map may be microscopically faster in some cases (when you're NOT making a lambda for the purpose, but using the same function in map and a listcomp). List comprehensions may be faster in other cases and most (not all) pythonistas consider them more direct and clearer. An example of the tiny speed advantage of map when using exactly the same function: $ python -mtimeit -s'xs=range(10)' 'ma
在某些情况下， map可能会在显微镜下更快（当您不为此目的制作lambda时，但在地图和listcomp中使用相同的功能时）。 在其他情况下，列表理解可能会更快，大多数（并非全部）pythonistas认为它们更直接，更清晰。 当使用完全相同的功能时，地图的速度优势很小的一个例子： $ python -mtimeit -s'xs=range(10)' 'map(hex, xs)' 100000 loops, best of 3: 4.86 usec per loop $ python -mtimeit -s'xs=range(10)' '[hex(x) for x
例如，需要多少内存来存储一百万（32位）整数列表？ alist = range(1000000) # or list(range(1000000)) in Python 3.0 "It depends." Python allocates space for lists in such a way as to achieve amortized constant time for appending elements to the list. In practice, what this means with the current implementation is... the list always has space allocated for a power-of-two number of element
例如，需要多少内存来存储一百万（32位）整数列表？ alist = range(1000000) # or list(range(1000000)) in Python 3.0 “这取决于。” Python以这种方式为列表分配空间，以实现将元素附加到列表的分摊恒定时间。 实际上，这对于当前实现意味着什么......该列表总是具有分配给两个幂数的元素的空间。 所以范围（1000000）实际上会分配一个足够容纳2 ^ 20个元素（〜104.5万）的列表。 这只是存储列表结构本身所需的空间（这是
I have a list containing a string and lists. I should show you; list_x = ['a', ['j', '1', 'x'], ['k', '2', 'y'], ['a', '3', 'hj'], 'd', ['b', '4', 'df'], ['c', '5', 'er'], ['d', '6', 'ty'], 'g', ['e', '7', 'hj'], ['f', '8', 'bv'], ['g', '9', 'sad'], 'j', ['h', '10', 'kj'], ['i', '11', 'nbv'], ['c', '12', 'uy'], 'n', ['d', '13', 'ipoas'], ['e', '14', 'fg']
我有一个包含一个字符串和列表的列表。 我应该告诉你; list_x = ['a', ['j', '1', 'x'], ['k', '2', 'y'], ['a', '3', 'hj'], 'd', ['b', '4', 'df'], ['c', '5', 'er'], ['d', '6', 'ty'], 'g', ['e', '7', 'hj'], ['f', '8', 'bv'], ['g', '9', 'sad'], 'j', ['h', '10', 'kj'], ['i', '11', 'nbv'], ['c', '12', 'uy'], 'n', ['d', '13', 'ipoas'], ['e', '14', 'fg'], ['f', '15',
I'm not sure how verbose I should go so please ask for elaboration if this is too terse. Is there a way to pack the for a,b,c in product(d['a'],d['b'],d['c']): in some syntactical sugar so I would only need to type mute variables a,b,c only once for this loop itself? Something like this may be? my_for_loop('a','b','c'): API_Call1(a) API_Call2(b,c) instead of
我不确定我应该走多远，所以如果这太简单了，请详细说明。 有没有办法将产品中的for a,b,c in product(d['a'],d['b'],d['c']):打包for a,b,c in product(d['a'],d['b'],d['c']):在某些语法糖中，所以我只需要键入mute变量a,b,c对于这个循环本身只有一次？ 像这样的东西可能是？ my_for_loop('a','b','c'): API_Call1(a) API_Call2(b,c) 代替 for a,b,c in product(d
I am parsing a list into a variable and another list with this script: b=[' 687.3774', ' 478.6', ' 47', ' 58', ' 96.90'] c,d=b[0],b[1:] It is always the first element that would be separated and this code works fine, however, it repeats the list b on the right hand side. This is not a problem but it does get annoying when my b is something big like line.replace('*',',').replace
我用这个脚本将一个列表解析成一个变量和另一个列表： b=[' 687.3774', ' 478.6', ' 47', ' 58', ' 96.90'] c,d=b[0],b[1:] 它始终是第一个被分开的元素，并且这段代码工作正常，但是，它重复右侧的列表b 。 这不是问题，但是当我的b有点像line.replace('*',',').replace(' ',',').split(',')时，它确实令人讨厌。 这看起来不像写下来的pythonic方式。 我已经阅读了这个论坛的一些
