iteration

How to produce combinations iteratively in R? [duplicate]

由 浪尽此生 提交于 2020-01-23 02:12:27
问题 This question already has an answer here : How do I find all possible subsets of a set iteratively in R? (1 answer) Closed 6 years ago . So I am currently using the following code to generate my combinations: combn(x,y) But the thing is that function stores all of the possible combinations. I dont want to store them, I just want to produce them through like a loop or something. It would be way more efficient for my program. Is there a way to generate combinations through a for loop rather

How to produce combinations iteratively in R? [duplicate]

由 蓝咒 提交于 2020-01-23 02:12:06
问题 This question already has an answer here : How do I find all possible subsets of a set iteratively in R? (1 answer) Closed 6 years ago . So I am currently using the following code to generate my combinations: combn(x,y) But the thing is that function stores all of the possible combinations. I dont want to store them, I just want to produce them through like a loop or something. It would be way more efficient for my program. Is there a way to generate combinations through a for loop rather

How do you iterate backward over circular buffer without a conditional?

由 萝らか妹 提交于 2020-01-22 17:33:28
问题 Iterating forward through a circular buffer without using a conditional is easy with the remainder operator... iterator = (iterator + 1) % buffer_size; I can't for the life of me figure out the reverse operation, iterating backward. 回答1: Does iterator = (iterator + buffer_size - 1) % buffer_size work for you? Go one less than all the way around. 回答2: Borealid's answer works. (note: iterator is set to 0 initially). Another solution is iterator = buffer_size - 1 - (buffer_size - iterator) %

How do you iterate backward over circular buffer without a conditional?

由 拜拜、爱过 提交于 2020-01-22 17:33:25
问题 Iterating forward through a circular buffer without using a conditional is easy with the remainder operator... iterator = (iterator + 1) % buffer_size; I can't for the life of me figure out the reverse operation, iterating backward. 回答1: Does iterator = (iterator + buffer_size - 1) % buffer_size work for you? Go one less than all the way around. 回答2: Borealid's answer works. (note: iterator is set to 0 initially). Another solution is iterator = buffer_size - 1 - (buffer_size - iterator) %

Pytorch框架学习（5）——DataLoder与Dataset

由 旧时模样 提交于 2020-01-22 02:56:50
Dataloder与DataSet 文章目录 Dataloder与DataSet 1. 人民币二分类 2. DataLoder与Dataset 2.1 DataLoder 2.2 DataSet 2.3 数据读取 1. 人民币二分类 今天主要分享机器学习训练过程的数据处理过程。 数据： 数据收集：Img，Label 数据划分：训练集，验证集，测试集 数据读取：DataLoder（Sampler：生成索引，DataSet：根据索引读取图片Img和标签Label） 数据预处理：transforms 2. DataLoder与Dataset 2.1 DataLoder torch.utils.data.DataLoder 功能：构建可迭代的数据装载器，训练时，每一个for循环就是从DataLoder中获取一个batchsize大小的数据 dataset：Dataset类，决定数据从哪读取及如何读取 batchsize：批大小 num_works：是否多进程读取数据 shuffle：每个epoch是否乱序 drop_last：当样本数不能被batchsize整除时，是否舍弃最后一批数据 注意 Epoch：所有训练样本都已输入到模型中，成为一个epoch Iteration：一批样本输入到模型中，称之为一个Iteration Batchsize：批大小

How can I ignore certain strings in my string centring function?

由 眉间皱痕 提交于 2020-01-22 00:20:14
问题 N.B: Directly connected to a problem I had a few years ago, but I'd like to resolve the first issue there which wasn't otherwise part of the question, so please don't flag it as a duplicate of my earlier question. I have a string centring function that centres the given string according to the given width (which is 113 characters): std::string center(std::string input, int width = 113) { return std::string((width - input.length()) / 2, ' ') + input; } I am using a game SDK in order to create

Iterate through 2 lists at once in Python

由 落爺英雄遲暮 提交于 2020-01-21 20:03:32
问题 I want to share/divide a specific amount to the members of two Python lists. The members of list1 will have double share each, while the members of list2 will have single share each, then print out the result, with name and value allocated. The following is my sample code: reminder = float(85000) list1 = ['Designer', 'Coder', 'Supervisor'] list2 = ['Artist', 'Attendant', 'Usher] for i,j in list1 and list2: print(i, (reminder/9)*2) print(j, (reminder/9)*1) When I ran the above code I got an

Iterate through 2 lists at once in Python

由 旧时模样 提交于 2020-01-21 20:03:23
问题 I want to share/divide a specific amount to the members of two Python lists. The members of list1 will have double share each, while the members of list2 will have single share each, then print out the result, with name and value allocated. The following is my sample code: reminder = float(85000) list1 = ['Designer', 'Coder', 'Supervisor'] list2 = ['Artist', 'Attendant', 'Usher] for i,j in list1 and list2: print(i, (reminder/9)*2) print(j, (reminder/9)*1) When I ran the above code I got an

Is it possible to turn a list into a nested dict of keys *without* recursion?

由 无人久伴 提交于 2020-01-21 06:59:27
问题 Supposing I had a list as follows: mylist = ['a','b','c','d'] Is it possible to create, from this list, the following dict without using recursion/a recursive function? { 'a': { 'b': { 'c': { 'd': { } } } } } 回答1: For the simple case, simply iterate and build, either from the end or the start: result = {} for name in reversed(mylist): result = {name: result} or result = current = {} for name in mylist: current[name] = {} current = current[name] The first solution can also be expressed as a

Python 3.x: Test if generator has elements remaining

由 笑着哭i 提交于 2020-01-20 18:30:47
问题 When I use a generator in a for loop, it seems to "know", when there are no more elements yielded. Now, I have to use a generator WITHOUT a for loop, and use next () by hand, to get the next element. My problem is, how do I know, if there are no more elements? I know only: next () raises an exception (StopIteration), if there is nothing left, BUT isn't an exception a little bit too "heavy" for such a simple problem? Isn't there a method like has_next () or so? The following lines should make
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