有没有可能拥有python 3.x代码并在其中执行2.x代码(例如函数调用)?
#Code written in python 3.x
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#This is the python 2.x part with Function2x() written in python 2.x
Function2x()
#End of python 2.x part
#Code written in python 3.x
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这会被认为是不好的做法吗?如果是这样,如果不能将代码从3.x移植到2.x,那么如何避免这个问题呢?
在apt-get更新之后,我将得到以下输出:
root@pratik--Notebook:/home/pratik# apt-get install python
Reading package lists... Done
Building dependency tree
Reading state information... Done
python is already the newest version (2.7.11-1).
The following packages were automatically installed and are no longer req
当在python2中运行时,我在try块本身中发现了这段检测异常的代码。
import sys
for i in range(3):
try:
if sys.exc_info()[1]:
print("Exception found")
else:
print("Exception not found")
raise Exception("Random exception")
except Exception as e:
我试图将Python代码(使用Tkinter)转换为Nuitka的可执行文件,但它给出了以下错误: $ nuitka --recurse-all srcfile.py --exe
Nuitka:WARNING:srcfile.py:3: Cannot find 'tkinter.ttk' as relative or absolute import. exe文件已创建并运行,但由于出现错误而停止: File "/home/abcd/srcfile.py", line 2, in <module>
import tkinter as tk
I
我需要pip包‘钻石’作为另一个包(humann3)的先决条件,但是有些东西不起作用。
当我尝试启动humann3时,它返回
CRITICAL ERROR: Can not call software version for diamond
所以我试着用"diamond -- version“检查我的钻石版本,但它返回
user@server:~$ diamond --version File
"/home/user/.local/bin/diamond", line 113
print "Diamond version %s" %
我想通过运行脚本从终端读取一个变量。这是我的script.py:
while True:
value = input('enter text: ')
if value == 'stop':
print('bye-bye')
break
else:
print('continue!')
然而,当我运行python script.py时,发生了一些奇怪的事情。如果我输入int-data (例如1,2,3),就没有问题。如果我输入'stop',我就会得到一个错误:
SyntaxErro
我对python很陌生,在阅读了一些运行的示例之后,我想尝试一下状态模型。
我从statsmodel网站上复制了以下示例,
#!/usr/bin/env python3
import numpy as np
import statsmodels.api as sm
spector_data = sm.datasets.spector.load()
spector_data.exog = sm.add_constant(spector_data.exog, prepend=False)
#Fit and summarize OLS model
mod = sm.OLS(spector_data
一些用Python2.7编写的程序抱怨ImportError: No module named 'urlparse'。因此,我需要安装模块,但我无法做到。该模块确实存在,例如在https://docs.python.org/2/library/urlparse.html中进行了描述。然而,无论是apt-get install还是pip install都无法找到一个名为urlparse、python-urlparse、urllib、python-urllib的模块. --我收到了类似于Could not find any downloads that satisfy the re
我从一些类似下面的书中得到了python代码,但它运行不正常。
# name.py
name = input('What is your first name? ')
print('Hello ' + name.capitalize() + '!')
结果是:
$ python name.py
What is your first name? jack
Traceback (most recent call last):
File "name.py", line 3, in <module>
name =
ERROR: Python 3 is not supported by the Google Cloud SDK. Please use a Python 2.x version that is 2.6 or greater.
If you have a compatible Python interpreter installed, you can use it by setting the CLOUDSDK_PYTHON environment variable to point to it.
我想我们应该问的第一个问题是,“谷歌从他们的客户身上赚了这么多钱,为什么他们不能雇一个人来确
我正在尝试遍历一个列表,并将条目与一个人输入的条目进行比较,然后做出决定。你知道我的逻辑哪里错了吗?
user = input("Enter your password: ")
passwords = ["pa$$w0rd", "password123", "scr1pt1ng", "F0r3n51c5", "123456"];
for n in passwords:
if n == user:
print("Found", n)
br