.wraps(func)
    482             def func_wrapper(*args, **kwargs):
--> 483                 return func(*args, **kwargs)
    484             # Avoid closing the file as long as the wrapper is alive,
    485             # see issue #18879.

TypeError: a bytes-like object is required, not 'str'

import csv
import shutil
from tempfile import NamedTemporaryFile 
filename = 'appendpyt2.csv' 
temp_file = NamedTemporaryFile(delete= False) 
with open(filename, 'rb')as csvfile, temp_file:
    reader =csv.DictReader(csvfile)
    filename = ["id", "name", "email", "amount", "sent"]
    writer= csv.DictWriter(temp_file, fieldnames = ["id", "name", "email","amout", "sent"])
    writer.writeheader()
    for row in reader:
        print(row)
        writer.writerow({
            "id": row["id"],
            "name": row["name"],
            "email":row["email"],
            "amout":"1234.56",
            "sent": ""
        })

1 回复 | 直到 7 年前

Mark Tolonen 7 年前

错误似乎来自于使用Python 3,但使用Python 2要求打开 csv 文件夹。如果使用Python 3,CSV文件不应以二进制模式打开,换行符参数应为空字符串。临时文件也默认为二进制模式,所以我已经覆盖了它。我还将以下内容用作输入文件,这是从代码中推导出来的,因为没有提供示例输入。

id,name,email
id1,name1,email1
id2,name2,email2

Python 3代码:

import csv
import shutil
from tempfile import NamedTemporaryFile 
filename = 'appendpyt2.csv' 
temp_file = NamedTemporaryFile(mode='w+',newline='',delete= False) 
with open(filename,newline='') as csvfile, temp_file:
    reader =csv.DictReader(csvfile)
    filename = ["id", "name", "email", "amount", "sent"]
    writer= csv.DictWriter(temp_file, fieldnames = ["id", "name", "email","amount", "sent"])
    writer.writeheader()
    for row in reader:
        print(row)
        writer.writerow({
            "id": row["id"],
            "name": row["name"],
            "email":row["email"],
            "amount":"1234.56",
            "sent": ""
        })

临时文件输出:

id,name,email,amount,sent
id1,name1,email1,1234.56,
id2,name2,email2,1234.56,