I came across this snippet of code, I understand it's a Cartesian product but if someone can break up this line for me [s+t for s in a for t in b] from the following code along with docs link for this kind of syntax. Apparently this for in syntax with s+t ??? is foreign for me and I'm new to python as well. Appreciate docs link so I can understand more about this syntax since there ar
我遇到了这段代码，我知道这是一个笛卡尔积，但如果有人能够为我分解这一行[s+t for s in a for t in b]下面的代码中的[s+t for s in a for t in b] ，以及这种语法的docs链接。 显然，这for in带语法s+t ??? 对我来说是陌生的，我也是python的新手。 欣赏文档链接，所以我可以更多地了解这种语法，因为我试图理解的循环中还有其他变体。 rows = 'ABCDEFGHI' cols = '123456789' def cross(a, b): return [s+t for s in
I have tested in Python 2.7, the two styles are the same. My confusion is, when reading first method to generate a list, I am always a bit confused if i%2 == 0 controls if we should execute the whole loop of i in range(100) , or i%2 == 0 is under loop of i in range(100) . I have the confusion maybe in the past I write Java and C++, thinking methods from there. Looking for advice how to read l
我已经在Python 2.7中测试过，这两种样式是相同的。 我的困惑是，当读第一个方法来生成一个列表，我总是有点困惑，如果i%2 == 0控制，如果我们应该执行i in range(100) ， i%2 == 0的整个循环，或者i%2 == 0是i in range(100)循环。 我有可能在过去编写Java和C ++，思考方法。 寻找如何阅读列表生成代码的建议，通常模式是[<something before loop> <the loop> <something after the loop>] ，在这种情况
This question already has an answer here: How to create a list with the characters of a string? 4 answers A string is an iterable, and a list can be constructed from an iterable. So all you need is a = list(a) print(a) Output: ['h', 'e', 'l', 'l', 'o'] Tried to do it differently Code: map(None,"sart")#for python 2.x #list(map(None,"sart")) fo
这个问题在这里已经有了答案： 如何使用字符串的字符创建列表？ 4个答案 一个字符串是一个可迭代的，并且一个列表可以由一个可迭代的构造而成。 所以你需要的是 a = list(a) print(a) 输出： ['你好'] 试图做不同的事情 码： map(None,"sart")#for python 2.x #list(map(None,"sart")) for python 3.x 输出： ['s', 'a', 'r', 't']
I'm experimenting with different ways to identify duplicate files, based on file content, by looping through the top level directory where folders AZ exist. Within folders AZ there is one additional layer of folders named after the current date. Finally, within the dated folders there are between several thousand to several million (<3 million) files in various formats. Using the scrip
我正在尝试使用不同的方法来根据文件内容识别重复文件，方法是循环访问文件夹AZ所在的顶级目录。 在文件夹AZ中，有一个额外的文件夹层以当前日期命名。 最后，在标有日期的文件夹中，有各种格式的数千到数百万（<300万）个文件。 使用下面的脚本，我能够在大约4个小时内处理大约800,000个文件。 然而，在一个大约1300万个文件的大型数据集上运行它时，它总是会破坏包含大约150万个文件的字母“I”。 鉴于我正在处理的数
Is there a way to vectorize an operation that takes several numpy arrays and puts them into a list of dictionaries? Here's a simplified example. The real scenario might involve more arrays and more dictionary keys. import numpy as np x = np.arange(10) y = np.arange(10, 20) z = np.arange(100, 110) print [dict(x=x[ii], y=y[ii], z=z[ii]) for ii in xrange(10)] I might have thousands or hund
有没有办法矢量化一个操作，它需要几个numpy数组并将它们放到一个字典列表中？ 这是一个简单的例子。 真正的场景可能涉及更多的数组和更多字典键。 import numpy as np x = np.arange(10) y = np.arange(10, 20) z = np.arange(100, 110) print [dict(x=x[ii], y=y[ii], z=z[ii]) for ii in xrange(10)] 我可能会在xrange调用中进行数千或数十万次迭代。 所有创建x ， y和z都是矢量化的（我的例子不像上面那么简单）。