In Python 3, the following returns a map object: map(lambda x: x**2, range(10)) If we want to turn this object into a list, we can just cast it as a list using list(mapobject) . However, I discovered through code golfing that *x, = mapobject makes x into a list. Why is this allowed in Python 3? This is an example of extended iterable unpacking, introduced into Python 3 by PEP 3132: This
在Python 3中，以下内容返回一个地图对象： map(lambda x: x**2, range(10)) 如果我们想把这个对象变成一个列表，我们可以使用list(mapobject)将它转换为列表。 不过，我通过代码打高尔夫发现了这一点 *x, = mapobject 使x成为一个列表。 为什么在Python 3中允许这样做？ 这是PEP 3132引入Python 3的扩展迭代解包的一个例子： 这个PEP建议更改为可迭代的解包语法，允许指定一个“全部收集”名称，该名称将被分配一个未分
I'm looking for the best way to do this using pythonexcelsqlgoogle sheets - I need to find all rows which fits to k values from list of n values. For example I have this table called Animals: | Name | mammal | move | dive | +----------+--------+--------+-------+ | Giraffe | 1 | 1 | 0 | | Frog | 0 | 1 | 1 | | Dolphin | 1 | 1 | 1 | | Sna
我正在寻找使用python excel sql google这样做的最佳方式 - 我需要从n个值列表中找到适合k值的所有行。 例如，我有这张名为Animals的表格： | Name | mammal | move | dive | +----------+--------+--------+-------+ | Giraffe | 1 | 1 | 0 | | Frog | 0 | 1 | 1 | | Dolphin | 1 | 1 | 1 | | Snail | 0 | 1 | 0 | | Bacteria | 0 | 0
I am creating a decorator that catches a raised error in it's target function, and allows the user to continue executing the script (bypassing the function) or drop out of the script. def catch_error(func): """ This decorator is used to make sure that if a decorated function breaks in the execution of a script, the script doesn't automatically crash. Instead, it gives you t
我正在创建一个装饰器，捕获目标函数中引发的错误，并允许用户继续执行脚本（绕过函数）或退出脚本。 def catch_error(func): """ This decorator is used to make sure that if a decorated function breaks in the execution of a script, the script doesn't automatically crash. Instead, it gives you the choice to continue or gracefully exit. """ def caught(*args): try:
I'm reading tab separated values from strings into an object like this: class Node(rect): def __init__(self, line): (self.id, self.x1, self.y1, self.x2, self.y2) = line.split('t') That works fine, but say I want to convert those x and y coordinates, which are read from the string line , to floats. What is the most pythonic way to do this? I imagine something like (self.id, fl
我正在从字符串读取tab分隔值到像这样的对象： class Node(rect): def __init__(self, line): (self.id, self.x1, self.y1, self.x2, self.y2) = line.split('t') 这工作正常，但说我想将这些从字符串line读取的x和y坐标转换为浮点数。 什么是最Python的方式来做到这一点？ 我想像一样 (self.id, float(self.x1), float(self.y1), float(self.x2), float(self.y2)) = line.split('t') 这当然不起作用。 有没
I have the following string: strIn = "Head,0.000235532,0.43656735" I would like to retrieve from this a variable type = "Head" and an array vals = [0.000235532,0.43656735] How can I do this using Python? I've considered using strIn[0],...,strIn[5] to get the type, but then I realized I'll get types of various lengths. Thanks. If you're using Python 3, you can do typ
我有以下字符串： strIn = "Head,0.000235532,0.43656735" 我想从这里获取一个变量type = "Head"和一个数组vals = [0.000235532,0.43656735] 我如何使用Python来做到这一点？ 我考虑过使用strIn [0]，...，strIn [5]来获取类型，但后来我意识到我会得到不同长度的类型。 谢谢。 如果你使用Python 3，你可以 type_, *vals = strIn.split(',') vals = [float(v) for v in vals] # Convert list of strings to l