python-3.6

问题 This is already solved by myself but I put this question for someone else. I think this kind of problem is better as much as we can. and there doesn't seem be in SOF. I updated spyder and PyQt5 . conda update spyder Collecting package metadata: done Solving environment: done ## Package Plan ## environment location: C:\Anaconda3 added / updated specs: - spyder The following packages will be downloaded: package | build ---------------------------|----------------- anaconda-custom | py36h363777c

问题 While installing mxnet package I am getting the error : cannot import name 'html5lib' & when I go for html5lib package installation I am getting the same error. The full error message can be seen in the image. I have installed python using Anaconda. 回答1: Try running conda update -f html5lib which will try to update your html5lib . I also encountered this after installing something ( tensorflow ) through conda that had to downgrade html5lib and that broke everything for pip . 回答2: Windows 10

问题 I just created a data type for my date data, which returns a datetime.datetime object Here is the code: import datetime class Date: def __new__(cls, dateTime, *args, **kwargs): return datetime.datetime.strptime(dateTime, "%Y-%m-%dT%H:%M:%S.%f%z") So everytime I give this class an ISO-8601 it should return the datetime object from the string... Python 3.7 Example: Date("2018-12-09T08:56:12.189Z") # Returns => datetime.datetime(2018, 12, 9, 8, 56, 12, 189000, tzinfo=datetime.timezone.utc) This

问题 I'm trying to get the text stored in the clipboard by just using ctypes in Python 3.6 . I tested a lot of solutions I found on Stack and GitHub, but they only work for Python 2 to Python 3.4 . This is the code you'll find almost everywhere: from ctypes import * def get_clipboard_text(): text = "" if windll.user32.OpenClipboard(c_int(0)): h_clip_mem = windll.user32.GetClipboardData(1) windll.kernel32.GlobalLock.restype = c_char_p text = windll.kernel32.GlobalLock(c_int(h_clip_mem)) windll

问题 I am using the following code in sympy : from sympy.physics.secondquant import F, Fd, NO, Commutator from sympy import symbols a, b, c, d = symbols("a,b,c,d") comm = NO(Commutator( Fd(a) * F(b), Fd(c) * F(d) ).doit()) print(comm) # gives me 0 Clearly, this should not be zero. Well, maybe I do not understand sympy . Here is what I want to calculate: [F†_a F_b, F†_c F_d] with not necessarily equal indices a, b, c and d. 来源： https://stackoverflow.com/questions/52227318/does-sympy-give-me-wrong

问题 total nexwbie here. I'm unsuccessfully trying to install the cv2 module for python but it doesn't work. I'm working with Python 3.6 (64bits) I typed the following commands in the cmd : C:\Users\leahj>C:\Users\leahj\AppData\Local\Programs\Python\Python36\Scripts\pip3 install cv2 Collecting cv2 Could not find a version that satisfies the requirement cv2 (from versions: ) No matching distribution found for cv2 C:\Users\leahj>C:\Users\leahj\AppData\Local\Programs\Python\Python36\Scripts\pip3

问题 While trying to figure out if a function is called with the @decorator syntax, we realized that inspect has a different behaviour when looking at a decorated class that inherits from a superclass. The following behaviour was found with CPython 3.6.2 under Windows 10. It was also reproduced in CPython 3.7.0 under Linux 64 bits. import inspect def decorate(f): lines = inspect.stack()[1].code_context print(f.__name__, lines) return f @decorate class Foo: pass @decorate class Bar(dict): pass

问题 I try to use RPi.GPIO with Python 3.6. I installed RPi.GPIO and it's working with Python 3.4, but not with Python 3.6 I get this Error: ModuleNotFoundError: No module named 'RPi' I immport the module in my script like this: import RPi.GPIO as GPIO 回答1: Add this line to the top of your *.py file: #!/usr/bin/env python3.6 Run these commands in your shell: sudo python3.6 -m pip install --upgrade pip setuptools wheel sudo python3.6 -m pip install RPi.GPIO This should fix the Problem. By this you

问题 I want to annotate a type of a variable in a for -loop. I tried this: for i: int in range(5): pass But it didn't work, obviously. What I expect is working autocomplete in PyCharm 2016.3.2. Pre-annotation like this: i: int for i in range(5): pass doesn't help. P.S. Pre-annotation works for PyCharm >= 2017.1 回答1: According to PEP 526, this is not allowed: In addition, one cannot annotate variables used in a for or with statement ; they can be annotated ahead of time, in a similar manner to

问题 I have a function that mimics range(). I am stuck at one point. I need to be able to make the first (x) and third (step) arguments optional, but the middle argument (y) mandatory. In the code below, everything works except the two commented out lines. If I am only passing in one argument, how do I construct the function to accept the single passed in argument as the mandatory (y) argument? I cannot do this: def float_range(x=0, y, step=1.0): Non-default parameter cannot follow a default