小编Sel*_*cuk的帖子

我读过这个问题

1）我安装了点子并执行

 pip install requests

并得到

Requirement already satisfied (use --upgrade to upgrade): requests in /usr/local/lib/python2.7/dist-packages/requests-2.9.1-py2.7.egg
Cleaning up...

2）我启动了python2 shell：

>>> from urllib.request import urlopen
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ImportError: No module named request

为什么我仍在捕获此异常？我做错了什么？

我想在我的 django 应用程序中包含全文搜索。我为此使用 whoosh-haystack。当我在已安装的应用程序中包含 whoosh 和 haystack 并执行命令./manage.py时，出现导入错误。任何人都可以解决这个问题。

INSTALLED_APPS = [
    'django.contrib.admin',
    'django.contrib.auth',
    'django.contrib.contenttypes',
    'django.contrib.sessions',
    'django.contrib.messages',
    'django.contrib.staticfiles',
    'whoosh',
    'haystack',
    'galleryApp',
    'galleryAdmin',
]

我得到的错误是

Traceback (most recent call last):
  File "./manage.py", line 10, in <module>
    execute_from_command_line(sys.argv)
  File "/usr/local/lib/python2.7/dist-packages/django/core/management/__init__.py", line 353, in execute_from_command_line
    utility.execute()
  File "/usr/local/lib/python2.7/dist-packages/django/core/management/__init__.py", line 327, in execute
    django.setup()
  File "/usr/local/lib/python2.7/dist-packages/django/__init__.py", line 18, in setup
    apps.populate(settings.INSTALLED_APPS)
  File "/usr/local/lib/python2.7/dist-packages/django/apps/registry.py", line 115, in populate
    app_config.ready()
  File "/usr/local/lib/python2.7/dist-packages/django/contrib/admin/apps.py", line 22, in ready
    self.module.autodiscover()
  File "/usr/local/lib/python2.7/dist-packages/django/contrib/admin/__init__.py", line 26, in autodiscover
    autodiscover_modules('admin', register_to=site) …

这里有一个用于生成密码的代码段,我有2个关于此的问题,请问你能分享一下如何理解？

1. urandom(6),来自urandom的帮助说,返回适合加密使用的n个随机字节,也就是说,它会返回6个字节,它是6个ASCII吗？

2. ord(c) ,获取上面的字节的十进制基数,为什么这里转移到十进制基数？

帮助urandom:

def urandom(n): # real signature unknown; restored from __doc__
    """
    urandom(n) -> str

    Return n random bytes suitable for cryptographic use.
    """
    return ""

Python脚本:

from os import urandom
letters = "ABCDEFGHJKLMNPRSTUVWXYZ"
password = "".join(letters[ord(c) % len(letters)] for c in urandom(6))

list_1 = ["a", "b", "c", "d", "e"]
list_2 = ["a", "c", "d"] 

我希望其中的元素索引list_1不在中list_2。我希望在这里

[1, 4]

就我而言，list_2是list_1所有元素的唯一子集。没有使用显式循环的更好方法吗？

I need to create a dictionary as shown below:

d = {'a': [2, 3, 4], 'b': [3, 4, 6], 'c': [2, 5, 9]}

I have keys in the form of a list

keys = ['a', 'b', 'c'] 

and values are present as separate lists

values = [[2, 3, 2], [3, 4, 5], [4, 6, 9]]. 

I've tried using

d(zip(keys, values))

but it returns a dictionary as

{'a': [2, 3, 2], 'b': [3, 4, 5], 'c': [4, 6, 9]}

Is there …

我有一个对象 ID 的映射：

mapping = {'id-1':'id-10', 'id-2':'id-14', 'id-3':'id-19'}

还有两个包含对象的列表，id是对象的一个属性，比如object_id1.id = 'id-1'。

list1 = [object_id1, object_id3, object_id4, object_id5]
list2 = [object_id19, object_id11, object_id14, object_id10]

我想要一个列表，其中包含一组中的那些相关对象。如

output = [(object_id1, object_id10), (object_id2, object_id14),(object_id3, object_id19)]

这可以在一个 for 循环中完成吗？

我有一个 python 函数，它返回具有以下结构的字典

{
   (int, int): {string: {string: int, string: float}}
}

我想知道如何使用类型提示来指定它。所以，这些位很清楚：

Dict[Tuple[int, int], Dict[str, Dict[str, # what comes here]]

但是，内部字典具有两个键的int和值类型。float我不知道如何注释

我有两个清单，

 l1 = [1,4,3,2,5]
 l2 = [4,3,2,10]

现在我想找到列表之间的公共元素，所以我使用以下代码，

list(set(l1).intersection(set(l2)))
>> [2, 3, 4]

但它改变了 的顺序l1，我不想改变顺序所以结果应该是，

>> [4, 3, 2]

寻找在不改变顺序的情况下执行此操作的最快方法？