I fully understand what is being passed to self in this example. I'm very confused on how it is being passed to self internally. Could someone help me understand? class Cars: def __init__(self, model, engine, doors): self.model = model self.engine = engine self.doors = doors tesla = Cars('Model S', 'Electric', 'Four door') ford = Cars('Mustang', 'v8', 'Two do
我完全理解这个例子中传递给self 东西 。 我很困惑它是如何在内部传递给self 。 有人能帮助我理解吗？ class Cars: def __init__(self, model, engine, doors): self.model = model self.engine = engine self.doors = doors tesla = Cars('Model S', 'Electric', 'Four door') ford = Cars('Mustang', 'v8', 'Two door') 初学者教程并没有涉及很多步骤，所以我会尽量简短但彻底。 我会尽量
Assume I have a function def multiply_by(x, multiplier): return x * multiplier How can I create a copy of that function and fix the multiplier in that function? multiply_by_5 = multiply_by? <-- here I need python magic such that multiply_by_5 would have only one argument x and the multiplier would be 5? So that multiply_by_5(2) 10 Is there a way in Python 2.7 to do that? 您可以
假设我有一个功能 def multiply_by(x, multiplier): return x * multiplier 如何创建该函数的副本并修复该函数中的乘数？ multiply_by_5 = multiply_by? <-- here I need python magic 这样multiply_by_5将只有一个参数x而乘数为5？ 以便 multiply_by_5(2) 10 Python 2.7有没有办法做到这一点？ 您可以使用带关键字参数的functools.partial ： >>> def multiply_by(x, multiplier): ... return x
Hej! I'm trying to write a function in Python for a polynomial p that is a linear combination of n basis functions phi_i. How can I define a function that is itself a sum of n other functions? I know that this works: phi1 = lambda x: x**2 phi2 = lambda x: x p = lambda x: phi1(x) + phi2(x) But if I try a loop like this: p = lambda x: 0 for i in range(0,n): p = lambda x: p(x)+phi[i](
HEJ！ 我试图用Python编写一个函数，用于多项式p，它是n个基函数phi_i的线性组合。 我怎样才能定义一个函数本身是其他n个函数的总和？ 我知道这是有效的： phi1 = lambda x: x**2 phi2 = lambda x: x p = lambda x: phi1(x) + phi2(x) 但是，如果我尝试这样的循环： p = lambda x: 0 for i in range(0,n): p = lambda x: p(x)+phi[i](x) 其中phi是我的基础函数列表，我创建了一个无限循环。 我检查了编写一个函数
I am uploading a large file with FTP using Python 3.4. I would like to be able to show the progress percentage while uploading the file. Here's my code: from ftplib import FTP import os.path # Init sizeWritten = 0 totalSize = os.path.getsize('test.zip') print('Total file size : ' + str(round(totalSize / 1024 / 1024 ,1)) + ' Mb') # Define a handle to print the percentage uploaded def han
我使用Python 3.4上传了一个带有FTP的大文件。 我希望能够在上传文件时显示进度百分比。 这是我的代码： from ftplib import FTP import os.path # Init sizeWritten = 0 totalSize = os.path.getsize('test.zip') print('Total file size : ' + str(round(totalSize / 1024 / 1024 ,1)) + ' Mb') # Define a handle to print the percentage uploaded def handle(block): sizeWritten += 1024 # this line fail becaus
def fib(n): if n == 1: return 0 if n == 2: return 1 return fib(n-2) + fib(n-1) def memo(f): cache = {} def memoized(n): if n not in cache: cache[n] = f(n) return cache[n] return memoized fib1 = memo(fib) This code runs really slow on my laptop, but if I change the name fib1 to fib, then everything works fine...anyone know the