I am new to programming and I got stuck with random number generation. I can simply generate random numbers using random function "randint" but could not generate set of random numbers. For instance i want to get 10 random numbers. from random import randint x = randint(1, 100) y = randint(1, 100) isFailedTest = (5<=x<=15) and (10<=y<=11) selected_test = [x,y] while (isFa
我是编程新手，并且被随机数生成卡住了。 我可以简单地使用随机函数“randint”生成随机数，但不能生成一组随机数。 例如，我想获得10个随机数。 from random import randint x = randint(1, 100) y = randint(1, 100) isFailedTest = (5<=x<=15) and (10<=y<=11) selected_test = [x,y] while (isFailedTest == False): 我可以一次生成1个随机数，但不能同时生成10个随机数。 这里1个数字表示二维数字例子（x，y
I have a list for example l = [10, 20, 30, 40, 50, 60] I need to increment the first n elements of the list given a condition. The condition is independent of the list. For example if n = 3 , the list l should become : l = [11, 21, 31, 40, 50, 60] I understand that I can do it with a for loop on each element of the list. But I need to do such operation around 150 million times. So, I am l
例如，我有一个列表 l = [10, 20, 30, 40, 50, 60] 我需要增加列表中前n元素的条件。 条件与列表无关。 例如，如果n = 3 ，则列表l应该变为： l = [11, 21, 31, 40, 50, 60] 我知道我可以在列表的每个元素上使用for循环。 但我需要做大约1.5亿次这样的操作。 所以，我正在寻找一个更快的方法来做到这一点。 任何帮助，高度赞赏。 提前致谢 您可以在列表顶部创建一个简单的数据结构，该结构存储每个增量操作的开始和
This question already has an answer here: Python list comprehension expensive 1 answer List comprehensions perform better here because you don't need to load the append attribute of the list and call it as a function!* Consider the following example : >>> import dis >>> def f1(): ... for i in range(5): ... l.append(i) ... >>> def f2(): ... [i for i
这个问题在这里已经有了答案： Python列表理解昂贵1个答案 列表推导在这里执行得更好，因为您不需要加载列表的append属性并将其作为函数调用！* 考虑下面的例子： >>> import dis >>> def f1(): ... for i in range(5): ... l.append(i) ... >>> def f2(): ... [i for i in range(5)] ... >>> dis.dis(f1) 2 0 SETUP_LOOP 33 (to 36)
This question already has an answer here: Are list-comprehensions and functional functions faster than “for loops”? 5 answers You can use the timeit library, or just use time.time() to time it yourself: >>> from time import time >>> def first(): ... ftime = time() ... _foo = [x ** 2 for x in range(1, 101)] ... print "First", time()-ftime ... >>> def s
这个问题在这里已经有了答案： 列表理解和功能函数比“for循环”更快吗？ 5个答案 您可以使用timeit库，或者只使用time.time()计时： >>> from time import time >>> def first(): ... ftime = time() ... _foo = [x ** 2 for x in range(1, 101)] ... print "First", time()-ftime ... >>> def second(): ... ftime = time() ... _foo = [] ... for x in range(1, 101)
This question already has an answer here: Are list-comprehensions and functional functions faster than “for loops”? 5 answers As you've written it, definitely go with the for loop. map will create an unnecessary list (of all None since that is presumably the return value of myfunction ) on python2.x -- and on python3.x, it won't call your function at all until you actually iterate o
这个问题在这里已经有了答案： 列表理解和功能函数比“for循环”更快吗？ 5个答案 正如你写的那样，绝对要使用for循环。 map会在python2.x上创建一个不必要的list （全部为None因为这可能是myfunction的返回值），并且在python3.x上它根本不会调用你的函数，直到你实际遍历结果为止！ 换句话说，只有当你真的想要迭代结果时才使用map 。 切勿将其用于功能的副作用。 为了向前兼容，不要假定map的返回值是一个list - 如果