This question already has an answer here: How do I pass a variable by reference? 23 answers Python passes references-to-objects by value. Python passes references-to-objects by value (like Java), and everything in Python is an object. This sounds simple, but then you will notice that some data types seem to exhibit pass-by-value characteristics, while others seem to act like pass-by-refer
这个问题在这里已经有了答案： 如何通过引用传递变量？ 23个答案 Python通过值传递引用到对象。 Python通过值传递引用到对象（如Java），而Python中的所有东西都是一个对象。 这听起来很简单，但您会注意到某些数据类型似乎具有传递值特性，而其他数据类似于传递引用......这是怎么回事？ 理解可变对象和不可变对象很重要。 一些对象，如字符串，元组和数字是不可变的。 在函数/方法内改变它们会创建一个新的实例，
I made a Java program that uses unicode escaped characters to break a multiline comment and hide some functionality. The program below prints "Hello Cruel World". I'm wondering if this is possible to do in Python (any version). If it is not possible, how is this prevented in the language? public static void main(String[] args) { print("Hello"); /* * u002Au002Fu0070
我制作了一个使用unicode转义字符的Java程序来打破多行注释并隐藏一些功能。 下面的程序打印出“你好，残酷的世界”。 我想知道这是否可以在Python（任何版本）中完成。 如果不可能，这在语言中如何被阻止？ public static void main(String[] args) { print("Hello"); /* * u002Au002Fu0070u0072u0069u006Eu0074u0028u0022u0043u0072u0075u0065u006Cu0022u0029u003Bu002Fu002A */ print("World"); } p
I made a very simple interactive console that I'd like to use in a complicated scraping application. It looks like this: #!/usr/bin/env python # -*- coding: utf-8 -*- import os, sys, codecs, code sys.__stdout__ = codecs.getwriter('utf8')(sys.__stdout__) sys.__stderr__ = codecs.getwriter('utf8')(sys.__stderr__) if 'DEBUG' in os.environ: import pdb import sys oeh = sys.excepth
我做了一个非常简单的交互式控制台，我想在复杂的抓取应用程序中使用。 它看起来像这样： #!/usr/bin/env python # -*- coding: utf-8 -*- import os, sys, codecs, code sys.__stdout__ = codecs.getwriter('utf8')(sys.__stdout__) sys.__stderr__ = codecs.getwriter('utf8')(sys.__stderr__) if 'DEBUG' in os.environ: import pdb import sys oeh = sys.excepthook def debug_exceptions(type, value,
I have recently started studying Python, but I couldn't find how to implement multi-line comments. Most languages have block comment symbols like /* */ I tried this in Python, but it throws an error, so this probably is not the correct way. Does Python actually have a multiline comment feature? You can use triple-quoted strings. When they're not a docstring (first thing in a clas
我最近开始学习Python，但是我找不到如何实现多行注释。 大多数语言都有块注释符号 /* */ 我在Python中试过这个，但它会抛出一个错误，所以这可能不是正确的方法。 Python实际上是否有多行注释功能？ 您可以使用三重引号的字符串。 当他们不是文档字符串（类/函数/模块中的第一件事）时，它们将被忽略。 ''' This is a multiline comment. ''' （确保适当缩进前导'''以避免IndentationError 。） Guido
I have a json file, such as the following: { "author":"John", "desc": "If it is important to decode all valid JSON correctly and speed isn't as important, you can use the built-in json module, orsimplejson. They are basically the same but sometimes simplej further along than the version of it that is included with distribution." //"birthday": "nothing" //I com
我有一个json文件，如下所示： { "author":"John", "desc": "If it is important to decode all valid JSON correctly and speed isn't as important, you can use the built-in json module, orsimplejson. They are basically the same but sometimes simplej further along than the version of it that is included with distribution." //"birthday": "nothing" //I comment this line