def fib(n): if n == 1: return 0 if n == 2: return 1 return fib(n-2) + fib(n-1) def memo(f): cache = {} def memoized(n): if n not in cache: cache[n] = f(n) return cache[n] return memoized fib1 = memo(fib) 此代码在我的笔记本电脑上运行速度非常慢，但如果我将名称更改为fib，那么一切正常......任何人都知道原因？ 谢谢！ fib递归到fi
I am trying to understand, how exactly variable binding in python works. Let's look at this: def foo(x): def bar(): print y return bar y = 5 bar = foo(2) bar() This prints 5 which seems reasonable to me. def foo(x): def bar(): print x return bar x = 5 bar = foo(2) bar() This prints 2, which is strange. In the first example python looks for the variable d
我想明白，python中的变量绑定是如何工作的。 我们来看看这个： def foo(x): def bar(): print y return bar y = 5 bar = foo(2) bar() 这打印5对我来说似乎是合理的。 def foo(x): def bar(): print x return bar x = 5 bar = foo(2) bar() 这打印2，这很奇怪。 在第一个例子中，python在执行期间查找变量，在第二个时候创建方法。 为什么这样？ 要清楚：这是非常酷，并按我的愿望
This question already has an answer here: Can you explain closures (as they relate to Python)? 13 answers What you're encountering is a closure. When you write part_string(my_indexes) you're creating a new function, and upon calling it, you use the old variables you gave to part_string together with the new variables given to function_instance . You may name the inner function wh
这个问题在这里已经有了答案： 你能解释关闭（因为它们与Python有关）吗？ 13个答案 你遇到的是封闭。 当您编写part_string(my_indexes)您正在创建一个新函数，并且在调用它时，将使用part_string的旧变量与赋予function_instance的新变量一起使用。 无论你想要什么，你都可以命名内部函数，并且没有约定。 （ obj在这里使用，但它可以是pie 。除了函数闭包（装饰器）的func外，没有其他约定。 如果你想传递两个变量
This question already has an answer here: Can you explain closures (as they relate to Python)? 13 answers What limitations have closures in Python compared to language X closures? 7 answers Closures in Python 6 answers
这个问题在这里已经有了答案： 你能解释关闭（因为它们与Python有关）吗？ 13个答案 与X语言关闭相比，Python有什么限制？ 7个答案 Python 6中的闭包答案
This question already has an answer here: Can you explain closures (as they relate to Python)? 13 answers the variable of the outer function is available only when that function is running on, ie it vanishes when my_outer_function ends. That's not entirely true. The variable is available in my_outer_function 's scope. my_inner_function has the scope of its own declarations and it
这个问题在这里已经有了答案： 你能解释关闭（因为它们与Python有关）吗？ 13个答案 外函数的变量只有在该函数运行时才可用，即当my_outer_function结束时它会消失。 这并非完全正确。 该变量在my_outer_function的作用域中可用。 my_inner_function具有其自己的声明范围及其父级范围。 my_inner_function引用的变量不在其自己的作用域内，因此这些引用在父域的作用域不再可用时作为闭包绑定到my_inner_function 。
How to convert real numpy array to int numpy array? Tried using map directly to array but it did not work. 使用astype方法。 >>> x = np.array([[1.0, 2.3], [1.3, 2.9]]) >>> x array([[ 1. , 2.3], [ 1.3, 2.9]]) >>> x.astype(int) array([[1, 2], [1, 2]]) Some numpy functions for how to control the rounding: rint, floor,trunc, ceil. depending how u wish to
如何将真正的numpy数组转换为int numpy数组？ 试图直接使用地图到数组，但它没有奏效。 使用astype方法。 >>> x = np.array([[1.0, 2.3], [1.3, 2.9]]) >>> x array([[ 1. , 2.3], [ 1.3, 2.9]]) >>> x.astype(int) array([[1, 2], [1, 2]]) 关于如何控制舍入的一些numpy函数：rint，floor，trunc，ceil。 取决于你希望如何绕浮点，向上，向下或到最近的整数。 >>> x =