Python TypeError: argument of type 'type' is not iterable 这个问题怎么解决

V2EX = way to explore

V2EX 是一个关于分享和探索的地方

For Existing Member Sign In

推荐学习书目

› Learn Python the Hard Way

Python Sites

› PyPI - Python Package Index

› http://diveintopython.org/toc/index.html

› Pocoo

值得关注的项目

› PyPy

› Celery

› Jinja2

› Read the Docs

› gevent

› pyenv

› virtualenv

› Stackless Python

› Beautiful Soup

› 结巴中文分词

› Green Unicorn

› Sentry

› Shovel

› Pyflakes

› pytest

Python 编程

› pep8 Checker

Styles

› PEP 8

› Google Python Style Guide

› Code Style from The Hitchhiker's Guide

This topic created in 2757 days ago, the information mentioned may be changed or developed.

self.is_back_test = params['is_back_test'] if "is_back_test" in params else False TypeError: argument of type 'type' is not iterable

这哥问题应该怎么解决

已经尝试过 self.is_back_test = params['is_back_test'] if "is_back_test" in params.keys() else False

使用的是 python3

8 replies • 2018-10-22 17:08:50 +08:00

conn4575

Oct 22, 2018 via Android

检查一下 params 是否是 None?

monkeylyf

Oct 22, 2018

和楼上意见一致。params 应该不是 dict。你可以 print(type(params)) 试试

Yourshell

Oct 22, 2018 via iPhone

你参数的类型不对啊，你先判断是不是可迭代对象啊，报错都这么明显了。

bany

Oct 22, 2018

```python
if isinstance(params, dcit):
return params.get('is_back_test', False)
else:
return False
```

laike9m

Oct 22, 2018 via Android

请贴上可以复现问题的代码谢谢

zcsnbb

Oct 22, 2018

@conn4575 找到问题了我忘了传参数感谢

zcsnbb

Oct 22, 2018

@monkeylyf 问题解决了感谢

zcsnbb

Oct 22, 2018

@bany 感谢！！！