如何在python中找到扩展名为.txt的目录中的所有文件？ You can use glob : import glob, os os.chdir("/mydir") for file in glob.glob("*.txt"): print(file) or simply os.listdir : import os for file in os.listdir("/mydir"): if file.endswith(".txt"): print(os.path.join("/mydir", file)) or if you want to traverse directory, use os.walk : import os for root, dirs, files in os.walk("/myd
如何在python中找到扩展名为.txt的目录中的所有文件？ 你可以使用glob ： import glob, os os.chdir("/mydir") for file in glob.glob("*.txt"): print(file) 或者干脆os.listdir ： import os for file in os.listdir("/mydir"): if file.endswith(".txt"): print(os.path.join("/mydir", file)) 或者如果你想遍历目录，使用os.walk ： import os for root, dirs, files in os.walk("/mydir"): for fil
Is there a better way to use glob.glob in python to get a list of multiple file types such as .txt, .mdown, and .markdown? Right now I have something like this: projectFiles1 = glob.glob( os.path.join(projectDir, '*.txt') ) projectFiles2 = glob.glob( os.path.join(projectDir, '*.mdown') ) projectFiles3 = glob.glob( os.path.join(projectDir, '*.markdown') ) Maybe there is a better way, but how ab
有没有更好的方式来使用Python中的glob.glob获取多种文件类型的列表，如.txt，.mdown和.markdown？ 现在我有这样的事情： projectFiles1 = glob.glob( os.path.join(projectDir, '*.txt') ) projectFiles2 = glob.glob( os.path.join(projectDir, '*.mdown') ) projectFiles3 = glob.glob( os.path.join(projectDir, '*.markdown') ) 也许有更好的办法，但如何： >>> import glob >>> types = ('*.pdf', '*
How to get the filename without the extension from a path in Python? I found out a method called os.path.basename to get the filename with extension. But even when I import os, I am not able to call it path.basename . Is it possible to call it as directly as basename? Getting the name of the file without the extension : import os print(os.path.splitext("path_to_file")[0]) As for your impo
如何在没有Python中的路径扩展的情况下获取文件名？ 我找到了一个名为os.path.basename的方法来获取带有扩展名的文件名。 但即使我输入os时，我也无法称之为path.basename 。 是否可以直接将它称为基本名称？ 获取没有扩展名的文件名称： import os print(os.path.splitext("path_to_file")[0]) 至于你的进口问题，你可以这样解决： from os.path import basename # now you can call it directly with basename print(
Can python detect OS and then contruct a if/else statement for File System. I would need to replace C:CobaltRCX in Fn string with the FileSys string. import os.path, csv from time import strftime if os.path.?????:## Windows FileSys = r"C:\working\" else: ##linux FileSys = r"\working\" y=(strftime("%y%m%d")) Fn = (r"C:\working\Setup%s.csv" %y) I usually just use this: import os
python可以检测操作系统，然后为文件系统构造一个if / else语句。 我需要用FileSys字符串替换Fn字符串中的C： CobaltRCX 。 import os.path, csv from time import strftime if os.path.?????:## Windows FileSys = r"C:\working\" else: ##linux FileSys = r"\working\" y=(strftime("%y%m%d")) Fn = (r"C:\working\Setup%s.csv" %y) 我通常只是使用这个： import os if os.name == 'nt': pass # Window
Given a path such as "mydir/myfile.txt" , how do I find the absolute filepath relative to the current working directory in Python? Eg on Windows, I might end up with: "C:/example/cwd/mydir/myfile.txt" >>> import os >>> os.path.abspath("mydir/myfile.txt") 'C:/example/cwd/mydir/myfile.txt' 如果它已经是绝对路径，它也可以工作： >>> import os >>> os.path
给定一个诸如"mydir/myfile.txt"类的路径，如何找到Python中相对于当前工作目录的绝对文件路径？ 例如在Windows上，我最终可能会得到： "C:/example/cwd/mydir/myfile.txt" >>> import os >>> os.path.abspath("mydir/myfile.txt") 'C:/example/cwd/mydir/myfile.txt' 如果它已经是绝对路径，它也可以工作： >>> import os >>> os.path.abspath("C:/example/cwd/mydir/myfile.tx