小编Thi*_*H S的帖子

启动新的scrapy项目时出错

我已经使用Scrapy网站提供的Ubuntu软件包安装了Scrapy.但是在开始Scrapy项目时

scrapy startproject test 
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我收到错误消息.

 Traceback (most recent call last):   File "/usr/bin/scrapy", line 5,
 in <module>
     from pkg_resources import load_entry_point   File "build/bdist.linux-x86_64/egg/pkg_resources/__init__.py", line 3084,
 in <module>
        File "build/bdist.linux-x86_64/egg/pkg_resources/__init__.py", line 3070, in _call_aside
        File "build/bdist.linux-x86_64/egg/pkg_resources/__init__.py", line 3097, in _initialize_master_working_set
        File "build/bdist.linux-x86_64/egg/pkg_resources/__init__.py", line 653, in _build_master
        File "build/bdist.linux-x86_64/egg/pkg_resources/__init__.py", line 666, in _build_from_requirements
        File "build/bdist.linux-x86_64/egg/pkg_resources/__init__.py", line 844, in resolve
      pkg_resources.ContextualVersionConflict: (pyasn1 0.1.7 (/usr/lib/python2.7/dist-packages),
 Requirement.parse('pyasn1>=0.1.8'), set(['pyasn1-modules']))
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请帮我解决这个错误.我正在运行Python 2.7.6

python scrapy

7
推荐指数
2
解决办法
8923
查看次数

如何明智地连接多个列表元素

输入:

l1 = ['a', '', '', '']
l2 = ['', 'b', '', '']
l3 = ['', '', 'c', '']
l4 = ['', '', '', 'd']
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预期产量:

['a', 'b', 'c', 'd']
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我试过了

list(map(str.__add__, l1, l2, l3, l4))

看起来str.__add__不接受两个以上的列表对象.

任何解决方法?

编辑:基于Jim Fasarakis-Hilliard的评论.

l1 = ['a', '1', '', '']
l2 = ['', 'b', '2', '']
l3 = ['', '', 'c', '']
l4 = ['', '', '', 'd']
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预期产量:

['a', '1b', '2c', 'd']
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谢谢

python list concatenation python-2.7 python-3.x

3
推荐指数
1
解决办法
613
查看次数

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python ×2

concatenation ×1

list ×1

python-2.7 ×1

python-3.x ×1

scrapy ×1